## Section5.8Optimization Problems

Many important applied problems involve finding the best way to accomplish some task. Often this involves finding the maximum or minimum value of some function: the minimum time to make a certain journey, the minimum cost for doing a task, the maximum power that can be generated by a device, and so on. Many of these problems can be solved by finding the appropriate function and then using techniques of calculus to find the maximum or the minimum value required.

Generally such a problem will have the following mathematical form: Find the largest (or smallest) value of $f(x)$ when $a\le x\le b\text{.}$ Sometimes $a$ or $b$ are infinite, but frequently the real world imposes some constraint on the values that $x$ may have.

Such a problem differs in two ways from the relative maximum and minimum problems we encountered when graphing functions: We are interested only in the function between $a$ and $b\text{,}$ and we want to know the largest or smallest value that $f(x)$ takes on, not merely values that are the largest or smallest in a small interval. That is, we seek not a relative maximum or minimum but a global (or absolute) maximum or minimum.

###### Guideline for Solving Optimization Problems.
1. Identify what is to be maximized or minimized and what the constraints are.

2. Draw a diagram (if appropriate) and label it.

3. Decide what the variables are and in what units their values are being measured in. For example, $A$ for area in square metres, $r$ for radius in inches, $C$ for cost in Euros. In other words, if the problem does not introduce these variables, you need to do so.

4. Write a formula for the function that is to be maximized or minimized.

5. Use the given constraint to express the formula from Step 4 in terms of a single variable, namely something like $f(x)$ (or $A(x)\text{,}$ $C(x)\text{,...,}$ whatever name is appropriate). Then identify the domain of this function, which is typically $\left[a,b\right]$ or $(a,b)\text{.}$

6. Find the critical points of $f\text{.}$ Compare all critical values and endpoints (or perhaps $\lim\limits_{x\to a^+} f(x)$ and $\lim\limits_{x\to a^-} f(x)$ or curve sketching if the interval is open) to determine the absolute extrema of $f\text{.}$

7. Provide your solution meaningfully, which includes unit(s).

You want to sell a certain number $n$ of items in order to maximize your profit. Market research tells you that if you set the price at $1.50, you will be able to sell 5000 items, and for every 10 cents you lower the price below$1.50 you will be able to sell another 1000 items. Suppose that your fixed costs ( “start-up costs” ) total $2000, and the per item cost of production ( “marginal cost” ) is$0.50.

Find the price to set per item and the number of items sold in order to maximize profit, and also determine the maximum profit you can get.

Solution

The first step is to convert the problem into a function maximization problem. Since we want to maximize profit by setting the price per item, we should look for a function $P(x)$ representing the profit when the price per item is $x\text{.}$ Profit is revenue minus costs, and revenue is number of items sold times the price per item, so we get $P=nx-2000-0.50n\text{.}$ The number of items sold is itself a function of $x\text{,}$ $n=5000+1000(1.5-x)/0.10\text{,}$ because $(1.5-x)/0.10$ is the number of multiples of 10 cents that the price is below 1.50. Now we substitute for $n$ in the profit function: \begin{align*} P(x)\amp =(5000+1000(1.5-x)/0.10)x-2000- 0.5(5000+1000(1.5-x)/0.10)\\ \amp =-10000x^2+25000x-12000 \end{align*} We want to know the maximum value of this function when $x$ is between 0 and $1.5\text{.}$ The derivative is $P'(x)=-20000x+25000\text{,}$ which is zero when $x=1.25\text{.}$ Since $P''(x)=-20000\lt 0\text{,}$ there must be a relative maximum at $x=1.25\text{,}$ and since this is the only critical value it must be a global maximum as well. (Alternately, we could compute $P(0)=-12000\text{,}$ $P(1.25)=3625\text{,}$ and $P(1.5)=3000$ and note that $P(1.25)$ is the maximum of these.) Thus the maximum profit is3625, attained when we set the price at $1.25 and sell 7500 items. A manufacturer determines that the daily average cost of producing $q$ units is \begin{equation*} \overline{C}(q) = 0.0001q^2-0.08q+65+\frac{5000}{q} \ \ \ q > 0 \end{equation*} Determine the number of units produced per day which minimizes the average cost. Solution We first note that the domain of the function $\overline{C}$ is the open interval $(0,\infty)\text{.}$ Calculate, \begin{equation*} \overline{C}'(q) = 0.0002q-0.08-\frac{5000}{q^2} \cdot \end{equation*} Then solving $C'(q)= 0$ gives $q = 500\text{,}$ which is the only critical point of $C\text{.}$ Next, \begin{equation*} \overline{C}''(q) = 0.0002+\frac{10,000}{q^3} \cdot \end{equation*} And so \begin{equation*} \overline{C}''(500) = 0.0002+\frac{10,000}{(500)^3} > 0\text{.} \end{equation*} Therefore, by the Second Derivative Test, $q=500$ is a relative minimum of $\overline{C}\text{.}$ Furthermore, $\overline{C}$ is concave upward everywhere, so the relative minimum of $\overline{C}$ must be the absolute maximum of $\overline{C}\text{.}$ The minimum cost is thus \begin{equation*} \overline{C}(500) = 0.0001(500)^2-0.08(500)+65+\frac{5000}{500} = 60 \end{equation*} or$60 per unit. The sketch of $\overline{C}$ follows.

It is estimated that the operating rate of a major manufacturer's factories over a certain 365-day period is given by

\begin{equation*} f(t)=100+\frac{800t}{t^2+90,000} \ \ \ 0\leq t\leq 365 \end{equation*}

percent. Determine the day on which the operating rate is maximized.

Solution

We wish to find the absolute maximum of $f$ on $[0,365]\text{.}$ We first calculate

\begin{equation*} \begin{split} f'(t) \amp = \frac{(t^2+90,000)(800)-800t(2t)}{(t^2+90,000)^2}\\ \amp = \frac{-800(t^2-90,000)}{(t^2+90,000)^2} \end{split} \end{equation*}

Therefore, $f'(t)=0 \implies t=-300$ or $300\text{.}$ So the only critical point of $f$ is $t=300$ (since $t=-300$ is outside the domain of $f$). Evaluating $f(t)$ at this critical point and both endpoints, we see

\begin{equation*} f(0) = 100, \ \ \ f(300)=101.33, \ \ \ f(365)=101.308 \end{equation*}

Thus, the manufacturing capacity operating rate was at a maximum after 300 days.

Find the largest rectangle (that is, the rectangle with largest area) that fits inside the graph of the parabola $\ds y=x^2$ below the line $y=a$ ($a$ is an unspecified constant value), with the top side of the rectangle on the horizontal line $y=a\text{;}$ see below.

Solution

We want to find the maximum value of some function $A(x)$ representing area. Perhaps the hardest part of this problem is deciding what $x$ should represent. The lower right corner of the rectangle is at $\ds (x,x^2)\text{,}$ and once this is chosen the rectangle is completely determined. So we can let the $x$ in $A(x)$ be the $x$ of the parabola $\ds f(x)=x^2\text{.}$ Then the area is

\begin{equation*} A(x)=(2x)(a-x^2)=-2x^3+2ax\text{.} \end{equation*}

We want the maximum value of $A(x)$ when $x$ is in $\ds [0,\sqrt{a}]\text{.}$ (You might object to allowing $x=0$ or $\ds x=\sqrt{a}\text{,}$ since then the “rectangle” has either no width or no height, so is not “really” a rectangle. But the problem is somewhat easier if we simply allow such rectangles, which have zero area.)

Setting $\ds 0=A'(x)=6x^2+2a$ we get $\ds x=\sqrt{a/3}$ as the only critical value. Testing this and the two endpoints, we have $\ds A(0)=A(\sqrt{a})=0$ and $\ds A(\sqrt{a/3})=(4/9)\sqrt{3}a^{3/2}\text{.}$ The maximum area thus occurs when the rectangle has dimensions $\ds 2\sqrt{a/3}\times (2/3)a\text{.}$

If you fit the largest possible cone inside a sphere, what fraction of the volume of the sphere is occupied by the cone? (Here by “cone” we mean a right circular cone, i.e., a cone for which the base is perpendicular to the axis of symmetry, and for which the cross-section cut perpendicular to the axis of symmetry at any point is a circle.)

Solution

Let $R$ be the radius of the sphere, and let $r$ and $h$ be the base radius and height of the cone inside the sphere. What we want to maximize is the volume of the cone: $\ds \pi r^2h/3\text{.}$ Here $R$ is a fixed value, but $r$ and $h$ can vary. Namely, we could choose $r$ to be as large as possible—equal to $R$—by taking the height equal to $R\text{;}$ or we could make the cone's height $h$ larger at the expense of making $r$ a little less than $R\text{.}$ See the cross-section depicted in the figure shown below. We have situated the picture in a convenient way relative to the $x$- and $y$-axes, namely, with the centre of the sphere at the origin and the vertex of the cone at the far left on the $x$-axis.

