## Section7.6Maxima and Minima

Now we want to deal with extrema of two-variable functions. Similar to single-variable functions, we distinguish between relative and absolute extrema in two-variable functions.

###### Definition7.42.Relative Extrema of a Two-Variable Function.

Let $z=f(x,y)$ be a function defined on a region $R \subseteq \mathbb{R}^2$ containing the point $(x_0,y_0)\text{.}$

Then $f$ has a relative maximum at $(x_0,y_0)$ if $f(x,y) \leq f(x_0,y_0)$ for all points $(x,y)$ that are sufficiently close to $(x_0,y_0)\text{.}$ The number $f(x_0,y_0)$ is the relative maximum value.

Then $f$ has a relative minimum at $(x_0,y_0)$ if $f(x,y) \geq f(x_0,y_0)$ for all points $(x,y)$ that are sufficiently close to $(x_0,y_0)\text{.}$ The number $f(x_0,y_0)$ is the relative minimum value.

In other words, if the point $(x_0,y_0,f(x_0,y_0))$ is the highest point on the graph of $f$ compared to nearby points, then $f$ has a relative maximum at $(x_0,y_0)\text{,}$ and similarly for a relative minimum. On the other hand, if the inequality holds for all points in the region $R\text{,}$ then we have an absolute minimum or maximum.

###### Definition7.43.Absolute Extrema of a Two-Variable Function.

Let $z=f(x,y)$ be a function defined on a region $R \subseteq \mathbb{R}^2$ containing the point $(x_0,y_0)\text{.}$

Then $f$ has an absolute maximum at $(x_0,y_0)$ if $f(x,y) \leq f(x_0,y_0)$ for all points $(x,y)$ in $R\text{.}$ The number $f(x_0,y_0)$ is the absolute maximum value.

Then $f$ has an absolute minimum at $(x_0,y_0)$ if $f(x,y) \geq f(x_0,y_0)$ for all points $(x,y)$ in $R\text{.}$ The number $f(x_0,y_0)$ is the absolute minimum value.

Suppose a surface given by $f(x,y)$ has a relative maximum at $(x_0,y_0,z_0)\text{;}$ geometrically, this point on the surface looks like the top of a hill. If we look at the cross-section in the plane $y=y_0\text{,}$ we will see a relative maximum on the curve at $(x_0,z_0)\text{,}$ and we know from single-variable calculus that ${\partial z\over\partial x}=0$ at this point. Likewise, in the plane $x=x_0\text{,}$ ${\partial z\over\partial y}=0\text{.}$ So if there is a relative maximum at $(x_0,y_0,z_0)\text{,}$ both partial derivatives at the point must be zero, and likewise for a relative minimum. Thus, to find relative maximum and minimum points, we need only consider those points at which both partial derivatives are 0. As in the single-variable case, it is possible for the derivatives to be 0 at a point that is neither a maximum or a minimum, so we need to test these points further.

###### Definition7.44.Critical Point of a Two-Variable Function.

Let $z=f(x,y)$ be a function defined on a region $R \subseteq \mathbb{R}^2$ containing the point $(x_0,y_0)\text{.}$

The point $(x_0,y_0)$ in $R$ is a critical point of $f$ if both $f_x(x_0,y_0)=0$ and $f_y(x_0,y_0)=0$ or at least one of the derivatives $f_x(x_0,y_0)$ and $f_y(x_0,y_0)$ does not exist.

You will recall that in the single-variable case, we examined three methods to identify maximum and minimum points; the most useful is the Second Derivative Test, though it does not always work. For functions of two variables there is also a Second Derivative Test; again it is by far the most useful test, though it doesn't always work.

Note: We interpret the Second Derivative Test as follows.

1. If the discriminant is positive at the point $(x_0,y_0)$ and $f_{xx}\lt 0\text{,}$ then the surface curves downward in all directions giving rise to a relative maximum.

2. If the discriminant is positive at the point $(x_0,y_0)$ and $f_{xx}>00\text{,}$ then the surface curves upward in all directions giving rise to a relative minimum.

3. If the discriminant is negative at the point $(x_0,y_0)\text{,}$ then the surface curves up in some directions and down in others. This behaviour creates a so-called saddle point at $(x_0,y_0,z_0)$ (see Figure 7.8).

Given the function $f(x,y)\text{,}$ the point $(x_0,y_0)$ is called a saddle point of $f$ if

1. there is at least one point $(x,y)$ arbitrarily close to $(x_0,y_0)$ for which $f(x,y) > f(x_0,y_0)\text{,}$ and

2. there is at least one point $(x,y)$ arbitrarily close to $(x_0,y_0)$ for which $f(x,y) \lt f(x_0,y_0)\text{.}$

###### Example7.47.Extrema on an Elliptic Paraboloid.

Verify that $f(x,y)=x^2+y^2$ has a minimum at $(0,0)\text{.}$

Solution

First, we compute all the needed derivatives:

The derivatives $f_x$ and $f_y$ are zero only at $(0,0)\text{.}$ Applying the Second Derivative Test there:

\begin{equation*} D(0,0)=f_{xx}(0,0)f_{yy}(0,0)-f_{xy}(0,0)^2= 2\cdot2-0=4>0\text{,} \end{equation*}

so there is a relative minimum at $(0,0)\text{,}$ and there are no other possibilities.

###### Example7.48.Saddle Point on a Hyperbolic Paraboloid.

Find all relative maxima and minima for $f(x,y)=x^2-y^2\text{.}$

Solution

The derivatives:

Again there is a single critical point, at $(0,0)\text{,}$ and

\begin{equation*} D(0,0)=f_{xx}(0,0)f_{yy}(0,0)-f_{xy}(0,0)^2= 2\cdot-2-0=-4\lt 0\text{,} \end{equation*}

so there is neither a maximum nor minimum at $(0,0,0)\text{,}$ but a saddle point. The surface is shown in Figure 7.8.

###### Example7.49.Finding Extrema.

Find all relative maxima and minima for $f(x,y)=x^4+y^4\text{.}$

Solution

The derivatives:

Again there is a single critical point, at $(0,0)\text{,}$ and

\begin{equation*} D(0,0)=f_{xx}(0,0)f_{yy}(0,0)-f_{xy}(0,0)^2= 0\cdot0-0=0\text{,} \end{equation*}

so we get no information. However, in this case it is easy to see that there is a minimum at $(0,0)\text{,}$ because $f(0,0)=0$ and at all other points $f(x,y)>0\text{.}$

Interactive Demonstration.

###### Example7.50.Finding Extrema.

Find all relative maxima and minima for $f(x,y)=x^3+y^3\text{.}$

Solution

The derivatives:

Again there is a single critical point, at $(0,0)\text{,}$ and

\begin{equation*} D(0,0)=f_{xx}(0,0)f_{yy}(0,0)-f_{xy}(0,0)^2= 0\cdot0-0=0\text{,} \end{equation*}

so we get no information. In this case, a little thought shows there is neither a maximum nor a minimum at $(0,0)\text{:}$ when $x$ and $y$ are both positive, $f(x,y)>0\text{,}$ and when $x$ and $y$ are both negative, $f(x,y)\lt 0\text{,}$ and there are points of both kinds arbitrarily close to $(0,0)\text{.}$ Alternately, if we look at the cross-section when $y=0\text{,}$ we get $f(x,0)=x^3\text{,}$ which does not have either a maximum or minimum at $x=0\text{.}$ In fact, $(0,0,0)$ is a saddle point.

Interactive Demonstration.

###### Example7.51.Maximizing Profits.

A certain company produces and sells entertainment units. The total weekly revenue and cost in dollars is given by

\begin{equation*} R(x,y) = -\frac{1}{4}x^2-\frac{3}{8}y^2-\frac{1}{4}xy+280x+260y+1000, \text{ and } \end{equation*}
\begin{equation*} C(x,y)= 160x + 160y + 6000 \end{equation*}

respectively, where $x$ denotes the number of fully assembled units and $y$ the number of kits produced and sold each week. Find the company's maximum profit per week, and the number of assembled units and kits that must be produced and sold to achieve this profit.

