## Section4.1Functions of Several Variables

Recall that a function $f: \mathbb{R}\to\mathbb{R}$ maps a single real value $x$ to a single real value $y\text{.}$ Such a function is referred to as a single-variable function and can be readily visualized in a two-dimensional coordinate system: above (or below) each point $x$ on the $x$-axis we graph the point $y\text{,}$ where of course $y=f(x)\text{.}$ By now, you have seen the graphs of many such functions. We now extend this visualization process to multi-variable functions, also referred to as functions of several variables.

In single-variable calculus we were concerned with functions that map the real numbers $\mathbb{R}$ to $\mathbb{R}\text{,}$ sometimes called “real functions of one variable”, meaning the “input” is a single real number and the “output” is likewise a single real number. Now we turn to functions of several variables, where several input variables are mapped to one value: functions $f\colon\mathbb{R}^n\to\mathbb{R}\text{.}$ We will deal primarily with $n=2$ and to a lesser extent $n=3\text{;}$ in fact many of the techniques we discuss can be applied to larger values of $n$ as well.

A function $f\colon\mathbb{R}^2\to\mathbb{R}$ maps a pair of values $(x,y)$ to a single real number. The three-dimensional coordinate system is a convenient way to visualize such functions: above (or below) each point $(x,y)$ in the $x$-$y$-plane we graph the point $(x,y,z)\text{,}$ where then $z=f(x,y)\text{.}$ In other words, in interpreting the graph of a function $f(x,y)\text{,}$ one often thinks of the value $z=f(x,y)$ of the function at the point $(x,y)$ as the height of the point $(x,y,z)$ on the graph of $f\text{.}$ If $f(x,y)>0\text{,}$ then the point $(x,y,z)$ is $f(x,y)$ units above the $x$-$y$-plane; if $f(x,y)\lt 0\text{,}$ then the point $(x,y,z)$ is $\rvert f(x,y)\rvert$ units below the $x$-$y$-plane.

In general, it is quite difficult to draw the graph of a function of two variables. But techniques have been developed that enable us to generate such graphs with minimum effort, using a computer. Still, it is valuable to be able to visualize relatively simple surfaces without such aids. It is often a good idea to examine the function on restricted subsets of the plane, especially lines. It can also be useful to identify those points $(x, y)$ that share a common $z$-value.

Before we introduce some special and some general three-dimensional graphs and their equations, let us first consider the domain of functions in two variables.

The variables $x$ and $y$ are called independent variables, and the variable $z\text{,}$ which is dependent on the values of $x$ and $y\text{,}$ is referred to as a dependent variable. As indicated already, the number $z=f(x,y)$ is called the value of $f$ at the point $(x,y)\text{.}$ Unless specified otherwise, the domain of the function $f$ will be taken to be the largest possible set for which the rule defining $f$ is meaningful. We will demonstrate this with several examples.

###### Example4.1. Domain of Two-Variable Functions.

Find the domain of each of the following functions.

1. $f(x,y)=\sqrt{1-x^2-y^2}\text{.}$

2. $g(x,y)=\dfrac{1}{\sqrt{1-x^2-y^2}}\text{.}$

3. $h(x,y)=\dfrac{1}{y-x^2}\text{.}$

Solution
1. The domain of $f$ is the set of points $(x,y)$ such that $1-x^2-y^2 \geq 0\text{.}$ We recognize $x^2+y^2=1$ as the equation of a circle of radius 1 centred at the origin, and so the domain of $f$ consists of all points which lie on or inside this circle (see below).

2. We now require that $1-x^2-y^2 > 0\text{.}$ The domain of $g$ therefore contains all points $(x,y)$ such that $x^2+y^2\lt 1\text{;}$ that is, all points which lie strictly inside the unit circle (see below).

3. We see that $h(x,y)$ is undefined for $x=y^2\text{.}$ The domain of $h$ therefore consists of all points in the $x$-$y$-plane except those which satisfy $y=\pm \sqrt{x}$ (see below).