Notice that the function we want to maximize, $\ds \pi r^2h/3\text{,}$ depends on two variables. This is frequently the case, but often the two variables are related in some way so that “really” there is only one variable. So our next step is to find the relationship and use it to solve for one of the variables in terms of the other, so as to have a function of only one variable to maximize. In this problem, the condition is apparent in the figure: the upper corner of the triangle, whose coordinates are $(h-R,r)\text{,}$ must be on the circle of radius $R\text{.}$ That is,

\begin{equation*} (h-R)^2+r^2=R^2\text{.} \end{equation*}

We can solve for $h$ in terms of $r$ or for $r$ in terms of $h\text{.}$ Either involves taking a square root, but we notice that the volume function contains $\ds r^2\text{,}$ not $r$ by itself, so it is easiest to solve for $\ds r^2$ directly: $\ds r^2=R^2-(h-R)^2\text{.}$ Then we substitute the result into $\ds \pi r^2h/3\text{:}$

\begin{equation*} \begin{split} V(h)\amp =\pi(R^2-(h-R)^2)h/3 \\ \amp =-{\pi\over3}h^3+{2\over3}\pi h^2R \end{split} \end{equation*}

We want to maximize $V(h)$ when $h$ is between 0 and $2R$ including the endpoints. Like in Example 5.8.4, we argue that zero volume at the endpoints of the closed interval makes the problem easier to solve Now we solve $\ds 0=f'(h)=-\pi h^2+(4/3)\pi h R\text{,}$ getting $h=0$ or $h=4R/3\text{.}$ We compute $V(0)=V(2R)=0$ and $\ds V(4R/3)=(32/81)\pi R^3\text{.}$ The maximum is the latter; since the volume of the sphere is $\ds (4/3)\pi R^3\text{,}$ the fraction of the sphere occupied by the cone is

\begin{equation*} {(32/81)\pi R^3\over (4/3)\pi R^3}={8\over 27}\approx 30\%\text{.} \end{equation*}

You are making cylindrical containers to contain a given volume. Suppose that the top and bottom are made of a material that is $N$ times as expensive (cost per unit area) as the material used for the lateral side of the cylinder.

Find (in terms of $N$) the ratio of height to base radius of the cylinder that minimizes the cost of making the containers.

Solution

Let us first choose letters to represent various things: $h$ for the height, $r$ for the base radius, $V$ for the volume of the cylinder, and $c$ for the cost per unit area of the lateral side of the cylinder; $V$ and $c$ are constants, $h$ and $r$ are variables. Now we can write the cost of materials:

\begin{equation*} c(2\pi rh)+Nc(2\pi r^2)\text{.} \end{equation*}

Again we have two variables; the relationship is provided by the fixed volume of the cylinder: $\ds V=\pi r^2h\text{.}$ We use this relationship to eliminate $h$ (we could eliminate $r\text{,}$ but it's a little easier if we eliminate $h\text{,}$ which appears in only one place in the above formula for cost). The result is

\begin{equation*} f(r)=2c\pi r{V\over\pi r^2}+2Nc\pi r^2={2cV\over r}+2Nc\pi r^2\text{.} \end{equation*}

We want to know the minimum value of this function when $r$ is in $(0,\infty)\text{.}$ We now set $\ds 0=f'(r)=-2cV/r^2+4Nc\pi r\text{,}$ giving $\ds r={\root 3 \of {V/(2N\pi)}}\text{.}$ Since $\ds f''(r)=4cV/r^3+4Nc\pi$ is positive when $r$ is positive, there is a relative minimum at the critical value, and hence a global minimum since there is only one critical value.

Finally, since $\ds h=V/(\pi r^2)\text{,}$

\begin{equation*} {h\over r}={V\over\pi r^3}={V\over \pi(V/(2N\pi))}=2N\text{,} \end{equation*}

so the minimum cost occurs when the height $h$ is $2N$ times the radius. If, for example, there is no difference in the cost of materials, the height is twice the radius (or the height is equal to the diameter).

Of all rectangles of area 100, which has the smallest perimeter?

Solution

First we must translate this into a purely mathematical problem in which we want to find the minimum value of a function. If $x$ denotes one of the sides of the rectangle, then the adjacent side must be $100/x$ (in order that the area be 100). So the function we want to minimize is

\begin{equation*} f(x)=2x+2{100\over x} \end{equation*}

since the perimeter is twice the length plus twice the width of the rectangle. Not all values of $x$ make sense in this problem: lengths of sides of rectangles must be positive, so $x>0\text{.}$ If $x>0$ then so is $100/x\text{,}$ so we need no second condition on $x\text{.}$

We next find $f'(x)$ and set it equal to zero: $\ds 0=f'(x)=2-200/x^2\text{.}$ Solving $f'(x)=0$ for $x$ gives us $x=\pm 10\text{.}$ We are interested only in $x>0\text{,}$ so only the value $x=10$ is of interest. Since $f'(x)$ is defined everywhere on the interval $(0,\infty)\text{,}$ there are no more critical values, and there are no endpoints. Is there a relative maximum, minimum, or neither at $x=10\text{?}$ The second derivative is $\ds f''(x)=400/x^3\text{,}$ and $f''(10)>0\text{,}$ so there is a relative minimum. Since there is only one critical value, this is also the global minimum, so the rectangle with smallest perimeter is the $10\times10$ square.

Suppose you want to reach a point $A$ that is located across the sand from a nearby road (see diagram below). Suppose that the road is straight, and $b$ is the distance from $A$ to the closest point $C$ on the road. Let $v$ be your speed on the road, and let $w\text{,}$ which is less than $v\text{,}$ be your speed on the sand. Right now you are at the point $D\text{,}$ which is a distance $a$ from $C\text{.}$ At what point $B$ should you turn off the road and head across the sand in order to minimize your travel time to $A\text{?}$

Solution

Let $x$ be the distance short of $C$ where you turn off, i.e., the distance from $B$ to $C\text{.}$ We want to minimize the total travel time. Recall that when traveling at constant velocity, time is distance divided by velocity.

You travel the distance $\ds \overline{DB}$ at speed $v\text{,}$ and then the distance $\ds \overline{BA}$ at speed $w\text{.}$ Since $\ds \overline{DB}=a-x$ and, by the Pythagorean Theorem, $\ds \overline{BA}=\sqrt{x^2+b^2}\text{,}$ the total time for the trip is

\begin{equation*} f(x)={a-x\over v}+{\sqrt{x^2+b^2}\over w}\text{.} \end{equation*}

We want to find the minimum value of $f$ when $x$ is between 0 and $a\text{.}$ As usual we set $f'(x)=0$ and solve for $x\text{:}$

\begin{equation*} \begin{split} 0=f'(x)=-{1\over v}+{x\over w\sqrt{x^2+b^2}}\cr\\ w\sqrt{x^2+b^2}=vx\cr\\ w^2(x^2+b^2) = v^2x^2\cr\\ w^2b^2=(v^2-w^2)x^2\cr\\ x={wb\over\sqrt{v^2-w^2}} \end{split} \end{equation*}

Notice that $a$ does not appear in the last expression, but $a$ is not irrelevant, since we are interested only in critical values that are in $[0,a]\text{,}$ and $\ds wb/\sqrt{v^2-w^2}$ is either in this interval or not. If it is, we can use the second derivative to test it:

\begin{equation*} f''(x) = {b^2\over (x^2+b^2)^{3/2}w}\text{.} \end{equation*}

Since this is always positive there is a relative minimum at the critical point, and so it is a global minimum as well.

If the critical value is not in $[0,a]$ it is larger than $a\text{.}$ In this case the minimum must occur at one of the endpoints. We can compute

\begin{align*} f(0)\amp =\ds{a\over v}+{b\over w}\\ f(a)\amp =\ds{\sqrt{a^2+b^2}\over w} \end{align*}

but it is difficult to determine which of these is smaller by direct comparison. If, as is likely in practice, we know the values of $v\text{,}$ $w\text{,}$ $a\text{,}$ and $b\text{,}$ then it is easy to determine this. With a little cleverness, however, we can determine the minimum in general. We have seen that $f''(x)$ is always positive, so the derivative $f'(x)$ is always increasing. We know that at $\ds wb/\sqrt{v^2-w^2}$ the derivative is zero, so for values of $x$ less than that critical value, the derivative is negative. This means that $f(0)>f(a)\text{,}$ so the minimum occurs when $x=a\text{.}$

So the upshot is this: If you start farther away from $C$ than $\ds wb/\sqrt{v^2-w^2}$ then you always want to cut across the sand when you are a distance $\ds wb/\sqrt{v^2-w^2}$ from point $C\text{.}$ If you start closer than this to $C\text{,}$ you should cut directly across the sand.

##### Exercises for Section 5.8.

Suppose the monthly profit a manufacturer realizes from selling $q$ units is given by

\begin{equation*} P(q)=-5q^2+1300q-15,000 \end{equation*}

dollars. What is the maximum monthly profit?

$69,500 per month when $q=130\text{.}$ Solution We first note that the domain of $P$ is $q \in [0,\infty)\text{.}$ We differentiate twice: \begin{equation*} P'(q) = -10q+1300, \ \ P''(q) = -10\text{.} \end{equation*} To find any critical points of $P\text{,}$ we first set \begin{equation*} P'(q) = 0 \implies q = 130 \in [0,\infty)\text{.} \end{equation*} Next, we look for any points where $P'(q)$ is undefined, which gives no solutions. Hence, the only critical point of $P$ is $q=130\text{.}$ Since $P''(130)\lt 0\text{,}$ by the Second Derivative Test for relative extrema, $P$ has a relative maximum at $(130, P(130)) = (130, 69500)\text{.}$ Furthermore, since $P$ is an upside down parabola, this relative maximum must be the absolute maximum. Hence, the maximum monthly profit is$69,500 per month. This maximum occurs when 130 units are produced.

Suppose the daily profit a manufacturer realizes from selling $q$ units is given by

\begin{equation*} P(q)=-0.2q^3+2q^2-1000 \end{equation*}

What is the maximum daily profit?