Solution

The weekly profits is given by

\begin{equation*} \begin{split} P(x,y) \amp = R(x,y) - C(x,y) \\ \amp = \left(-\frac{1}{4}x^2-\frac{3}{8}y^2-\frac{1}{4}xy+280x+260y+1000\right)-\left(160x + 160y + 6000\right) \\ \amp = -\frac{1}{4}x^2-\frac{3}{8}y^2-\frac{1}{4}xy+120x+110y-5000, \end{split} \end{equation*}

where $x \geq 0$ and $y \geq 0\text{.}$

To find the relative maximum profit per week, we calculate the critical points by setting the first partial derivatives equal to zero:

\begin{equation*} P_x = -\frac{1}{2}x - \frac{1}{4}y+120 = 0 \end{equation*}
\begin{equation*} P_y = -\frac{3}{4}y - \frac{1}{4}x + 100 = 0 \end{equation*}

We solve for $y$ by multiplying the second equation by 2 and subtracting it from the first equation, which yields

\begin{equation*} \frac{5}{4}y -80 = 0 \implies y = 64\text{.} \end{equation*}

Substituting this value of $y$ into the first equation we obtain

\begin{equation*} -\frac{1}{2}x -\frac{1}{4}(64)+120 = 0 \implies x + 32 - 240 = 0 \implies x = 208\text{.} \end{equation*}

Therefore, the only critical point of $P$ is $(208,64)\text{.}$

We now apply the Second Derivative Test to confirm that the weekly maximum profit occurs at $(208,64)\text{.}$ For this we need the second partial derivatives of $P\text{:}$

\begin{equation*} P_{xx}=-\frac{1}{2}, \ \ P_{xy}=-\frac{1}{4}, \ \ P_{yy}=-\frac{3}{4} \cdot \end{equation*}

Therefore, the discriminant is given by

\begin{equation*} D(x,y) = \left(-\frac{1}{2}\right)\left(-\frac{3}{4}\right)-\left(-\frac{1}{4}\right)^2 = \frac{3}{8}-\frac{1}{16} = \frac{5}{16} > 0 \end{equation*}

for all $(x,y)$ in the domain of $P$ and, in particular, $D(208,64) > 0\text{.}$ Since $P_{xx}(208,64)=-\frac{1}{2} \lt 0\text{,}$ the point $(208,64)$ leads to a relative maximum of $P\text{.}$ Hence, the maximizing number of assembled units is 208 units and the maximizing number of kits is 64 kits, whose production and sale leads to a weekly profit of