$29.63 per day Solution The domain of the profit function must be $[0,\infty)\text{.}$ We now differentiate: \begin{equation*} P'(q) = \diff{}{q}\left(-0.2q^3+2q^2-1000\right)= -0.6q^2+4q\text{.} \end{equation*} To find the critical points of $P\text{,}$ we set $P'(q)=0\text{:}$ \begin{equation*} -0.6q^2+4q = 0 \implies q = 0, \ \ \frac{20}{3}\text{.} \end{equation*} Since $q=0$ is an endpoint, it cannot be a critical point. Hence, the profit function has one critical point at $q=20/3\text{.}$ We further notice that there are no points where $P'(q)$ DNE, and so this gives no additional critical points. We now compare profits: \begin{equation*} P(0) = -1000, \ \ P(20/3) = 29.63 \end{equation*} We further notice that the profit is decreasing for all $q > 20/3\text{.}$ We conclude that the absolute maximum profit the manufacturer realizes is$29.63 per day when the level of production is set to approximately 7 units.

A manufacturer determines that the total cost $C(q)$ of manufacturing $q$ units per day is given by

\begin{equation*} C(q)=400+4q+0.0001q^2 \end{equation*}

dollars. If each unit is sold at

\begin{equation*} p=10-0.0004q \end{equation*}

dollars, what is the daily level of production that maximizes the daily profit?

6000 units/day

Solution

We first construct the daily profit as a function of $q\text{.}$

\begin{equation*} \begin{split} P(q)\amp =pq-C(q)\\ \amp =\left(10-0.0004q\right)q-\left(400+4q+0.0001q^2\right)\\ \amp =-0.0005q^2+6q-400. \end{split} \end{equation*}

To determine the domain of $P\text{,}$ we first notice that we require both the quantity demanded $q$ and the unit price $p$ to be positive:

\begin{equation*} q \geq 0, \ \ p= 10-0.0004q \geq 0\text{.} \end{equation*}

Therefore, the domain is $[0, 25000]\text{.}$

Next, we find the interior critical points of $P$ by computing

\begin{equation*} \begin{split} P'(q)\amp = 0\\ -0.001q+6 \amp = 0, \end{split} \end{equation*}

which gives a single solution, $q=6000$ (note that this point is in the domain of $P$). Since there are no points $q$ in the domain of $P$ for which $P'(q)$ does not exist, this must be the only critical point.

We have defined the profit function on a closed interval, and so to determine the absolute maximum, we can compare the value of the profit function at all critical points and endpoints:

\begin{equation*} P(0) = -400, \ P(6000)=17600, \ P(25000) =-162900\text{.} \end{equation*}

So the absolute maximum is attained when $q=6000\text{.}$

Therefore, with a production level of $6000$ units per day, the profit is maximized at $17,600. A certain company has a weekly fixed cost of$10,000, and a variable production cost of

\begin{equation*} V(q)=0.001q^3-q^2+500q \end{equation*}

dollars per unit. If the revenue from selling $q$ units per week is

\begin{equation*} R(q)=-3q^2+1500q\text{,} \end{equation*}

what is the level of production that will maximize the weekly profit?

Solution

We first construct the weekly profit function, $P(q)\text{:}$

\begin{equation*} \begin{split} P(q) \amp = R(q) - V(q) - 10000 \\ \amp = \left(-3q^2+1500q\right) - \left(0.001q^3-q^2+500q\right)-10000\\ \amp = -0.001q^3-2q^2+1000q-10000 \end{split} \end{equation*}

where $q \geq 0\text{.}$ Now calculate

\begin{equation*} P'(q)= -0.003 q^2 - 4 q + 1000\text{.} \end{equation*}

Since $P'(q)$ DNE has no solutions, we set $P'(q) = 0$ to find any critical points.

Using the quadratic formula, we find two solutions, $q = 215.25\text{,}$ and $q=-1548.58\text{.}$ We reject the second root (since it is not in the domain of $P$).

We now compare the profit at the critical point and at the endpoints:

\begin{equation*} P(0)=-10,000 \ \ P(215.25) = 102612, \ \ \lim_{q \to \infty} P(q) = - \infty\text{.} \end{equation*}

Therefore, $q \approx 215$ is the absolute maximum of $P\text{.}$

Therefore, the profit is maximal when the level of production is at approximately $215$ units per week.

Suppose the total monthly cost of manufacturing $q$ units is given by

\begin{equation*} C(q) = 0.5q^2-50q+15,000 \end{equation*}

dollars.

1. Determine the average cost function $\overline{C}\text{.}$

The average cost function $\overline{C}$ is given by

\begin{equation*} \overline{C}(q) = \frac{C(q)}{q} = \frac{0.5q^2-50q+15000}{q} =0.5q-50+\frac{15,000}{q} \ \ \ (q > 0)\text{.} \end{equation*}
2. Determine the level of production that results in the smallest average production cost.

$q=173$
Solution

We now wish to minimize $\overline{C}(q)\text{.}$ Set

\begin{equation*} \overline{C}'(q) = 0 \implies c\text{,} \end{equation*}

which gives two solutions

\begin{equation*} q = \pm 173.205\text{.} \end{equation*}

Reject the negative root, and evaluate whether the remaining critical point is a relative maximum or minimum.

\begin{equation*} \overline{C}''(q) = \frac{30,000}{q^3}\text{,} \end{equation*}

and in particular, $\overline{C}''(173.205) > 0\text{.}$ By the Second Derivative Test, $q=173.205$ is a relative minimum; furthermore, since $\overline{C}$ is concave up for all $q > 0\text{,}$ we conclude that this is the absolute minimum. Therefore, the level of production that results in the minimal average production cost is about $173$ units.

3. Determine the level of production for which the average cost is equal to the marginal cost.

$q=173$
Solution

The marginal cost is given by

\begin{equation*} C'(q) = q-50\text{.} \end{equation*}

Setting $C'(q) = \overline{C}(q)\text{,}$ we find

\begin{equation*} q-50 = 0.5q-50+\frac{15,000}{q} \end{equation*}
\begin{equation*} 0.5-\frac{15000}{q^2} = 0 \end{equation*}

Recognizing this equation from part (b), and rejecting the negative solution as before, we see that the marginal cost is equal to the average cost when $q \approx 173\text{.}$

4. What can you deduce from your results?

The average cost is minimized when $\overline{C} = C'(q)\text{;}$ that is, when it is equal to the marginal cost.

Given the following demand equation,

\begin{equation*} p=\sqrt{300-0.5q}\text{,} \end{equation*}

where $p$ is unit price and $q$ is the number of units manufactured per week, how many units should be manufactured and sold each week in order to maximize the total revenue?

400 units

Solution

We first determine the domain of the demand equation:

\begin{equation*} 300-0.5q \geq 0 \implies q \leq 600\text{.} \end{equation*}

We also know that $q$ must be a positive number. Therefore, the revenue function is given by

\begin{equation*} R(q) = pq = q\sqrt{300-0.5q}, \ \ 0 \leq q \leq 600\text{.} \end{equation*}

We now look for critical points of the revenue function. We compute

\begin{equation*} R'(q) = \sqrt{300-0.5q} + \frac{-q}{4\sqrt{300-0.5q}}\text{.} \end{equation*}

The only point in the domain of $R$ for which $R'$ is undefined is when $q=600\text{.}$ Since this is a boundary point, it cannot be a critical point. We therefore need ot solve $R'(q) = 0\text{:}$

\begin{equation*} \sqrt{300-0.5q} - \frac{q}{4\sqrt{300-0.5q}} = 0 \implies q = 400\text{.} \end{equation*}

To determine the absolute maximum revenue, we compare

\begin{equation*} R(0) = 0, \ \ R(400) = 4000, \ \ R(600) = 0\text{.} \end{equation*}

We conclude that in order to maxmize revenue, $400$ units should be manufactured and sold.

We define the average revenue as

\begin{equation*} \overline{R}(q)=\frac{R(q)}{q} \ \ \ q > 0\text{.} \end{equation*}

Show that if $R(q)$ is concave downward, then the maximum average revenue occurs when $\overline{R}(q)=R'(q)\text{.}$

By the Second Derivative Test, the critical point does give a maximum revenue.

Solution

We define

\begin{equation*} \overline{R}(q) = \frac{R(q)}{q} \end{equation*}

for $q > 0\text{.}$ We first differentiate the average revenue:

\begin{equation*} \overline{R}'(q) = \frac{qR'(q) - R(q)}{q^2}\text{.} \end{equation*}

Notice that $\overline{R}'(q)$ is undefined when $q=0\text{,}$ but since this point is not in the domain of $\overline{R}(q)\text{,}$ we reject this as a critical point. Therefore, the only critical point will occur when

\begin{equation*} \overline{R}'(q) = 0 \implies qR'(q) - R(q) = 0 \implies R'(q) = \frac{R(q)}{q}\text{.} \end{equation*}

Notice that this last equality can be written as

\begin{equation*} R'(q) = \frac{R(q)}{q} = \overline{R(q)}\text{.} \end{equation*}

Hence, the critical point of the average revenue function satisfies

\begin{equation*} \overline{R(q)} = R'(q)\text{.} \end{equation*}

Differentiating once more, we see that

\begin{equation*} \overline{R}''(q) = \frac{\left(qR''(q)+R'(q)-R'(q)\right)-\left(qR'(q)-R(q)\right)}{q^4} = \frac{q^3R''(q)-qR'(q)+R(q)}{q^4}\text{.} \end{equation*}

Therefore, when $qR'(q) = q\text{,}$ we see that

\begin{equation*} \overline{R}'' = \frac{q^3R''(q)}{q^4} = \frac{R''(q)}{q}\text{.} \end{equation*}

Since $R(q)$ is concave down everywhere and $q>0\text{,}$ this means that the average revenue function is concave down when $\overline{R(q)} = R'(q)\text{.}$ By the Second Derivative Test, this critical point is a relative maximum.

The gross domestic product (GDP) of a certain country following a national crisis (at $t \equiv 0$) is approximated by

\begin{equation*} G(t)=-0.4t^3+4.8t^2+20 \ \ \ 0 \leq t \leq 12 \end{equation*}

where $G(t)$ is measured in billions of dollars. When during this time period is the GDP at its highest?

$G$ maximal at $t=8\text{.}$

Solution

We compute

\begin{equation*} G'(t) = -1.2t^2+9.6t \ \ \text{ and } \ \ G''(t) = -2.4t + 9.6\text{.} \end{equation*}

We now solve for the critical points of $G\text{:}$

\begin{equation*} \begin{split} G'(t) \amp = 0 \\ -1.2t^2 + 9.6 t \amp = 0 \\ t(-1.2t + 9.6) \amp = 0\\ t \amp =0, 8 \end{split} \end{equation*}

Since $t=0$ is an endpoint, it is not a critical point. Furthermore, since $G'(t)$ is defined for all $t$ in the domain of $G\text{,}$ we conclude that there $G$ has a single critical point at $t=8\text{.}$

We now compare the value of $G$ at the critical point and both boundary points:

\begin{equation*} G(0) = 20, \ \ G(8) = 122.4, \ \ G(12) = 20 \end{equation*}

Therefore, the GDP is at its highest after 8 years.

Suppose the amount of money in an account is given by

\begin{equation*} a(t)=-0.01t^4+0.5t^3+3.8t^2+12.6t+1200 \ \ \ 0\leq t \leq 55 \end{equation*}

thousands of dollars over 55 years. Determine the year during which the value of the account is maximal.

$a(t)$ maximal at $t=42.181\text{.}$

Solution

We differentiate:

\begin{equation*} a'(t) = -0.04 t^3 + 1.5 t^2 + 7.6 t + 12.6\text{.} \end{equation*}

Setting $a'(t)=0$ gives $t=42.1814$ as the only real solution. This is in the domain of $a(t)$ and is therefore a critical point of $a\text{.}$ Setting $a'(t)$ DNE gives no solutions.

We now compare:

\begin{equation*} a(0) = 1200, \ \ a(42.1814) = 14,360.7, \ \ a(55) = 5069.25\text{.} \end{equation*}

We conclude that the value of the account is maximal between the 42nd and 43rd years.

Over a time period of 5 years, it is shown that the number $N$ of independently owned bakeries is given by

\begin{equation*} N(t)=2+8.82t-7.73t^2+2.08t^3-0.175t^4 \ \ \ 0\leq t \leq 5 \end{equation*}

(in millions of bakeries). Determine the absolute extrema of the function $N$ on the interval $[0,5]$ and interpret your results.

$N$ has an absolute minimum of 0.86 million bakeries at about $x=3.1$ and an absolute maximum of 5.1 million bakeries at about $x=0.8\text{.}$

Solution

We differentiate with respect to $t\text{:}$

\begin{equation*} N'(t) = -0.7 t^3 + 6.24 t^2 - 15.46 t + 8.82 \end{equation*}

To find the critical points, we set $N'(t)=0\text{.}$ This gives three solutions: $t=0.812897\text{,}$ $t=3.0978\text{,}$ $t=5.00359\text{.}$ In considering the domain of $N\text{,}$ we reject the last solution. So we compare:

\begin{equation*} N(0) = 2, \ N(0.812897) = 5.10264, \ N(3.0978) = 0.860355\text{.} \end{equation*}

We conclude that over the 5 year period, the number of bakeries was maximal after about $0.8$ years, and minimal after abut $3$ years.

Suppose

\begin{equation*} C(q)=0.02q^3-0.03q^2+20q+300 \end{equation*}

is the daily cost function for a certain producer where $q$ is measured in units of thousands. Determine the level of production that minimizes the daily average cost per unit.

Solution

We first construct the average cost function.

\begin{equation*} \overline{C}(q) = \frac{C(q)}{q} = 0.02q^2-0.03q+20+\frac{300}{q}\text{,} \end{equation*}

for $q > 0\text{.}$ Differentiating, we find that

\begin{equation*} \overline{C}'(q) = 0.04q-0.03 -\frac{300}{q^2}\text{.} \end{equation*}

Since there are no points in the domain of $C$ for which $\overline{C}'(q)$ is undefined, it remains to set $\overline{C}'(q)=0\text{:}$

\begin{equation*} 0.04q-0.03 - \frac{300}{q^2} = 0 \implies q = 19.8276\text{.} \end{equation*}

We consider the following sign diagram for $\overline{C}'(q)\text{:}$

Since the average cost is decreasing on the interval $(0,86.9784)\text{,}$ and increasing from $(86.9784,\infty)\text{,}$ we conclude that the absolute minimal average cost occurs when about 20,000 units are produced.

Find the dimensions of the rectangle of largest area having fixed perimeter $100\text{.}$

$25\times 25$

Solution

Suppose we have the following rectangle:

Then the area is given by

\begin{equation*} A(x,y) = xy\text{,} \end{equation*}

and the perimeter is

\begin{equation*} P(x,y) = 2x+2y=100 \end{equation*}

where we have that $x,y \geq 0\text{.}$

We will use the constraint on the perimeter to write $y$ as a function $x$ :

\begin{equation*} 2x+2y = 100 \implies y = 50-x \end{equation*}

Therefore, we can write the area as a function of only one variable:

\begin{equation*} A(x) = x(50-x) = 50x-x^2 \end{equation*}

with domain $0\leq x\leq 50\text{.}$ Differentiating, we find

\begin{equation*} A'(x) = 50-2x\text{.} \end{equation*}

This means that $x=25$ is the only critical point. Now compare:

\begin{equation*} A(0) = 0, \ A(25) = 625, \ A(50) = 0\text{.} \end{equation*}

We conclude that the area is maximized when the dimensions are $25$x$25\text{.}$

Find the dimensions of the rectangle of largest area having fixed perimeter $P\text{.}$

$P/4\times P/4$

Solution

Suppose we have the following rectangle:

Then the area is given by

\begin{equation*} A(x,y) = xy\text{,} \end{equation*}

and the perimeter (for some fixed constant $P$) is

\begin{equation*} P= 2x+2y \end{equation*}

where we have that $x,y \geq 0\text{.}$

We will use the constraint on the perimeter to write $y$ as a function $x$ :

\begin{equation*} 2x+2y = P \implies y = \frac{P}{2}-x \end{equation*}

Therefore, we can write the area as a function of only one variable:

\begin{equation*} A(x) = x\left(\frac{P}{2}-x\right) = \frac{P}{2}x-x^2 \end{equation*}

with domain $0\leq x\leq \frac{P}{2}\text{.}$ Differentiating, we find

\begin{equation*} A'(x) = \frac{P}{2}-2x\text{.} \end{equation*}

This means that $x=P/4$ is the only critical point. Now compare:

\begin{equation*} A(0) = 0, \ A(P/4) = \frac{P^2}{16} \ A(P/2) = 0\text{.} \end{equation*}

We conclude that the area is maximized when the dimensions are $\dfrac{P}{4}$x$\dfrac{P}{4}\text{.}$

A box with square base and no top is to hold a volume $100\text{.}$ Find the dimensions of the box that requires the least material for the five sides. Also find the ratio of height to side of the base.

$\ds w=l=2\cdot 5^{2/3}\text{,}$ $\ds h=5^{2/3}\text{,}$ $\ds h/w=1/2$

Solution

Suppose the base of the box has side length$l$ and the height of the box is $h\text{.}$ Then the volume of the box is

\begin{equation*} V(l,h) = l^2 h\text{,} \end{equation*}

and the surface area is

\begin{equation*} S(l,h) = 4lh + l^2\text{.} \end{equation*}

If the volume of the box is constrained, then we can write

\begin{equation*} 100= l^2h \implies h = \frac{100}{l^2}\text{.} \end{equation*}

And so we can write

\begin{equation*} S(l) = \frac{400}{l} + l^2\text{,} \end{equation*}

where $l > 0\text{.}$

Differentiating, we find

\begin{equation*} S'(l) = 2l - \frac{400}{l^2}\text{.} \end{equation*}

Therefore, the critical point of the surface area is $l=2\cdot 5^{2/3}\text{.}$

We construct the following sign diagram for $S'\text{:}$

Therefore, the surface area is decreasing for $l\in(0,2\cdot 5^{2/3})$ and increasing for $l \in (2\cdot 5^{2/3},\infty)\text{.}$ We conclude that the surface area is minimised with dimensions $2\cdot 5^{2/3}$ x $2\cdot 5^{2/3}$ x $5^{2/3}\text{.}$

A box with square base is to hold a volume $200\text{.}$ The bottom and top are formed by folding in flaps from all four sides, so that the bottom and top consist of two layers of cardboard. Find the dimensions of the box that requires the least material. Also find the ratio of height to side of the base.