\begin{equation*} P(208,64)=-\frac{1}{4}(208)^2-\frac{3}{8}(64)^2-\frac{1}{4}(208)(64)+120(208)+110(64)-5000 = \10,680\text{.} \end{equation*} ###### Example7.52.Optimizing Surface. A rectangular container with an open top needs to have a capacity of 10 m$^3$ and be constructed of a thin sheet of metal. What are the dimensions of the box if the amount of metal is minimized? Solution Let $x\text{,}$ $y$ and $z$ be the length, width and height of the box respectively, measured in metres. Then the volume of the box is \begin{equation*} xyz = 10 \implies z = \frac{10}{xy}\text{.} \end{equation*} Hence, the surface area of the box is \begin{equation*} \begin{split} A(x,y,z) \amp = 2xz + 2yz + xy \\ A(x,y) \amp = 2x \left(\frac{10}{xy}\right) + 2y \left(\frac{10}{xy}\right) + xy \\ \amp = \frac{20}{y} + \frac{20}{x} + xy, \end{split} \end{equation*} with domain $\{(x,y)\in \R^2 | x>0, y>0\}\text{.}$ We now find the critical points of $A\text{:}$ \begin{equation*} A_x= - \frac{20}{x^2} + y, \text{ and } A_y = -\frac{20}{y^2}+x\text{.} \end{equation*} Therefore, \begin{equation*} \begin{split} A_x = 0 \text{ and } A_y=0 \amp \implies y=\frac{20}{x^2} \text{ and } x=\frac{20}{y^2}\\ \amp \implies y = \frac{20}{20/y^2} \implies 20y = y^4 \\ \amp \implies y(20-y^3) = 0 \end{split} \end{equation*} Now, $y=0 \notin D^{y}_A\text{,}$ but \begin{equation*} y=20^{1/3} \in D^y_A \implies x = \frac{20}{(20^{1/3})^2} = 20^{1/3} \in D^x_A\text{.} \end{equation*} Hence, $(20^{1/3},20^{1/3})$ is the only critical point of $A\text{.}$ Since the domain of $A$ is unbounded, we use the Second Derivative Test to classify this critical point. The second-order partial derivatives are \begin{equation*} A_{xx} = \frac{40}{x^3}, A_{yy}=\frac{40}{y^3}, A_{xy}=1\text{,} \end{equation*} and so the discriminant is \begin{equation*} D\left(20^{1/3},20^{1/3}\right)= \frac{40}{\left(20^{1/3}\right)^3} \cdot \frac{40}{\left(20^{1/3}\right)^3} -1 = 3 > 0\text{.} \end{equation*} Since \begin{equation*} A_{xx}\left(20^{1/3},20^{1/3}\right) = \frac{40}{\left(20^{1/3}\right)^3} = 2 > 0\text{,} \end{equation*} the critical point yields a relative minimum with a value of \begin{equation*} z=\dfrac{10}{(20^{1/3})(20^{1/3})} = \dfrac{20^{1/3}}{2}\text{.} \end{equation*} Is this an absolute minimum? It is, but it is difficult to see this analytically; physically and graphically it is clear that there is a minimum, in which case it must be at the single critical point. Thus, the dimensions of the box which minimize the amount of metal are width $=20^{1/3} \approx 2.7$ m, length = $20^{1/3} \approx 2.7$ m and height = $\frac{20^{1/3}}{2} \approx 1.4$ m. Recall that when we did single-variable absolute maximum and minimum problems, the easiest cases were those for which the variable could be limited to a finite closed interval, for then we simply had to check all critical values and the endpoints. The previous example is difficult because there is no finite boundary to the domain of the problem—both $w$ and $l$ can be in $(0,\infty)\text{.}$ As in the single-variable case, the problem is often simpler when there is a finite boundary. As in the case of single-variable functions, this means that the absolute maximum and minimum values must occur at a critical point or on the boundary; however, in the two-variable case the boundary is a curve, not merely two endpoints. ###### Guideline for Finding Absolute Extrema. Suppose that $z=f(x,y)$ is a continuous function on a closed and bounded region $R \subseteq \mathbb{R}^2\text{.}$ 1. List the interior points of $R$ that are critical points of $f\text{,}$ which may lead to relative extrema of $f\text{.}$ Evaluate $f$ at these points. 2. List the boundary points of $R\text{,}$ which may lead to relative extrema of $f\text{.}$ Evaluate $f$ at these points. 3. Compare the maximum and minimum values from steps 1 and 2, the largest value is the absolute maximum and the smallest value is the absolute minimum. We will demonstrate the above guideline with two examples. ###### Example7.54.Finding Absolute Extrema. Find the absolute extrema of \begin{equation*} f(x,y)=xy-x^3y^2 \end{equation*} on the region bounded by $0 \leq x\leq 1, 0 \leq y \leq 1\text{.}$ Solution We apply the guideline for finding absolute extrema. We begin by determining the boundary points: Notice that the constraint $0 \leq x\leq 1\text{,}$ $0 \leq y \leq 1$ determines a square region bounded by the lines $x=0\text{,}$ $y=0\text{,}$ $x=1$ and $y=1$ with vertices $(0,0)\text{,}$ $(0,1)\text{,}$ $(1,0)$ and $(1,1)\text{.}$ When $x=0\text{,}$ then $f(0,y)=0\text{.}$ Similarly, when $y=0\text{,}$ then $f(x,0)=0\text{.}$ On the edge $x=1, 0\leq y \leq 1\text{:}$ \begin{equation*} a(y) = f(1,y) = y-y^2 \implies a'(y)=1-2y=0 \implies y = 0.5\text{.} \end{equation*} Hence, $(1,0.5)$ is the only critical point of $a$ on this edge. On the edge $y=1, 0 \leq x \leq 1\text{:}$ \begin{equation*} b(x)=f(x,1) = x-x^3 \implies b'(x)=1-3x^2=0 \implies x= \frac{1}{\sqrt{3}} \end{equation*} Hence, $\left(\frac{1}{\sqrt{3}},1\right)$ is the only critical point of $b$ on this edge. Next, we determine the interior points: To find the critical points of $f\text{,}$ we begin by computing the first-order partial derivatives of $f\text{:}$ \begin{equation*} f_x=y-3x^2y^2, f_y = x-2x^3y \end{equation*} We set the derivatives equal to zero and solve for $x$ and $y\text{:}$ \begin{equation*} \begin{gathered} f_x=y-3x^2y^2=0, f_y=x-2x^3y=0 \\ \implies y(1-3x^2y)=0, x(1-2x^2y)=0 \\ \implies y = 0 \text{ or } 1-3x^2y=0, x=0 \text{ or } 1-2x^2y = 0 \end{gathered} \end{equation*} If $y=0\text{,}$ we cannot have $2x^2y=1\text{,}$ so we must have $x=0\text{,}$ which yields the point $(0,0)\text{.}$ If $3x^2y=1\text{,}$ we cannot have $x=0\text{,}$ so we must have $2x^2y=1$ But then $x^2y=\frac{1}{3}=\frac{1}{2}\text{,}$ which is impossible. Hence, the only critical point on the square is $(0,0)\text{.}$ Lastly, we compare values of $f$ at all interior and boundary points we have found: \begin{equation*} \begin{split} \amp f(x,0)=0 \\ \amp f(0,y)=0 \\ \amp f(0,0)=f(0,1)=f(1,0)=0\\ \amp f(1,1) = 1\cdot 1 - 1^3 \cdot 1^2 = 0\\ \amp f\left(1,\frac{1}{2}\right)=1\cdot \frac{1}{2} - 1^3\cdot\left(\frac{1}{2}\right)^2=\frac{1}{4} = 0.25\\ \amp f\left(\frac{1}{\sqrt{3}},1\right)=\frac{1}{\sqrt{3}}\cdot 1 - \left(\frac{1}{2}\right)^3\cdot 1^2=\frac{2}{3\sqrt{3}} \approx 0.385 \end{split} \end{equation*} Thus, the absolute maximum occurs at $\left(\dfrac{1}{\sqrt{3}},1\right)$ with a value of $\dfrac{2}{3\sqrt{3}}\text{,}$ and the absolute minimum occurs along the lines $x=0\text{,}$ $0\leq y\leq 1$ and $y=0\text{,}$ $0 \leq x \leq 1$ as well as the point $(1,1)\text{.}$ ###### Example7.55.Optimizing Volume of a Box. The length of the diagonal of a box is to be one metre; find the maximum possible volume. Solution If the box is placed with one corner at the origin, and sides along the axes, the length of the diagonal is $\ds\sqrt{x^2+y^2+z^2}\text{,}$ and the volume is \begin{equation*} V=xyz=xy\sqrt{1-x^2-y^2}\text{.} \end{equation*} Clearly, $x^2+y^2\le 1\text{,}$ so the domain we are interested in is the quarter of the unit disk in the first quadrant as shown below. To find the interior points that lead to relative extrema, we calculate the derivatives: \begin{align*} V_x\amp ={y-2yx^2-y^3\over\sqrt{1-x^2-y^2}}\\ V_y\amp ={x-2xy^2-x^3\over\sqrt{1-x^2-y^2}} \end{align*} If these are both 0, then $x=0$ or $y=0\text{,}$ or $x=y=1/\sqrt3\text{.