$\ds \root 3\of {100}\times\root 3\of {100}\times 2\root 3\of {100}\text{,}$ $h/s=2$

Solution

Suppose the dimensions of the base of the box are $l$x$l\text{,}$ and the height of the box is $h\text{.}$ The volume is then

\begin{equation*} V(l,h) = l^2 h\text{,} \end{equation*}

and the surface area (noting that the top and bottom are doubled) is

\begin{equation*} S(l,h) = 4lh + 2l^2+2l^2\text{.} \end{equation*}

Since the volume is constrained at $V=200\text{,}$ we can write

\begin{equation*} h = \frac{200}{l^2}\text{,} \end{equation*}

which means that

\begin{equation*} S(l) = \frac{800}{l} + 4l^2\text{,} \end{equation*}

where $l > 0\text{.}$

Differentiating $S$ with respect to $l\text{,}$ we compute

\begin{equation*} S'(l) = 8x - \frac{800}{x^2}\text{.} \end{equation*}

Therefore,

\begin{equation*} S'(l) = 0 \implies x^3 = 100 \implies x = 10^{2/3}\text{.} \end{equation*}

We construct the following sign diagram for $S'\text{:}$

We conclude that the surface area is minimized with dimensions $10^{2/3}$ x $10^{2/3}$ x $2\cdot 10^{2/3}\text{.}$

A box with square base and no top is to hold a volume $V\text{.}$ Find (in terms of $V$) the dimensions of the box that requires the least material for the five sides. Also find the ratio of height to side of the base. (This ratio will not involve $V\text{.}$)

$\ds w=l=2^{1/3}V^{1/3}\text{,}$ $\ds h=V^{1/3}/2^{2/3}\text{,}$ $h/w=1/2$

Solution

Suppose the base of the box has side length$l$ and the height of the box is $h\text{.}$ Then the volume of the box is

\begin{equation*} V(l,h) = l^2 h\text{,} \end{equation*}

and the surface area is

\begin{equation*} S(l,h) = 4lh + l^2\text{.} \end{equation*}

If the volume of the box is constrained, then we can write

\begin{equation*} V= l^2h \implies h = \frac{V}{l^2}\text{.} \end{equation*}

And so we can write

\begin{equation*} S(l) = \frac{4V}{l} + l^2\text{,} \end{equation*}

where $l > 0\text{.}$

Differentiating, we find

\begin{equation*} S'(l) = 2l - \frac{4V}{l^2}\text{.} \end{equation*}

Therefore, the critical point of the surface area is when $l = (2V)^{1/3}\text{.}$

We construct the following sign diagram for $S'\text{:}$

Therefore, the surface area is minimised with dimensions $(2V)^{1/3}$ x $(2V)^{1/3}$ x $h\text{,}$ where

\begin{equation*} h = \frac{V}{(2V)^{2/3}}\text{.} \end{equation*}

This gives a ratio

\begin{equation*} \frac{h}{l} = \frac{1}{2}\text{.} \end{equation*}

You have $100$ feet of fence to make a rectangular play area alongside the wall of your house. The wall of the house bounds one side. What is the largest size possible (in square feet) for the play area?

$1250$ square feet

Solution

We draw the following diagram, where the blue lines indicate the fencing ($x$ and $y$ are measured in feet):

The area will then be

\begin{equation*} A(x,y) = xy\text{,} \end{equation*}

with perimeter

\begin{equation*} P(x,y) = 2x + y\text{.} \end{equation*}

Since there is a fixed amount of fencing material, we require

\begin{equation*} 100 = 2x+y \implies y = 100-2x\text{.} \end{equation*}

Hence, we can write the area as a function of $x$ only:

\begin{equation*} A(x) = x(100-2x)\text{,} \end{equation*}

for $0 \leq x \leq 50\text{.}$ Now differentiate:

\begin{equation*} A'(x) = 100-4x\text{,} \end{equation*}

and so $x = 25$ gives the only critical point. Now compare:

\begin{equation*} A(0) = 0, \ A(25) = 1250, \ A(50) = 0\text{.} \end{equation*}

We conclude that 1250 ft$^2$ is the maximal area.

You have $l$ feet of fence to make a rectangular play area alongside the wall of your house. The wall of the house bounds one side. What is the largest size possible (in square feet) for the play area?

$\ds l^2/8$ square feet

Solution

We draw the following diagram, where the blue lines indicate the fencing ($x$ and $y$ are measured in feet):

The area will then be

\begin{equation*} A(x,y) = xy\text{,} \end{equation*}

with perimeter

\begin{equation*} P(x,y) = 2x + y\text{.} \end{equation*}

Since there is a fixed amount of fencing material, we require

\begin{equation*} l = 2x+y \implies y = l-2x\text{.} \end{equation*}

Hence, we can write the area as a function of $x$ only:

\begin{equation*} A(x) = x(l-2x)\text{,} \end{equation*}

for $0 \leq x \leq l/2\text{.}$ Now differentiate:

\begin{equation*} A'(x) = l-4x\text{,} \end{equation*}

and so $x = l/4$ gives the only critical point. Now compare:

\begin{equation*} A(0) = 0, \ A(l/4) = \frac{l^2}{8}, \ A(50) = 0\text{.} \end{equation*}

We conclude that $\dfrac{l^2}{8}$ ft$^2$ is the maximal area.

Marketing tells you that if you set the price of an item at $10 then you will be unable to sell it, but that you can sell 500 items for each dollar below$10 that you set the price. Suppose your fixed costs total $3000, and your marginal cost is$2 per item. What is the most profit you can make?

$5000 Solution Let $q$ be the number of units in question. If our fixed costs total$3000 and our marginal cost is $2 per unit, then our total cost function $C(q)$ (in dollars) is given by \begin{equation*} C(q)=3000+2q\text{.} \end{equation*} Now, let $p$ be the unit price (in dollars). Then marketing tells us that we can sell $q=500$ units for each dollar below$10 that we set the price. Let $0 \le h \le 10\text{.}$ Then

\begin{equation*} p= 10-h \implies q=500h\text{,} \end{equation*}

for some $h \in [0,10]\text{.}$ We need to construct our revenue $R$ as a function of $q\text{,}$ and so we let $h=q/500$ (where we rearranged the second equation above). Then,

\begin{equation*} R(q)=pq=(10-h)q=\left(10-\frac{q}{500}\right)q = 10q-\frac{q^2}{500} \ \ \ (0\leq q \leq 5000)\text{.} \end{equation*}

Finally, we construct our profit function $P$ as a function of $q$ units sold.

\begin{equation*} P(q) = R(q) - C(q) = \left(10q-\frac{q^2}{500}\right)-\left(3000+2q\right) = 8q-\frac{q^2}{500}-3000\text{,} \end{equation*}

where $0\leq q \leq 5000\text{.}$

Now, find the critical points of $P\text{.}$

\begin{equation*} \begin{split} P'(q) \amp = 0 \\ 8-\frac{2q}{500} \amp = 0, \end{split} \end{equation*}

which gives $q=2000$ as our only interior critical point. Comparing

\begin{equation*} P(0)=-3000, \ \ P(2000)= 5000, \ \ \text{ and } \ \ P(5000)=-13,000\text{,} \end{equation*}

we conclude that the profit is at an absolute maximum when we sell $2000$ units. Since $q=2000\text{,}$ this means that $h = 4$ and hence $p = 10-4 = 6.$ Therefore, each unit is sold at $6 per unit. The profit yielded will be$5000.

Find the area of the largest rectangle that fits inside a semicircle of radius $10$ (one side of the rectangle is along the diameter of the semicircle).

$100$

Solution

We first draw a sketch of the scenario. The shaded area is the area to be optimized.

Therefore,

\begin{equation*} A(w,h) = w \cdot h\text{.} \end{equation*}

We wish to write area as a function of $w$ only. So let

\begin{equation*} h = h(w) = \sqrt{100-\left(\frac{w}{2}\right)^2}\text{.} \end{equation*}

Hence,

\begin{equation*} A(w) = w \sqrt{100 -\frac{w^2}{4}}\text{,} \end{equation*}

where $0 \leq w \leq 200\text{.}$

We can now differentiate with respect to $w\text{:}$

\begin{equation*} A'(w) = \diff{}{w} w \sqrt{100-\frac{w^2}{4}} = \frac{200-w^2}{400-w^2} \end{equation*}

And so $A'(w)=0$ when $w= \pm \sqrt{2} 100$ and $A'(w)$ is undefined when $w = \pm 200\text{.}$ Since the domain of $A$ is $0 \leq w \leq 200\text{,}$ we conclude that there is only one interior critical point at $w = \sqrt{2} 100\text{.}$ We now compare the value of $A$ at this critical point with value of $A$ at the boundary points:

\begin{equation*} A(0) = 0, \ \ A(\sqrt{2} 100) = 100, \ \ A(200) = 0 \end{equation*}

Therefore, the maximum area of the rectangle is $100\text{.}$

Find the area of the largest rectangle that fits inside a semicircle of radius $r$ (one side of the rectangle is along the diameter of the semicircle).

$\ds r^2$

Solution

We first draw a sketch of the scenario. The shaded area is the area to be optimized.

Therefore,

\begin{equation*} A(w,h) = w \cdot h\text{.} \end{equation*}

We wish to write area as a function of $w$ only. So let

\begin{equation*} h = h(w) = \sqrt{r^2-\left(\frac{w}{2}\right)^2}\text{.} \end{equation*}

Hence,

\begin{equation*} A(w) = w \sqrt{r^2 -\frac{w^2}{4}}\text{,} \end{equation*}

where $0 \leq w \leq 2r\text{.}$

We can now differentiate with respect to $w\text{:}$

\begin{equation*} A'(w) = \diff{}{w} w \sqrt{r^2-\frac{w^2}{4}} = \frac{2r^2-w^2}{4r^2-w^2} \end{equation*}

And so $A'(w)=0$ when $w= \pm \sqrt{2} r$ and $A'(w)$ is undefined when $w = \pm 2r\text{.}$ Since the domain of $A$ is $0 \leq w \leq 2r\text{,}$ we conclude that there is only one interior critical point at $w = \sqrt{2} r\text{.}$ We now compare the value of $A$ at this critical point with value of $A$ at the boundary points:

\begin{equation*} A(0) = 0, \ \ A(\sqrt{2} r) = \sqrt{2} r \sqrt{\frac{r^2}{2}} = r^2, \ \ A(2r) = 0 \end{equation*}

Therefore, the maximum area of the rectangle is $r^2\text{.}$

For a cylinder with surface area $50\text{,}$ including the top and the bottom, find the ratio of height to base radius that maximizes the volume.

$h/r=2$

Solution

Let $r$ be the radius and $h$ be the height. Then the volume of the cylinder is

\begin{equation*} V(r,h) = \pi r^2 h\text{,} \end{equation*}

and the total surface area (including top and bottom) is

\begin{equation*} S(r,h) = 2\pi rh+2\pi r^2\text{.} \end{equation*}

Constraining the surface area allows us the write $h$ as a function of $r\text{:}$

\begin{equation*} 50 = 2\pi rh+2\pi r^2 \implies h = \frac{50-2\pi r^2}{2\pi r}\text{.} \end{equation*}

Hence, as a function of the radius, we write

\begin{equation*} V(r) = \pi r^2 \left(\frac{50-2\pi r^2}{2\pi r}\right) = r(25-\pi r^2)\text{.} \end{equation*}

Notice that the domain of $V(r)$ is $0 \lt r \leq \frac{5}{\sqrt{\pi}}\text{.}$

We can now differentiate the volume with respect to $r\text{:}$

\begin{equation*} V'(r) = 25 - 3\pi r^2\text{.} \end{equation*}

Setting $V'(r) = 0$ gives $r = \pm \frac{5}{\sqrt{3\pi}}\text{.}$ We reject the negative solution, and so are left with only one critical point. To determine the absolute maximum, we compare values:

\begin{equation*} V(0) = 0, \ V\left(\frac{5}{\sqrt{3\pi}}\right) = \frac{250}{3\sqrt{3\pi}}, \ V\left(\frac{5}{\sqrt{\pi}}\right) = 0\text{.} \end{equation*}

We conclude that the maximum volume is achievable with a height to radius ratio of

\begin{equation*} \frac{h}{r} = 2\text{.} \end{equation*}

For a cylinder with given surface area $S\text{,}$ including the top and the bottom, find the ratio of height to base radius that maximizes the volume.

$h/r=2$

Solution

Let $r$ be the radius and $h$ be the height. Then the volume of the cylinder is

\begin{equation*} V(r,h) = \pi r^2 h\text{,} \end{equation*}

and the total surface area (including top and bottom) is

\begin{equation*} S(r,h) = 2\pi rh+2\pi r^2\text{.} \end{equation*}

Constraining the surface area allows us the write $h$ as a function of $r\text{:}$

\begin{equation*} S = 2\pi rh+2\pi r^2 \implies h = \frac{S-2\pi r^2}{2\pi r}\text{.} \end{equation*}

Hence, as a function of the radius, we write

\begin{equation*} V(r) = \pi r^2 \left(\frac{S-2\pi r^2}{2\pi r}\right) = r(S/2-\pi r^2)\text{.} \end{equation*}

Notice that the domain of $V(r)$ is $0 \lt r \leq \frac{\sqrt{S/2}}{\sqrt{\pi}}\text{.}$

We can now differentiate the volume with respect to $r\text{:}$

\begin{equation*} V'(r) = \frac{S}{2} - 3\pi r^2\text{.} \end{equation*}

Setting $V'(r) = 0$ gives $r = \pm \frac{\sqrt{S/2}}{\sqrt{3\pi}}\text{.}$ We reject the negative solution, and so are left with only one critical point. To determine the absolute maximum, we compare values:

\begin{equation*} V(0) = 0, \ V\left(\frac{\sqrt{S/2}}{\sqrt{3\pi}}\right) = \frac{\sqrt{S}}{3\sqrt{3\pi}}, \ V\left(\frac{5}{\sqrt{\pi}}\right) = 0\text{.} \end{equation*}

We conclude that the maximum volume is achievable with a height to radius ratio of

\begin{equation*} \frac{h}{r} = 2\text{.} \end{equation*}

You want to make cylindrical containers to hold 1 liter using the least amount of construction material. The side is made from a rectangular piece of material, and this can be done with no material wasted. However, the top and bottom are cut from squares of side $2r\text{,}$ so that $\ds 2(2r)^2=8r^2$ of material is needed (rather than $\ds 2\pi r^2\text{,}$ which is the total area of the top and bottom). Find the dimensions of the container using the least amount of material, and also find the ratio of height to radius for this container.

$r=5\text{,}$ $h=40/\pi\text{,}$ $h/r=8/\pi$

Solution

We make the following diagrams illustrating the amount of construction material required per container. For the top and bottom, we need:

And for the side of the cylinder, we need:

Suppose $r$ and $h$ are measured in cm. Then the total surface area of construction material (in cm$^2$) is

\begin{equation*} S(h,r) = 8r^2 + 2\pi r h\text{.} \end{equation*}

We further have the constraint on the volume: (note that 1L = 1000 cm$^3$)

\begin{equation*} V(h,r) = \pi r^2 h = 1000\text{.} \end{equation*}

Therefore, we write (for $r \neq 0$)

\begin{equation*} h = \frac{1000}{\pi r^2}\text{,} \end{equation*}

and so we can write the surface area as a function of radius only:

\begin{equation*} S(r) = 8r^2 + \frac{2000}{r}, \ \ \ r \neq 0\text{.} \end{equation*}

Differentiating with respect to $r\text{,}$ we find

\begin{equation*} S'(r) = 16r - \frac{2000}{r^2}\text{.} \end{equation*}

Notice that $S'(r)$ is undefined at $r=0\text{,}$ but this point is not in the domain of $S$ and so is not a critical point. Setting $S'(r)=0$ gives $r=5$ as the only critical point.

Notice that

\begin{equation*} S''(r) = 16 + \frac{4000}{r^3}\text{,} \end{equation*}

and so $S''(r) > 0$ for all $r > 0\text{.}$

Since $S$ is concave up on its entire domain, and has a single critical point at $r=5\text{,}$ this must be the absolute minimum. We conclude that the amount of construction material is minimized when $r=5$ cm and $h= \dfrac{40}{\pi}$ cm. This gives a ratio of

\begin{equation*} \frac{h}{r} = \frac{40}{\pi 5 } = \frac{8}{\pi}\text{.} \end{equation*}

You want to make cylindrical containers of a given volume $V$ using the least amount of construction material. The side is made from a rectangular piece of material, and this can be done with no material wasted. However, the top and bottom are cut from squares of side $2r\text{,}$ so that $\ds 2(2r)^2=8r^2$ of material is needed (rather than $\ds 2\pi r^2\text{,}$ which is the total area of the top and bottom). Find the optimal ratio of height to radius.

$8/\pi$

Solution

We make the following diagrams illustrating the amount of construction material required per container. For the top and bottom, we need:

And for the side of the cylinder, we need:

Suppose $r$ and $h$ are measured in some units. Then the total surface area of construction material (in units$^2$) is

\begin{equation*} S(h,r) = 8r^2 + 2\pi r h\text{.} \end{equation*}

We further have the constraint on the volume:

\begin{equation*} V(h,r) = \pi r^2 h = V\text{,} \end{equation*}

where $V$ is some positive, nonzero constant and is measured in units$^3\text{.}$ Therefore, we write (for $r \neq 0$)

\begin{equation*} h = \frac{V}{\pi r^2}\text{,} \end{equation*}

and so we can write the surface area as a function of radius only:

\begin{equation*} S(r) = 8r^2 + \frac{2V}{r}, \ \ \ r \neq 0\text{.} \end{equation*}

Differentiating with respect to $r\text{,}$ we find

\begin{equation*} S'(r) = 16r - \frac{2V}{r^2}\text{.} \end{equation*}

Notice that $S'(r)$ is undefined at $r=0\text{,}$ but this point is not in the domain of $S$ and so is not a critical point. Setting $S'(r)=0$ gives $r=\frac{1}{2}\sqrt[3]{V}$ as the only critical point.

Notice that

\begin{equation*} S''(r) = 16 + \frac{4V}{r^3}\text{,} \end{equation*}

and so $S''(r) > 0$ for all $r > 0\text{.}$

Since $S$ is concave up on its entire domain, and has a single critical point at $r=\frac{1}{2}\sqrt[3]{V}\text{,}$ this must be the absolute minimum. We conclude that the amount of construction material is minimized when $r=\frac{1}{2}\sqrt[3]{V}$ cm and $h= \dfrac{4\sqrt[3]{V}}{\pi}$ cm. This gives a ratio of

\begin{equation*} \frac{h}{r} = \frac{4/\pi}{1/2} = \frac{8}{\pi}\text{.} \end{equation*}

Given a right circular cone, you put an upside-down cone inside it so that its vertex is at the center of the base of the larger cone and its base is parallel to the base of the larger cone. If you choose the upside-down cone to have the largest possible volume, what fraction of the volume of the larger cone does it occupy? Let $H$ and $R$ be the height and base radius of the larger cone, and let $h$ and $r$ be the height and base radius of the smaller cone. Hint

Use similar triangles to get an equation relating $h$ and $r\text{.}$

$4/27$

Solution

We make a sketch of the scenario described:

Both cones are right circular cones, so if we let $H$ and $R$ be the height and radius of the outer (larger) cone, respectively, with $h\text{,}$ $r$ being the height and radius of the inner (smaller) cone, we must have:

Then, by similar triangles,

\begin{equation*} \frac{R}{r} = \frac{H}{H-h} \end{equation*}

for $0 \lt r \lt R$ and $0 \lt h \lt H\text{.}$ Notice that if $h=H$ we would require $r=0\text{,}$ and similarly if $r=R$ we would require $h=0\text{.}$ Similarly, taking either $r$ or $h$ equal to zero would result in a volume of zero.