}$ The boundary of the domain is composed of three curves: $x=0$ for $y\in[0,1]\text{;}$ $y=0$ for $x\in[0,1]\text{;}$ and $x^2+y^2=1\text{,}$ where $x\ge0$ and $y\ge0\text{.}$ In all three cases, the volume $xy\sqrt{1-x^2-y^2}$ is 0. Comparing the relative extrema, the maximum occurs at the only critical point $(1/\sqrt3,1/\sqrt3,\sqrt3/\sqrt3)\text{.}$ The surface $xy\sqrt{1-x^2-y^2}$ on its domain is shown below. Interactive Demonstration. ##### Exercises for Section 7.6. Classify the critical points of the following functions: 1. $\displaystyle f(x,y)=x^2+4y^2-2x+8y-1$ Answer minimum at $(1,-1)$ Solution The first-order partial derivatives are \begin{equation*} f_x = 2x-2, \ \ \text{ and } \ \ f_y= 8y+8\text{.} \end{equation*} To find the critical points of $f\text{,}$ we set both $f_x=0$ and $f_y=0\text{:}$ \begin{equation*} f_x = 0 \implies x = 1, \ \text{ and } f_y = 0 \implies y = -1\text{.} \end{equation*} This gives exactly one critical point at $(1,-1)\text{.}$ To classify this critical point, we construct the discriminant: \begin{equation*} D(x,y) = f_{xx}f_{yy} - \left(f_{xy}\right)^2 = 2(8)-0 = 16 > 0\text{.} \end{equation*} Since the discriminant is positive everywhere on its domain, and $f_{xx} > 0\text{,}$ the point $(1,-1)$ must be a relative minimum. 2. $\displaystyle f(x,y)=x^2-y^2+6x-10y+2$ Answer none Solution The first-order partial derivatives of $f$ are \begin{equation*} f_x = 2x+6, \ \ \text{ and } f_y = -2y-10\text{.} \end{equation*} To find the critical points of $f\text{,}$ we set both $f_x=0$ and $f_y=0\text{:}$ \begin{equation*} f_x = 0 \implies x = -3, \ \ \text{ and } f_y = 0 \implies y=-5\text{.} \end{equation*} Therefore, there is only one critical point at $(-3,-5)\text{.}$ To classify this critical point, we construct the discriminant: \begin{equation*} D(x,y) = f_{xx}f_{yy} - \left(f_{xy}\right)^2 = 2(-2)-0 = -4 \lt 0\text{.} \end{equation*} Since the discriminant is negative everywhere in its domain, the function $f$ has no relative extrema. 3. $\displaystyle f(x,y)=xy$ Answer none Solution The first-order partial derivatives of $f$ are \begin{equation*} f_x = y, \ \ \text{ and } \ \ f_y = x\text{.} \end{equation*} To find the critical points of $f\text{,}$ we set both $f_x=0$ and $f_y=0\text{:}$ \begin{equation*} f_x = 0 \implies y=0, \ \ \text{ and } f_y = 0 \implies x=0\text{.} \end{equation*} Thus, the point $(0,0)$ is the only critical point of $f\text{.}$ To classify this critical point, we construct the discriminant: \begin{equation*} D(x,y) = f_{xx}f_{yy} - \left(f_{xy}\right)^2 = 0\text{.} \end{equation*} Therefore, the discriminant tells us nothing about the behaviour of $f$ at this critical point. Instead, we notice that the function $f(x,y) = xy$ is increasing in some directions as we move away from $(0,0)\text{,}$ but decreasing in others. Therefore this must be a saddle. 4. $\displaystyle f(x,y)=9+4x-y-2x^2-3y^2$ Answer maximum at $(1,-1/6)$ Solution We find the first partial derivatives of $f(x,y)$ to be \begin{equation*} f_x = 4 - 4x, \ \ \text{ and } \ \ f_y = -1-6y\text{.} \end{equation*} To find the critical points of $f\text{,}$ we set both $f_x=0$ and $f_y=0\text{:}$ \begin{equation*} f_x = 0 \implies x=1, \ \ \text{ and } f_y = 0 \implies y=-1/6\text{.} \end{equation*} Therefore, the point $(1, -1/6)$ is the only critical point of $f\text{.}$ To classify this critical point, we construct the discriminant: \begin{equation*} D(x,y) = f_{xx}f_{yy} - \left(f_{xy}\right)^2 = -4(-6) - 0 > 0\text{.} \end{equation*} Since the discriminant is positive and $f_{xx} \lt 0$ everywhere, we conclude that the critical point $(1,-1/6)$ is a relative maximum. 5. $\displaystyle f(x,y)=x^2+4xy+y^2-6y+1$ Answer none Solution We find the first partial derivatives of $f(x,y)$ to be \begin{equation*} f_x=2x+4y, \ \ \text{ and } \ \ f_y=4x+2y-6\text{.} \end{equation*} To find the critical points of $f\text{,}$ we set $f_x=0$ and $f_y=0\text{:}$ \begin{equation*} 2x+4y = 0 \implies x=-2y\text{,} \end{equation*} so then \begin{equation*} 4x+2y-6=0 \implies 4(-2y)+2y-6 = 0 \implies y=-1\text{,} \end{equation*} which in turn gives $x=2\text{.}$ Therefore $(2,-1)$ is the only critical point. To classify this point, we construct the discriminant, \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 = (2)(2)-4^2 \lt 0 \end{equation*} everywhere and in particular at $(2,-1)\text{.}$ Thus, $f$ has no relative extrema. 6. $\displaystyle f(x,y)=x^2-xy+2y^2-5x+6y-9$ Answer minimum at $(2,-1)$ Solution The first-order partial derivatives of $f$ are \begin{equation*} f_x = 2x - y - 5, \ \ \text{ and } \ \ f_y = -x+4y+6\text{.} \end{equation*} To find the critical points of $f\text{,}$ we set both $f_x=0$ and $f_y=0\text{:}$ \begin{equation*} f_x = 0 \implies y = 2x-5\text{.} \end{equation*} Plugging this equation into $f_y = 0\text{,}$ we find \begin{equation*} -x + 4(2x-5)+6 = 0 \implies x = 2\text{.} \end{equation*} When $x=2\text{,}$ we must have $y=2(2)-5) = -1\text{.}$ Hence, there is excatly one critical point at $(2,-1)\text{.}$ To classify this point, we construct the discriminant, \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 =2(4)-(-1) = 9 > 0\text{,} \end{equation*} which is positive everyone on the domain of $f\text{.}$ Since we further have that $f_{xx} > 0$ everywhere, $f$ must have a relative minimum at the point $(2,-1)\text{.}$ 7. $\displaystyle f(x,y)=xy(5x+y-15)$ Answer minimum at $(1,5)$ Solution The first-order partial derivatives of $f$ are \begin{equation*} f_x = 10xy + y^2 - 15y, \ \ \text{ and } \ \ f_y = 5x^2+2xy-15x\text{.} \end{equation*} To find the critical points of $f\text{,}$ we set both $f_x=0$ and $f_y=0\text{.}$ We first notice by inspection setting $x=0$ satisfies $f_y=0$ automatically, and that $f_x (0,y) = y(y-15)\text{.}$ Hence, $(0,0)$ and $(0,15)$ must be critical points. Additionally, setting $y=0$ automatically satisfies $f_z = 0\text{,}$ and $f_y(x,0) = 5x(x-3)\text{.}$ Thereofore, $(3,0)$ is another critical point. Now assume that $x, y \neq 0\text{:}$ \begin{equation*} f_x = 0 \implies y= 15-10x\text{.} \end{equation*} Therefore, we need \begin{equation*} 5x + 2 (15-10x) - 15 = 0 \implies x = 1\text{.} \end{equation*} When $x=1\text{,}$ we must have $y=15-10(1) = 5\text{.}$ Hence, $f$ has four critical points at $(0,0)\text{,}$ $(0,15)\text{,}$ $(3,0)$ and $(1,5)\text{.}$ To classify these points, we construct the discriminant, \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 = (10y)(2x)-(10x+2y-15)^2\text{.} \end{equation*} We now classify the four critical points:  Crit. pt $(x,y)$ $D(x,y)$ $f_{xx}(x,y)$ Conclusion $(0,0)$ $\lt 0$ - saddle $(0,15)$ $\lt 0$ - saddle $(3,0)$ $\lt 0$ - saddle $(1,5)$ $>0$ $>0$ rel. min 8. $\displaystyle f(x,y)=x^2y-xy^2+4xy-4x^2-4y^2$ Answer maximum at $(0,0)$ Solution The first-order partial derivatives of $f$ are \begin{equation*} f_x = 2xy-y^2+4y-8x, \ \ \text{ and } \ \ f_y = x^2-2xy + 4x-8y\text{.} \end{equation*} To find the critical points of $f\text{,}$ we set both $f_x=0$ and $f_y=0\text{.}$ By inspection, we see that the point $(0,0)$ is a solution to this system of equations. To find any other critical point(s), we now assume that $x, y \neq 0\text{:}$ \begin{equation*} f_x = 0 \implies (2y-8) x = y^2-4y\text{.} \end{equation*} If $y=4\text{,}$ we see that we both $x=8$ and $x=-4$ satistfy the system of equations. Thus, $(8,4)$ and $(-4,4)$ are two additional critical points. We now assume that $y\neq 4\text{,}$ and let $x= y\frac{y-4}{2y-8}\text{.}$ Then the condition $f_y = 0$ becomes: \begin{equation*} f_y = 0 \implies \left(\frac{y^2-4y}{2y-8}\right)^2 - 2y \frac{y^2-4y}{2y-8} + 4 \frac{y^2-4y}{2y-8} - 8y = 0\text{.} \end{equation*} Since $y\neq 0\text{,}$ we can divide through by a factor of $x\text{:}$ \begin{equation*} y\left(\frac{y-4}{2y-8}\right) - 2y + 4 - 8\left(\frac{2y-8}{y-4}\right) = 0 \implies y = -8\text{.} \end{equation*} When $y=-8\text{,}$ we need $x=-4\text{.}$ In total, $f$ has 4 critical points: $(0,0)\text{,}$ $(8,4)\text{,}$ $(-4,4)$ and $(-8,-4)\text{.}$ To classify these critical points, we construct the discriminant, \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 = (2y-8)(-2x-8) - (2x-2y+4)^2\text{.} \end{equation*} We now classify these critical points:  Crit. pt $(x,y)$ $D(x,y)$ $f_{xx}(x,y)$ Conclusion $(0,0)$ $>0$ $\lt 0$ rel. max $(8,4)$ $\lt 0$ - saddle $(-4,4)$ $\lt 0$ - saddle $(-8,-4)$ $\lt 0$ - saddle 9. $\displaystyle f(x,y)=e^{-x^3/3+x-y^2}$ Answer saddle at $(-1,0)\text{,}$ maximum at $(1,0)$ Solution The first-order partial derivatives of $f$ are \begin{equation*} f_x = \left(-x^2+1\right)e^{-\frac{x^3}{3} + x - y^2} \ \ \text{ and } \ \ f_y = -2y e^{-\frac{x^3}{3} + x - y^2}\text{.} \end{equation*} To find the critical points of $f\text{,}$ we set both $f_x=0$ and $f_y=0\text{.}$ We see that \begin{equation*} f_x = 0 \implies x = \pm 1, \ \ \text{ and } f_y = 0 \implies y = 0\text{.} \end{equation*} Thus, $f$ has two critical points: $(1, 0)$ and $(-1,0)\text{.}$ We now construct the discriminant: \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 = e^{-\frac{x^3}{3} + x - y^2}\bigl(2(x^4-2x^2-2x+1)(2y^2-1)-(2(x^2-1)y)\bigr)\text{.} \end{equation*} We now classify these critical points:  Crit. pt $(x,y)$ $D(x,y)$ $f_{xx}(x,y)$ Conclusion $(1,0)$ $>0$ $\lt 0$ rel. max $(-1,0)$ $\lt 0$ - saddle 10. $\displaystyle f(x,y)=2x^2+2xy+2y^2-6x$ Answer minimum at $(2,-1)$ Solution The first-order partial derivatives of $f$ are \begin{equation*} f_x = 4x+2y-6, \ \ \text{ and } \ \ f_y = 2x+4y\text{.} \end{equation*} To find the critical points of $f\text{,}$ we set both $f_x=0$ and $f_y=0\text{.}$ We see that \begin{equation*} f_y = 0 \implies x = -2y\text{,} \end{equation*} and so \begin{equation*} f_x = 0 \implies 4(-2y)+2y-6 = 0 \implies y=-1\text{.} \end{equation*} When $y=-1\text{,}$ we need $x=2\text{.}$ Thus, $f$ has one critical point at $(2,-1)\text{.}$ To classify this critical point, we construct the discriminant: \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 = (4)(4) - 2 = 14 > 0\text{.} \end{equation*} Since the discriminant and $f_{xx}$ are both positive everywhere on the domain of $f\text{,}$ we conclude that $(2,-1)$ must be a relative minimum of $f\text{.}$ 11. $\displaystyle f(x,y)=x^4+y^4-4xy$ Answer minimum at $(-1,-1)\text{,}$ maximum at $(1,1)\text{,}$ saddle at $(0,0)$ Solution The first-order partial derivatives of $f$ are \begin{equation*} f_x = 4x^3-4y \ \ \text{ and } \ \ f_y = 4y^3-4x\text{.} \end{equation*} To find the critical points of $f\text{,}$ we set both $f_x=0$ and $f_y=0\text{.}$ We first observe that $(0,0)$ must be a critical point. We now assume that $x,y \neq 0\text{.}$ Then: \begin{equation*} f_x = 0 \implies y = x^3\text{,} \end{equation*} and so \begin{equation*} f_y = 0 \implies x^9 = x\text{.} \end{equation*} Thus, $x = \pm 1\text{.}$ When $x = -1\text{,}$ we need $y = (-1)^3 = -1\text{,}$ and when $x=1\text{,}$ we need $y=1^3 = 1\text{.}$ Hence, there are three critical points in total: $(0,0)\text{,}$ $(1,1)$ and $(-1,-1)\text{.}$ We now construct the discriminant: \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 = (12x^2)(12y^2) - 16\text{.} \end{equation*} Using the discriminant, we can classify these critical points:  Crit. pt $(x,y)$ $D(x,y)$ $f_{xx}(x,y)$ Conclusion $(0,0)$ $\lt 0$ - saddle $(1,1)$ $>0$ $>0$ rel. min $(-1,-1)$ $>0$ $>0$ rel. min 12. $\displaystyle f(x,y)=x^2y-2xy^2+3xy+4$ Answer minimum at $(-1,1/2)\text{,}$ saddle at $(-3,0)\text{,}$ $(0,0)$ and $(0,3/2)\text{.}$ Solution The first-order partial derivatives of $f$ are \begin{equation*} f_x = 2xy - 2y^2 + 3y \ \ \text{ and } \ \ f_y = x^2-4xy + 3x\text{.} \end{equation*} To find the critical points of $f\text{,}$ we set both $f_x=0$ and $f_y=0\text{.}$ We first observe that taking $y=0$ satisfies $f_x = 0$ automatically. Since $f_y(x,0) = x(x+3)\text{,}$ we see that $(0,0)$ and $(-3,0)$ must be critical points. Furthermore, setting $x=0$ satisfies $f_y = 0$ automcatically, and $f_x(0,y) = y(-2y+3)\text{.}$ Therefore, $(0, 3/2)$ is another critical point. To find any remaining critical points, we assume that $x,y \neq 0\text{.}$ Then: \begin{equation*} f_x = 0 \implies 2x - 2y + 3 = 0 \implies x = y - \frac{3}{2}\text{.} \end{equation*} Then, \begin{equation*} f_y = 0 \implies (y-3/2) - 4y + 3 = 0 \implies y = \frac{1}{2}\text{.} \end{equation*} Hence, $(-1,1/2)$ is another critical point of $f\text{.}$ We now construct the discriminant: \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 = (2y)(-4x)-(2x-4y+3)^2\text{.} \end{equation*} Using the discriminant, we can classify these critical points:  Crit. pt $(x,y)$ $D(x,y)$ $f_{xx}(x,y)$ Conclusion $(0,0)$ $\lt 0$ - saddle $(-3,0)$ $\lt 0$ - saddle $(0,3/2)$ $\lt 0$ - saddle $(-1,1/2)$ $>0$ $>0$ rel. min Find the absolute maximum and minimum points of $f=x^2+3y-3xy$ over the region bounded by $y=x\text{,}$ $y=0\text{,}$ and $x=2\text{.}$ Answer $f(2,2)=-2\text{,}$ $f(2,0)=4$ Solution The triangular region bounded by the lines $y=x\text{,}$ $y=0$ and $x=2$ is shown below. We first find the first-order partial derivatives of $f\text{:}$ \begin{gather*} f_x = 2x-3y\\ f_y = 3-3x \end{gather*} We now find and classify any relative extrema. Since \begin{equation*} f_y = 0 \implies x = 1, \ \ \text{ and } f_x = 0 \implies 2x = 3y\text{,} \end{equation*} we see that $(1, 3/2)$ is the only point which satisfies both $f_x = f_y= 0\text{.}$ This is located inside the bounded region, and so it is a critical point of $f\text{.}$ The boundary of the region is composed of three lines: $y=0$ for $x \in [0,2]\text{,}$ $x=2$ for $y \in [0,2]$ and $y=x$ for $x \in [0,2]\text{.}$ And so we have the following boundary points: $(0,0)\text{,}$ $(2,2)$ and $(2,0)\text{.}$ Over the first line, we have \begin{equation*} f(x,0) = a(x) = x^2\text{,} \end{equation*} which gives no additional critical points. Over the second line, we have \begin{equation*} f(x,x) = b(x) = -2x^2+3x\text{.} \end{equation*} Therefore, the point $(3/4, 3/4)$ is an aditional critical point. Finally, over the last boundary line, \begin{equation*} f(2, y) = c(y) = 4-3y\text{,} \end{equation*} which gives no additional critical points. We now compare the values of $f(x,y)$ at all of the identified critical and boundary points:  $f(x,y)$ $(0,0)$ 0 $(2,2)$ -2 $(2,0)$ 4 $(3/4, 3/4)$ 1.125 $(1,3/2)$ 1 Therefore, $f$ has an absolute maximum of 4 at $(2,0)$ and an absolute minimum of -2 at $(2,2)\text{.}$ A six-sided rectangular box is to hold $1/2$ cubic meter; what shape should the box be to minimize surface area? Answer a cube $1/\root 3 \of {2}$ on a side A certain company produces and sells flat screen TVs. The total weekly revenue and cost in dollars is given by \begin{equation*} R(x,y)=-0.2x^2-0.25y^2-0.2xy+225x+135y+1500, \ \ \text{ and } \end{equation*} \begin{equation*} C(x,y) = 125x + 45y + 5500 \end{equation*} respectively, where $x$ denotes the number of deluxe units and $y$ the number of standard units produced and sold each week. Find the company's maximum profit per week, and the number of deluxe and standard units that must be produced and sold to achieve this profit. Answer $200$ deluxe units and $100$ standard units;10,500