$H$ and $R$ are fixed and nonzero, so we can write

\begin{equation*} r = \frac{R(H-h)}{H}\text{.} \end{equation*}

The volume of the inner cone is

\begin{equation*} V(r,h) = \pi r^2 \frac{h}{3}\text{,} \end{equation*}

and by using the above relation, we have for $0 \lt h \lt H\text{,}$

\begin{equation*} V(h) = \pi \left(\frac{R(H-h)}{H}\right)^2 \frac{h}{3} = \frac{\pi R^2}{3H^2} (H-h)^2h\text{.} \end{equation*}

Now differentiate with respect to $h\text{:}$

\begin{equation*} V'(h) = \frac{\pi R^2}{3H^2} \left(3h^2-4hH+H^2\right)\text{.} \end{equation*}

Solving $V'(h) = 0$ gives

\begin{equation*} h = \frac{H}{3}, \text{ or } h = H\text{.} \end{equation*}

The solution $h=H$ is not in the domain of $V\text{,}$ and so we get one critical point at $h = \frac{H}{3}\text{.}$ The corresponding radius is $r = \frac{2R}{3}\text{.}$ Hence, the maximum volume of the inner cone is

\begin{equation*} V_{in} = \frac{4\pi}{3^4} R^2 H\text{.} \end{equation*}

Since the volume of the outer cone is

\begin{equation*} V_{out} = \pi R^2 \frac{H}{3}\text{,} \end{equation*}

we have

\begin{equation*} \frac{V_{in}}{V_{out}} = \frac{4}{27}\text{.} \end{equation*}

A container holding a fixed volume is being made in the shape of a cylinder with a hemispherical top. (The hemispherical top has the same radius as the cylinder.) Find the ratio of height to radius of the cylinder which minimizes the cost of the container if (a) the cost per unit area of the top is twice as great as the cost per unit area of the side, and the container is made with no bottom; (b) the same as in (a), except that the container is made with a circular bottom, for which the cost per unit area is 1.5 times the cost per unit area of the side.

(a) 2, (b) $7/2$

Solution

The container is made up of two parts:

Therefore, the total volume of the container (for some fixed, nonzero constant $V$) is

\begin{equation*} V = \pi r^2 h +\frac{2}{3} \pi r^3\text{.} \end{equation*}

This means that we can write

\begin{equation*} h = \frac{V-\frac{2}{3}\pi r^3}{\pi r^2}\text{,} \end{equation*}

where $0 \lt r \lt \sqrt[3]{\frac{3V}{2\pi}}\text{.}$

1. If the container is to be made with no bottom, then the total surface area is

\begin{equation*} S(r,h) = 2\pi r h + 2\pi r^2\text{.} \end{equation*}

If the cost of the top material is twice the cost of the side material, then we want to minimize the cost function:

\begin{equation*} C(r,h) = 2\pi rh + 4\pi r^2\text{.} \end{equation*}

Using the relation $h=h(r)\text{,}$ we write (for $r>0$)

\begin{equation*} C(r) = \frac{2V}{r} + \frac{16}{3} \pi r^2\text{.} \end{equation*}

Differentiating with respect to $r$ gives

\begin{equation*} C'(r) = -\frac{2V}{r^2} + \frac{32}{3}\pi r\text{.} \end{equation*}

We set $C'(r) = 0$ to find one critical point (remember that $r=0$ is not in the domain of $C$):

\begin{equation*} r = \sqrt[3]{\frac{3V}{16\pi}}\text{.} \end{equation*}

At this value of $r\text{,}$ the corresponding height is

\begin{equation*} h = \frac{V-\frac{2}{3}\pi r^3}{\pi r^2} = \frac{7V}{8\pi \left(\frac{3V}{16\pi}\right)^{2/3}}\text{.} \end{equation*}

This gives a ratio of

\begin{equation*} \frac{r}{h} = \frac{3V \cdot 8\pi}{16 \pi \cdot 7V} = \frac{3}{14}\text{.} \end{equation*}
2. Now if the container is to be made with a circular bottom, the total surface area is

\begin{equation*} S(r,y) = 2\pi r h + 2\pi r^2 + \pi r^2\text{.} \end{equation*}

Hence, if the material cost for the top is twice the cost of the side material, and the cost for the bottom is 1.5 times to cost of the side material, we can construct the following cost function:

\begin{equation*} C(r,h) = (1) \cdot 2\pi r h + (2) \cdot 2\pi r^2 + (1.5) \cdot \pi r^2 = 2\pi rh +\frac{11}{2} \pi r^2\text{.} \end{equation*}

So as a function of $r$ only, we find

\begin{equation*} C(r) = \frac{2V}{r} + \frac{25}{6} \pi r^2\text{.} \end{equation*}

Now differentiate:

\begin{equation*} C'(r) = -\frac{2V}{r^2} + \frac{25}{3} \pi r\text{.} \end{equation*}

Setting $C'(r) = 0$ gives

\begin{equation*} r = \sqrt[3]{\frac{6V}{25\pi}}\text{.} \end{equation*}

At this value of $r\text{,}$ the corresponding value of $h$ is

\begin{equation*} h = \frac{13V}{25\pi \left(\frac{6V}{25\pi}\right)^{2/3}}\text{.} \end{equation*}

This gives a ratio of

\begin{equation*} \frac{r}{h} = \frac{6}{13}\text{.} \end{equation*}

A piece of cardboard is 1 meter by $1/2$ meter. A square is to be cut from each corner and the sides folded up to make an open-top box. What are the dimensions of the box with maximum possible volume?

$\ds{\sqrt3\over6}\times{\sqrt3\over6}+{1\over2}\times {1\over4}-{\sqrt3\over 12}$

Solution

The piece of cardboard is to be cut as follows:

This results in a box with dimensions $(1-2x) \times (0.5-2x) \times x\text{.}$ The volume of this box is then

\begin{equation*} V(x) = (1-2x)(0.5-2x)x = 4 x^3 - 3 x^2 + 0.5 x\text{,} \end{equation*}

for $0 \leq x \leq 0.25\text{.}$ We differentiate:

\begin{equation*} V'(x) = 12 x^2 - 6 x + 0.5\text{,} \end{equation*}

and so $V'(x) = 0$ for $x= 1/4 \pm 1/\sqrt{48}\text{.}$ Only one of these critical points is in the domain of $V\text{:}$

\begin{equation*} x = \frac{1}{4} - \frac{\sqrt{3}}{12}\text{.} \end{equation*}

Now compare:

\begin{equation*} V(0) = 0, \ \ V\left(\frac{1}{4} - \frac{\sqrt{3}}{12}\right) \approx 0.024, \ \ V(0.25) = 0\text{.} \end{equation*}

We conclude that the dimensions which maximize the volume of the box are

\begin{equation*} \frac{\sqrt{3}}{6} \ \text{m} \ \times \frac{\sqrt{3}}{6} + \frac{1}{2}\ \text{m} \ \times \frac{1}{4} - \frac{\sqrt{3}}{12}\ \text{m}\text{.} \end{equation*}

(a) A square piece of cardboard of side $a$ is used to make an open-top box by cutting out a small square from each corner and bending up the sides. How large a square should be cut from each corner in order that the box have maximum volume? (b) What if the piece of cardboard used to make the box is a rectangle of sides $a$ and $b\text{?}$

(a) $a/6\text{,}$ (b) $\ds (a+b-\sqrt{a^2-ab+b^2})/6$

Solution

We start with the general case, where we have the following scenario: The piece of cardboard is to be cut as follows:

Assume that $b \leq a\text{.}$ Then $0 \leq x \leq \frac{b}{2}\text{.}$ The volume of the resulting box is therefore

\begin{equation*} V(x) = x (a-2x) (b - 2x)\text{.} \end{equation*}

Differentiating, we find

\begin{equation*} V'(x) = a(b-4x)+4x(3x-b)\text{.} \end{equation*}

Now set $V'(x) = 0\text{,}$ which gives

\begin{equation*} x = \frac{1}{6} \bigl(\pm \sqrt{a^2 - a b + b^2} + a + b\bigr)\text{.} \end{equation*}
1. If we have $a=b\text{,}$ then

\begin{equation*} x = \frac{1}{6} \bigl(\pm \sqrt{b^2 - b^2 + b^2} + b + b\bigr)= \frac{1}{6} \bigl( 2b \pm b\bigr)\text{.} \end{equation*}

Hence, $x=\frac{b}{6}$ is the only critical point. We compare:

\begin{equation*} V(0) = 0, \ \ V(b/6) = \frac{2b^3}{27}, \ \ V(b/2) = 0\text{.} \end{equation*}

We conclude that $x = \frac{b}{6}$ is the optimal size of the square.

2. Now let $a \neq b\text{.}$ Since $x \leq b/2\text{,}$ and we took $b \lt a$ we see that

\begin{equation*} x = \frac{1}{6} \bigl(-\sqrt{a^2 - a b + b^2} + a + b\bigr) \end{equation*}

is the only critical point. Since

\begin{equation*} V(0) = 0, \ \ \text{ and } \ \ V(b/2) = 0\text{,} \end{equation*}

we conclude that this choice of $x$ must maximize the volume.

A window consists of a rectangular piece of clear glass with a semicircular piece of colored glass on top; the colored glass transmits only $1/2$ as much light per unit area as the the clear glass. If the distance from top to bottom (across both the rectangle and the semicircle) is 2 meters and the window may be no more than 1.5 meters wide, find the dimensions of the rectangular portion of the window that lets through the most light.