Solution

First, construct the profit function $P(x,y)\text{.}$

\begin{equation*} \begin{split} P(x,y)\amp =R(x,y)-C(x,y) \\ \amp = -0.2x^2-0.25y^2-0.2xy+225x+135y+1500-\left(125x+90y+5500\right)\\ \amp =-0.2x^2-0.25y^2-0.2xy+100x+90y-4000, \ \ \text{ for } x,y \geq 0. \end{split} \end{equation*}

So the first partial derivatives of $P$ are

\begin{equation*} P_x=-0.4x-0.2y+100, \ \ \text{ and } \ \ P_y=-0.5y-0.2x+90\text{.} \end{equation*}

To find the critical points of $P\text{,}$ we set $P_x=0$ and $P_y=0$ and solve the resulting system of equations.

\begin{equation*} \begin{split} -0.4x-0.2y \amp =-100\\ -0.2x-0.5y \amp =-90 \end{split} \end{equation*}

Subtract twice the second equation from the first equation. This gives

\begin{equation*} -0.2y+y=(-100+180)=80 \implies y=100\text{.} \end{equation*}

And so

\begin{equation*} -0.4x-0.2(100) = -100 \implies x=200\text{.} \end{equation*}

This gives one critical point, $(100,200)\text{.}$ To classify this critical point, we construct the discriminant, $D\text{.}$

\begin{equation*} \begin{split} D(x,y)\amp =P_{xx}P_{yy}-P_{xy}^2\\ \amp =\left(-0.4\right)\left(-0.5\right)-\left(-0.2\right)^2\\ \amp =0.16, \end{split} \end{equation*}

for all $x,y > 0\text{.}$ Therefore, since $D(100,200)>0$ and $P_{xx}(100,200)\lt 0\text{,}$ this point is a relative maximum with $P(100,200)=\dollar12,950\text{.}$

When $x=0\text{,}$ we see that the profit takes a maximum of $4100, while when $y=0\text{,}$ the maximum profit is$8500. The remaining boundary point is at $(0,0)\text{,}$ where the resulting profit is -$4000. Hence, the absolute maximum occurs when 100 deluze units and 200 stnadard units are produced. A certain company produces and sells hand-made wooden living room tables. The total monthly revenue and cost in dollars is given by \begin{equation*} R(x,y)=-0.005x^2-0.003y^2-0.002xy+31x+22y, \ \ \text{ and } \end{equation*} \begin{equation*} C(x,y)=17x+10y+200 \end{equation*} respectively, where $x$ denotes the number of finished tables and $y$ the number of unfinished tables produced and sold each month. Find the company's maximum profit per month, and the number of finished and unfinished tables that must be produced and sold to achieve this profit. Answer 1071 finished units and 1643 unfinished units;$17,157.14

Solution

First, construct the profit function $P(x,y)\text{.}$

\begin{equation*} \begin{split} P(x,y)\amp =R(x,y)-C(x,y) \\ \amp = -0.005x^2-0.003y^2-0.002xy+31x+22y-\left(17x+10y+200\right)\\ \amp = -0.005x^2-0.003y^2-0.002xy+14x+12y-200,\ \ \text{ for } x,y \geq 0. \end{split} \end{equation*}

So the first partial derivatives of $P$ are

\begin{equation*} P_x= -0.01 x - 0.002 y + 14, \ \ \text{ and } \ \ P_y= -0.002 x - 0.006 y + 12\text{.} \end{equation*}

To find the critical points of $P\text{,}$ we set $P_x=0$ and $P_y=0$ and solve the resulting system of equations.

\begin{equation*} P_x = 0 \implies y = 7000-5x\text{,} \end{equation*}

and so

\begin{equation*} P_y = 0 \implies -0.0002x - 0.006 (7000-5x) + 12 = 0 \implies x \approx 1071.43\text{.} \end{equation*}

Therefore, there is one critical point at $(1071, 1643)\text{.}$

We now investigate the behaviour of the profit along the boundary lines $x=0$ and $y=0\text{:}$

\begin{equation*} P(0,y) = a(y) = -0.003y^2+12y-200\text{,} \end{equation*}

And so setting $a'(y) = 0$ gives a critical point at $(0,2000)\text{.}$ Now, since

\begin{equation*} P(x,0) = b(x) = -0.005x^2+14x - 200\text{,} \end{equation*}

we see an additional critical point at $(1400,0)\text{.}$

We now compare the profit at each of these critical and boundary points:

 $P(x,y)$ $(0,0)$ -200 $(0,2000)$ 11,800 $(1400,0)$ 9,600 $(1071,1643)$ 17,157

Therefore, in order to maximize profit, the company should produce 1071 finished tables and 1643 unfinished tables.

The post office will accept packages whose combined length and girth is at most 130 inches. (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.) What is the largest volume that can be sent in a rectangular box?

$65/3\times 65/3\times 130/3$

Solution

Let $l\text{,}$ $w$ and $h$ be the length, width and height of the box in inches, respectively. Then the volume of the package is given by

\begin{equation*} V = l \times w \times h, \qquad l,w,h \geq 0 \end{equation*}

and the girth is given by

\begin{equation*} g = 2w + 2h\text{.} \end{equation*}

Therefore, we require $l + g = l + 2w + 2h \leq 130\text{.}$ Let $l = 130-2w-2h\text{.}$ Then

\begin{equation*} V(w,h) = (130-2w-2h)\cdot w \cdot h\text{,} \end{equation*}

and we can now find the first partial derivatives with respect to $w$ and $h\text{:}$

\begin{equation*} V_w = \frac{\partial}{\partial w} \left((130-2w-2h)\cdot w \cdot h\right) = -2h(2w + h - 65) \end{equation*}
\begin{equation*} V_h = \frac{\partial}{\partial h} \left((130-2w-2h)\cdot w \cdot h\right) = -2w (w + 2h - 65) \end{equation*}

We now need to find the points where both $V_w$ and $V_h$ are equal to zero. We see that $h=0$ and $w=0$ is a solution, but this is not an interior critical point because it is a boundary point. Instead, we set

\begin{equation*} 2w + h -65 = 0 \qquad \text{ and } \qquad w + 2h - 65 = 0\text{.} \end{equation*}

Let $h = 65 - 2w\text{.}$ Then

\begin{equation*} w + 2h - 65 = w + 2 (65-2w)- 65 = 65 - 3w = 0\text{,} \end{equation*}

and so $w = \dfrac{65}{3}$ and $h = \dfrac{65}{3}\text{.}$ Now, since

\begin{equation*} V(0,0) = 0 \text{ and } V\left(\frac{65}{3},\frac{65}{3}\right) = \frac{549,250}{27}\text{,} \end{equation*}

we conclude that the maximum volume of the box is $V = \dfrac{549,250}{27}\text{,}$ with dimensions of $\dfrac{130}{3} \times \dfrac{65}{3} \times \dfrac{65}{3}\text{.}$

The bottom of a rectangular box costs twice as much per unit area as the sides and top. Find the shape for a given volume that will minimize cost.

It has a square base, and is one and one half times as tall as wide. If the volume is $V$ the dimensions are $\root 3 \of {2V/3}\times \root 3 \of {2V/3}\times \root 3\of {9V/4}\text{.}$

Solution

Let $l\text{,}$ $w$ and $h$ be the length, width and height of the box in inches, respectively. Then the volume of the package is given by

\begin{equation*} V = l \times w \times h, \qquad l,w,h \geq 0 \end{equation*}

and the area of the bottom is given by

\begin{equation*} A_{bot} = l \times w\text{.} \end{equation*}

The cost to construct the box is then proportional to

\begin{equation*} C(l,w,h) = A_{top} + A_{sides} + 2\times A_{bot} = lw + 2lh + 2wh + 2lw = 3lw + 2lh + 2wh\text{.} \end{equation*}

We now wish to find the minimum cost for any given volume, $V\text{.}$ We notice that, for $V>0\text{,}$ we require all of $l,w,h > 0\text{.}$ So this minimum must occur at an interior critical point.

We use the constraint to write the cost as a function of two variables:

\begin{equation*} C(l,h) = 3\frac{V}{h} + 2lh + 2\frac{V}{l}\text{.} \end{equation*}

Now compute the first-order partial derivatives of $C\text{:}$

\begin{equation*} C_l = 2h - 2\frac{V}{l^2}, \ \ \text{ and } \ \ C_h = -3\frac{V}{h^2} + 2l\text{.} \end{equation*}

Then the system of equations we need to solve is:

\begin{equation*} \begin{split} 2h - 2\frac{V}{l^2} \amp = 0 \\ -3\frac{V}{h^2} + 2l \amp = 0 \end{split} \end{equation*}

Therefore, we let

\begin{equation*} h = \frac{V}{l^2} \implies l = \sqrt[3]{\frac{2V}{3}}, \ \ \text{ and } \ \ h = \sqrt[3]{\frac{9V}{4}}\text{.} \end{equation*}

Finally, we find the resulting $w = \sqrt[3]{\frac{2V}{3}}\text{.}$ Hence, for any given volume, the dimensions which minimize the cost are

\begin{equation*} \sqrt[3]{\frac{2V}{3}} \times \sqrt[3]{\frac{2V}{3}} \times \sqrt[3]{\frac{9V}{4}}\text{.} \end{equation*}

Using the methods of this section, find the shortest distance from the origin to the plane $x+y+z=10\text{.}$

$\sqrt{100/3}$

Solution

We wish to find the shortest distance from the origin $(0,0,0)$ to the plane $x+y+z=10\text{.}$ We first find an expression for the distance from the origin to any point:

\begin{equation*} d(x,y,z) = \sqrt{x^2+y^2+z^2}\text{.} \end{equation*}

Therefore, we wish to minimize $d$ subject to the constraint $x+y+z = 10\text{.}$ Using this constraint, we write $d$ as a function of two variables:

\begin{equation*} d(x,y) = \sqrt{x^2+y^2 + (10 - x - y)^2}\text{.} \end{equation*}

The first-order partial derivatives are given by

\begin{equation*} d_x = \frac{2 x - 2 (-x - y + 10))}{2 \sqrt{x^2 + (-x - y + 10)^2 + y^2}}, \ \ \text{ and } \ \ d_y = \frac{2 y - 2 (-x - y + 10)}{2 \sqrt{x^2 + (-x - y + 10)^2 + y^2}}\text{.} \end{equation*}

To find any interior critical points, we set both $d_x=0$ and $d_y=0\text{,}$ and so

\begin{equation*} 4x+2y-20=0, \ 4y +2x-20 = 0 \implies x = y = \frac{10}{3}\text{.} \end{equation*}

Therefore, the point $\left(\frac{10}{3}, \frac{10}{3}, \frac{10}{3}\right)$ is the only critical point. Since the distance grows without bound as we move away from the origin, this must be the absolute minimum.

Hence, the absolute minimum distance from the orgin to the point is

\begin{equation*} d\left( \frac{10}{3}, \frac{10}{3}, \frac{10}{3}\right) = \frac{10}{\sqrt{3}}\text{.} \end{equation*}

Using the methods of this section, find the shortest distance from the point $(x_0,y_0,z_0)$ to the plane $ax+by+cz=d\text{.}$ You may assume that $c\not=0\text{;}$ use of Sage or similar software is recommended.

$|ax_0+by_0+cz_0-d|/\sqrt{a^2+b^2+c^2}$

Solution

We wish to find the shortest distance from the origin $(x_0,y_0,z_0)$ to the plane $ax+by+cz=d\text{.}$ We first find an expression for the distance from the origin to any point:

\begin{equation*} f(x,y,z) = \sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}\text{.} \end{equation*}

Therefore, we wish to minimize $d$ subject to the constraint $ax+by+cz=d\text{.}$ Using this constraint, we write $d$ as a function of two variables:

\begin{equation*} f(x,y) = \sqrt{(x-x_0)^2+(y-y_0)^2 + \left(\frac{1}{c} (d - ax - by)\right)^2}\text{.} \end{equation*}

The first-order partial derivatives are given by

\begin{equation*} f_x = \frac{2 (x - x_0) - (2 a (-a x - b y + d))/c^2}{2 \sqrt{(-a x - b y + d)^2/c^2 + (x - x_0)^2 + (y - y_0)^2}}\text{,} \end{equation*}

and

\begin{equation*} f_y = \frac{2 (y - y_0) - (2 a (-a x - b y + d))/c^2}{2 \sqrt{(-a x - b y + d)^2/c^2 + (x - x_0)^2 + (y - y_0)^2}}\text{.} \end{equation*}

To find any interior critical points, we set both $f_x=0$ and $f_y=0\text{,}$ and so

\begin{equation*} 2 (x - x_0) - (2 a (-a x - b y + d))/c^2 = 0, \ \text{ and } \ 2 (y - y_0) - (2 a (-a x - b y + d))/c^2 = 0\text{.} \end{equation*}

The solution to this system of equations gives one critical point.

Using software, we find that the minimum distance is $\frac{ax_0 + by_0 + cz_0 - d}{\sqrt{a^2+b^2+c^2}}\text{.}$

A trough is to be formed by bending up two sides of a long metal rectangle so that the cross-section of the trough is an isosceles trapezoid. If the width of the metal sheet is 2 meters, how should it be bent to maximize the volume of the trough?

The sides and bottom should all be $2/3$ meter, and the sides should be bent up at angle $\pi/3\text{.}$

Solution

A labelled cross-section of the trough is shown below:

Then the problem is to maximize the area of this trapezoid (since this will maximize the volume of the trough for any given length) such that $a+a+b = 2$ metres. We first find an expression for the area, which is the area of the rectangular portion plus the area of the two right traingles:

\begin{equation*} A(a,b,h)= b\times h + h \times \sqrt{a^2-h^2}\text{.} \end{equation*}

We now use the constraint to write the area as a function of two variables:

\begin{equation*} A(a,h) = h (2-2a + \sqrt{a^2-h^2}\text{.} \end{equation*}

The first-order partial derivatives of $A$ are

\begin{equation*} A_a = \frac{ah}{\sqrt{a^2-h^2}}-2h, \ \ \text{ and } \ \ A_h = -\frac{h^2}{\sqrt{a^2 - h^2}} + \sqrt{a^2 - h^2} -2a+2\text{.} \end{equation*}

To find any critical points, we set $A_a = A_h = 0$ (we will assume that all of $a,h > 0$):

\begin{equation*} A_a = 0 \implies a=2\sqrt{a^2-h^2} \implies h^2 = \frac{3}{4} a^2\text{,} \end{equation*}

and so plugging this condition into $A_h = 0$ gives

\begin{equation*} -2a+2 + \sqrt{\frac{1}{4}a^2} - \frac{3a^2}{4\sqrt{\frac{1}{4}a^2}} = -2a+2+\frac{a}{2} - \frac{3a}{2} = 0 \implies a = \frac{2}{3}\text{.} \end{equation*}

The corresponding value of $h$ is therefore

\begin{equation*} h = \sqrt{\frac{3}{4}\left(\frac{2}{3}\right)^2} = \frac{1}{\sqrt{3}}\text{.} \end{equation*}

Finally, we compute the corresponding value for $b\text{:}$

\begin{equation*} b = 2 - 2\left(\frac{2}{3}\right) = \frac{2}{3}\text{.} \end{equation*}

We conclude that the sides and bottom should be $2/3$ m long, and the height of the trough should be $1/\sqrt{3}$ m (this corresponds to an angle of $\pi/3$).

Given the three points $(1,4)\text{,}$ $(5,2)\text{,}$ and $(3,-2)\text{,}$ $\ds(x-1)^2+(y-4)^2+(x-5)^2+(y-2)^2+(x-3)^2+(y+2)^2$ is the sum of the squares of the distances from point $(x,y)$ to the three points. Find $x$ and $y$ so that this quantity is minimized.

$(3,4/3)$

Solution

We wish to find the point $(x,y)$ which minimizes the sum of squares distance to the three given points. Let

\begin{equation*} S(x,y) = (x-1)^2 + (y-4)^2 + (x-5)^2 + (y-2)^2+(x-3)^2+(y+2)^2\text{.} \end{equation*}

Now compute the first-order partial derivatives of $s\text{:}$

\begin{equation*} S_x = 2(x-1)+2(x-5)+2(x-3) = 6x-18, \ \ \text{ and } \ \ S_y = 2(y-4)+2(y-2)+2(y+2)=6y-8\text{.} \end{equation*}

To find the critical point(s) of $S\text{,}$ we set both $S_x=0$ and $S_y=0\text{.}$ This gives exactly one critical point at $(3, 4/3)\text{.}$

We can use the Second Derivative Test to classify this critical point:

\begin{equation*} D(x,y) = 6(6) - 0 = 36 > 0\text{,} \end{equation*}

and $f_{xx}(x,y) > 0\text{.}$ Hence, this critical point is a relative minimum. Since it is the only critical point, and the domain of $S$ is unbounded, it must be the absolute minimum. We illustrate the solution below.