$1.5$ meters wide by $1.25$ meters tall

A window consists of a rectangular piece of clear glass with a semicircular piece of colored glass on top. Suppose that the colored glass transmits only $k$ times as much light per unit area as the clear glass ($k$ is between $0$ and $1$). If the distance from top to bottom (across both the rectangle and the semicircle) is a fixed distance $H\text{,}$ find (in terms of $k$) the ratio of vertical side to horizontal side of the rectangle for which the window lets through the most light.

If $k\le 2/\pi$ the ratio is $(2-k\pi)/4\text{;}$ if $k\ge 2/\pi\text{,}$ the ratio is zero: the window should be semicircular with no rectangular part.

Solution

We make the following diagram, where $0 \leq w \leq 1.5$ m:

Therefore, the surface area of the window is the surface area of the top part:

\begin{equation*} \frac{\pi w^2}{4}\text{,} \end{equation*}

plus the surface area of the rectangular part:

\begin{equation*} w \left(2-\frac{w}{2}\right)\text{.} \end{equation*}

So if the top part only transmits half as much light as the rectangular part, we can construct the following function which is to be maximized:

\begin{equation*} I(w) = w \left(2-\frac{w}{2}\right) + \frac{\pi w^2}{8}\text{,} \end{equation*}

where $0 \leq w \leq 1.5\text{.}$ We differentiate:

\begin{equation*} I'(w) = \frac{1}{4} \left(\pi-4\right) w + 2\text{.} \end{equation*}

So $I'(w) = 0$ when $w = \frac{8}{4-\pi}\text{.}$ This is not in the domain of $I\text{,}$ and so there are no critical points. So we compare:

\begin{equation*} I(0) = 0, \ I(1.5) \approx 2.8\text{.} \end{equation*}

We conclude that the amount of light let in is maximized when the width of the window is $1.5$ m. We make the following diagram:

Therefore, the surface area of the window is the surface area of the top part:

\begin{equation*} \frac{\pi w^2}{4}\text{,} \end{equation*}

plus the surface area of the rectangular part:

\begin{equation*} w \left(H-\frac{w}{2}\right)\text{.} \end{equation*}

So if the top part only transmits only $0 \leq k \leq 1$ times as much light as the rectangular part, we can construct the following function which is to be maximized:

\begin{equation*} I(w) = w \left(H-\frac{w}{2}\right) + \frac{k\pi w^2}{4}\text{,} \end{equation*}

where $0 \leq w \leq 2H\text{.}$ We differentiate:

\begin{equation*} I'(w) = H - w + \frac{k\pi w}{2}\text{.} \end{equation*}

We set $I'(w)=0$ to find any critical points:

\begin{equation*} H - w + \frac{k\pi w}{2} = 0 \implies w = \frac{H}{1-\frac{k\pi}{2}}\text{.} \end{equation*}

Therefore, if $k \geq 2/\pi\text{,}$ there is no critical point. But for $0 \leq k \lt \frac{2}{\pi}\text{,}$ there is a critical point at $w = \dfrac{H}{1-\frac{k\pi}{2}}\text{..}$

Let's first consider $k \geq 2/\pi\text{:}$ then we compare

\begin{equation*} I(0)= 0, \ \ I(2H) = \frac{k\pi H^2}{4}\text{.} \end{equation*}

Therefore, the window which lets in the most amount of light is a semicircle (i.e. there should be no rectangular part).

Now if $0 \leq k \lt 2/\pi\text{,}$ we find that the light intensity is maximized when $w = \dfrac{H}{1-\frac{k\pi}{2}}\text{.}$

You are designing a poster to contain a fixed amount $A$ of printing (measured in square centimeters) and have margins of $a$ centimeters at the top and bottom and $b$ centimeters at the sides. Find the ratio of vertical dimension to horizontal dimension of the printed area on the poster if you want to minimize the amount of posterboard needed.

$a/b$

Solution

We make a sketch of the scenario described:

The total area of the poster board is therefore

\begin{equation*} f(x,y) = (x+2b)\times(y+2a)\text{.} \end{equation*}

We want to minimize $f$ subject to the constraint $A=x\times y\text{.}$ We use this constraint to write $f$ as a function of $x$ only:

\begin{equation*} f(x) = \left(x+2b\right)\left(\frac{A}{x}+2a\right)\text{.} \end{equation*}

Now differentiate:

\begin{equation*} f'(x) = 2a - \frac{2Ab}{x^2}\text{.} \end{equation*}

Therefore, solving $f'(x)=0$ gives

\begin{equation*} x = \pm \sqrt{\frac{Ab}{a}}\text{.} \end{equation*}

Since $x$ is a side-length, we reject the negative solution. Therefore, there is a single critical point at $x=\sqrt{Ab/a}\text{.}$ Since

\begin{equation*} f''(x) = \frac{4AB}{x^3}\text{,} \end{equation*}

which is positve for all $x>0\text{,}$ we conclude that this critical point is the absolute minimum.

Therefore, the area of the poster board is minimized with a ratio of

\begin{equation*} \frac{x}{y} = \frac{\sqrt{Ab/a}\sqrt{Ab/a}}{A} = \frac{b}{a}\text{.} \end{equation*}

What fraction of the volume of a sphere is taken up by the largest cylinder that can be fit inside the sphere?

$\ds 1/\sqrt3\approx 58\%$

Solution

Consider the following cross-section of a cylinder of radius $r_c$ and height $h$ inside a sphere of radius $r_s\text{:}$

Then we see that, in order for the cylinder to fit inside the sphere, we must have

\begin{equation*} \left(\frac{h}{2}\right)^2 +r_c^2 = r_s^2\text{.} \end{equation*}

The volume of the cylinder is

\begin{equation*} V_c(r_c,h) = \pi r_c^2 h\text{,} \end{equation*}

and so we can use the constraint on $r_c$ to write

\begin{equation*} V_c(h) = \pi \left(r_s^2-\frac{h^2}{4}\right) h\text{.} \end{equation*}

This formula is only valid for $0 \leq h \leq 2r_s\text{.}$

Differentiating, we find

\begin{equation*} V_c'(h) = \pi \left(r_s^2-\frac{3h^2}{4}\right)\text{.} \end{equation*}

Setting $V_c'(h) = 0\text{,}$ we find

\begin{equation*} h=\frac{2}{\sqrt{3}}r_s \end{equation*}

to be the only critical point of $V_c$ (we reject the negative solution). We now compare the value of $V_c$ at this critical point and at the boundary points:

\begin{equation*} V_c (0) = 0, \ V_c\left(\frac{2}{\sqrt{3}}r_s\right) = \frac{4\pi}{3\sqrt{3}} r_s^3, \ V_c(2r_s) = 0\text{.} \end{equation*}

Therefore, the maximum possible volume of the cylinder is $\pi r_s^2\text{.}$ The ratio of the volume of this cylinder to the volume of the sphere is therefore

\begin{equation*} \frac{V_c}{V_s} = \frac{1}{\sqrt{3}}\text{.} \end{equation*}

The U.S. post office will accept a box for shipment only if the sum of the length and girth is at most 108 in. (Girth is the maximum distance around the package perpendicular to the length; for a rectangular box, the length is the largest of the three dimensions.) Find the dimensions of the largest acceptable box with square front and back.

$18\times18\times36$

Solution

Let $l$ be the length of the box and $w$ be the width of the box in inches. If the box has square front and back faces, then the volume of this box must be

\begin{equation*} V(l,g) = l \times w \times w\text{,} \end{equation*}

where we require $l + 4w \leq 108\text{.}$ In order to maximize the volume, let

\begin{equation*} w = \frac{108-l}{4} = 27-\frac{l}{4}\text{.} \end{equation*}

Then we can rewrite the volume as a function of length only:

\begin{equation*} V(l) = l \left(27-\frac{l}{4}\right)^2, \ \ \text{ for } 0 \leq l \leq 108 \end{equation*}

We now differentiate with respect to $l\text{:}$

\begin{equation*} V'(l) = \frac{3}{16}(l-108)(l-36)\text{,} \end{equation*}

which gives one interior critical point at $l=36\text{.}$ We now compare volumes:

\begin{equation*} V(0) = 0,\ \ V(36) = 11,664, \ \ V(108) = 0 \end{equation*}

Hence, the volume of the box is maximal with dimensions $36 \times 18 \times 18$ in.

Find the dimensions of the lightest cylindrical can containing 0.25 liter (=250 cm${}^3$) if the top and bottom are made of a material that is twice as heavy (per unit area) as the material used for the side.

$\ds r=5/(2\pi)^{1/3}\approx 2.7\hbox{ cm}\text{,}$ $\ds h=5\cdot2^{5/3}/\pi^{1/3}=4r\approx 10.8 \hbox{ cm}$

A conical paper cup is to hold $1/4$ of a liter. Find the height and radius of the cone which minimizes the amount of paper needed to make the cup. Use the formula $\ds \pi r\sqrt{r^2+h^2}$ for the area of the side of a cone.

$\ds h={750\over\pi}\left({2\pi^2\over 750^2}\right)^{1/3}\text{,}$ $\ds r=\left({750^2\over 2\pi^2}\right)^{1/6}$

A conical paper cup is to hold a fixed volume of water. Find the ratio of height to base radius of the cone which minimizes the amount of paper needed to make the cup. Use the formula $\ds \pi r\sqrt{r^2+h^2}$ for the area of the side of a cone, called the lateral area of the cone.

$\ds h/r=\sqrt2$
$1/2$
How are your answers to Problem 5.8.19 affected if the cost per item for the $x$ items, instead of being simply $2, decreases below$2 in proportion to $x$ (because of economy of scale and volume discounts) by 1 cent for each 25 items produced?