Suppose that $f(x,y)=x^2+y^2+kxy\text{.}$ Find and classify the critical points, and discuss how they change when $k$ takes on different values.

Solution

We compute the first-order partial derivatives of $f(x,y) = x^2+y^2+kxy\text{,}$ for some $k\in \mathbb{R}\text{:}$

\begin{equation*} f_x = 2x+ky, \ \ \text{ and } \ \ f_y = 2y+kx\text{.} \end{equation*}

To find the critical point(s) of $f\text{,}$ we set both $f_x=0$ and $f_y=0\text{:}$

\begin{equation*} f_x = 0 \implies x = -\frac{ky}{2}\text{,} \end{equation*}

and so

\begin{equation*} f_y = 0 \implies 2y + k\left(-\frac{ky}{2}\right) = 0\text{.} \end{equation*}

Hence, the only critical point is at $(0,0)\text{.}$ To classify this critical point, we use the Second Derivative Test:

\begin{equation*} D(0,0) = 2(2) -k^2 = 4-k^2\text{,} \end{equation*}

and

\begin{equation*} f_{xx}(0,0) = 2\text{.} \end{equation*}

As a function of $k\text{,}$ the discriminant is therefore an upside down parabola:

We therefore have three cases:

1. For $k \in (-2,2)\text{,}$ we have $D(0,0) > 0$ and $f_{xx}(0,0) > 0$ and so the point $(0,0)$ is a relative minimum.

2. For $k\in(-\infty,-2)\cup(2,\infty)\text{,}$ then $D(0,0)\lt 0$ and so the point $(0,0)$ is a saddle point.

3. For $k=-2,2\text{,}$ then $D(0,0)=0$ and the Second Derivative Test is inconclusive. Further investigation of the surfaces $z=x^2+y^2-2xy$ and $z=x^2+y^2+2xy$ reveals that $(0,0)$ is a relative minimum in both cases.

Find the shortest distance from the point $(0,b)$ to the parabola $y=x^2\text{.}$

$|b|$ if $b\le1/2\text{,}$ otherwise $\ds\sqrt{b-1/4}$

Solution

We want to find the shortest distance from the point $(0,b)$ to the parabola $y=x^2\text{.}$ We first construct the distance function:

\begin{equation*} d(x,y) = \sqrt{x^2+(y-b)^2}\text{.} \end{equation*}

We now use the constraint that $y=x^2$ to write the distance as a function of $x$ only:

\begin{equation*} d(x) = \sqrt{x^2+(x^2-b)^2}\text{.} \end{equation*}

Differentiating, we find

\begin{equation*} d'(x) = \frac{-2bx+2x^3+x}{\sqrt{x^2+(x^2-b)^2}}\text{.} \end{equation*}

To find any critical point(s), we set

\begin{equation*} d'(x) = 0 \implies -2bx+2x^3+x = 0 \implies x=0, \pm \sqrt{\frac{2b-1}{2}}\text{.} \end{equation*}

To classify these three critical points, we will use the First Derivative Test for single-variable functions. We compute:

\begin{equation*} d''(x) = \frac{-2 b^3 + 6 b^2 x^2 + b^2 - 6 b x^4 + 2 x^6 + 3 x^4}{(b^2 - 2 b x^2 + x^4 + x^2)^(3/2)}\text{,} \end{equation*}

We find that

\begin{equation*} d''(0) = \frac{1-2b}{\sqrt{b^2}} \end{equation*}
\begin{equation*} d''(\sqrt{(2b-1)/2)}) = \frac{8b-4}{\sqrt{4b-1}} \end{equation*}
\begin{equation*} d''(-\sqrt{(2b-1)/2)}) = \frac{8b-4}{\sqrt{4b-1}} \end{equation*}

We see that there are three cases:

1. $b > 1/2\text{:}$ $x=0$ is the relative minimum. Since this is the only relative minimum on an unbounded domain, it must be the absolute minimum. This point corresponds to a distance of $|b|\text{.}$

2. $b \lt 1/2\text{:}$ $x = \pm \sqrt{\frac{2b-1}{2}}$ are relative minima. These points both correspond to a distance of $\sqrt{b-1/4}\text{,}$ which is the absolute minimum distance in this case.

3. $b=1/2\text{:}$ the Second Derivative Test is inconclusive. However, using the First Derivative Test, we conclude that in this case, $|d|$ is the minimum distance.

Find the shortest distance from the point $(0,0,b)$ to the paraboloid $z=x^2+y^2\text{.}$

$|b|$ if $b\le1/2\text{,}$ otherwise $\ds\sqrt{b-1/4}$

Solution

We want to find the shortest distance from the point $(0,0,b)$ to the paraboloid $z=x^2+y^2\text{.}$ Due to symmetry, this problem reduces to Exercise 7.6.13. Therefore, if

1. $b \geq 1/2\text{:}$ $x=0$ is the relative minimum. Since this is the only relative minimum on an unbounded domain, it must be the absolute minimum. This point corresponds to a distance of $|b|\text{.}$

2. $b \lt 1/2\text{:}$ $x = \pm \sqrt{\frac{2b-1}{2}}$ are relative minima. These points both correspond to a distance of $\sqrt{b-1/4}\text{,}$ which is the absolute minimum distance in this case.

Consider the function $f(x,y)=x^3-3x^2y+y^3\text{.}$

1. Show that $(0,0)$ is the only critical point of $f\text{.}$

Solution

The first partial derivatives of $f$ are

\begin{equation*} f_x = \diffp{}{x} \left(x^3-3x^2y+y^3\right) = 3x^2-6xy\text{,} \end{equation*}

and

\begin{equation*} f_y=\diffp{}{y}\left(x^3-3x^2y+y^3\right) = -3x^2+3y^2\text{.} \end{equation*}

To find the critical points of $f\text{,}$ we need to solve the system of equations

\begin{equation*} f_x = 3x^2-6xy = 0, \ \ \text{ and } \ \ f_y= -3x^2 + 3y^2 =0\text{.} \end{equation*}

From the second equation, let

\begin{equation*} 3x^2=3y^2 \implies y = \pm x\text{.} \end{equation*}

Now substitute these values of $y$ into the first equation,

\begin{equation*} \begin{array}{ccc} 3x^2-6x(-x)= 0 \amp \text{ or } \amp 3x^2-6x(x)=0\\ 8x^2 = 0 \amp \amp -3x^2=0 \end{array} \end{equation*}

Therefore, $x=0$ is the only solution, which implies that $y=0\text{.}$ Thus, $(0,0)$ is our only critical point.

2. Show that the Discriminant Test is inconclusive for $f\text{.}$

Solution

We first find the second partial derivatives of $f\text{.}$

\begin{equation*} f_{xx} = \diffp{}{x} \left(3x^2-6xy\right) = 6x-6y \end{equation*}
\begin{equation*} f_{yy} = \diffp{}{y} \left(-3x^2+ 3y^2\right)= 6y \end{equation*}
\begin{equation*} f_{xy} = \diffp{}{y} \left(3x^2-6xy\right)=-6x = f_{yx} \end{equation*}

Using the formula for the discriminant, we get

\begin{equation*} D=f_{xx}f_{yy}-f_{xy}^2 = 36y\left(x-y\right)+36x^2\text{.} \end{equation*}

So at the critical point $(0,0)\text{,}$ $D=0$ and the test is inconclusive.

3. Determine the cross-sections of $f$ obtained by setting $y=kx$ for various values of $k\text{.}$ Solution

We show cross-sections of $f$ where $y=kx$ for $k=-2,0,2$ below.

4. What kind of critical point is $(0,0)\text{?}$

Solution

The above cross-sections indicate that the critical point $(0,0,0)$ is a saddle point of $f\text{.}$ We justify this statement with the following surface plot:

Find the volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid $2x^2+72y^2+18z^2=288\text{.}$

$\ds 1024/\sqrt3$