###### Example4.2. Domain of Two-Variable Functions.

A manufacturer produces a model X and a model Y, and determines that the unit prices of these two products are related. Let $q_x$ be the weekly quantity demanded of model X, and let $q_y$ be the weekly quantity demanded of model Y. The unit price of model X is found to be

\begin{equation*} p_x=500-q_x-\frac{1}{3}q_y\text{,} \end{equation*}

and the unit price of model Y is found to be

\begin{equation*} p_y=200-\frac{1}{3}q_x-\frac{1}{5}q_y\text{.} \end{equation*}
1. Determine the revenue function $R(q_x,q_y)\text{.}$

2. Sketch the domain of $R\text{.}$

Solution
1. Selling $q_x$ units of model X yields a revenue of $q_xp_x$ dollars per week, and selling $q_y$ units of model Y yields a revenue of $q_yp_y$ dollars per week. We therefore construct the revenue function $R$ as

\begin{equation*} \begin{split} R(q_x,q_y) \amp = q_xp_x + q_yp_y \\ \amp = q_x\left(500-q_x-\frac{1}{3}q_y\right)+q_y\left(200-\frac{1}{3}q_x-\frac{1}{5}q_y\right) \\ \amp = \frac{1}{15}\left(-15q_x^2-10q_xq_y+7500q_x-3q_y^2+3000q_y\right), \end{split} \end{equation*}

dollars per week.

2. The domain of $R$ is all points $(q_x,q_y)$ in the plane such that $q_x\text{,}$$q_y\text{,}$$p_x$ and $p_y \geq 0\text{.}$ That is, where

\begin{equation*} \begin{array}{c} q_x \geq 0, \\[1ex] q_y \geq 0, \\[1ex] 500-q_x-\frac{1}{3}q_y \geq 0 \iff q_y \leq 1500 - 3q_x \\[1ex] 200-\frac{1}{3}q_x-\frac{1}{5}q_y \geq 0 \iff q_y \leq 1000 - \frac{5}{3}q_x. \end{array} \end{equation*}

We find the domain by graphing the two lines given by $p_x=0$ and $p_y=0$ in the first quadrant of the $q_x$-$q_y$-plane. The domain of $R$ is the region which is bounded by the lines $p_x=0\text{,}$ $p_y=0\text{,}$ $q_x=0$ and $q_y=0$ for which the desired inequality holds. The final solution is given by the shaded region (including the boundary) in the figure below.

Let us now consider some functions that we can describe and graph with the knowledge we have attained so far.

###### Example4.3. Plane.

Describe and graph the function $f(x,y)=3x+4y-5\text{.}$

Solution

We first write this as $z = 3x+4y-5$ and then $3x+4y-z = 5\text{,}$ which is the equation of a plane. However, we can also determine that this equation represents a plane from the following analysis: If we hold $x$ constant, then the equation simplifies to that of a straight line in the $y$-$z$-plane. Generally speaking, if we choose any point on the graph of $f\text{,}$ for example from $f(0,0)= -5\text{,}$ this function grows linearly in every direction in the same way that a linear function behaves as can be seen below.

To graph the plane, we alternately let $x\text{,}$ then $y\text{,}$ then $z$ be equal to zero, which leads to a linear equation in the $y$-$z$-, $x$-$z$-, and $x$-$y$-planes respectively. These lines represent the intersection of the functions' graph with each of the coordinate planes in the three-dimensional coordinate system and form part of the plane surface as shown below.

Interactive Demonstration. Click to move the graphic below investigating the multivariate function $f(x,y)=3x+4y-5\text{.}$

###### Example4.4. Sphere.

Describe and graph the equation $x^2+y^2+z^2=4\text{.}$

Solution

The equation $x^2+y^2+z^2=4$ represents a sphere of radius 2 and centre $(0,0,0)$ as shown below. We cannot write this in the form $z=f(x,y)\text{,}$ since for each $x$ and $y$ in the disk $x^2+y^2\lt 4$ there are two corresponding points on the sphere, namely one above and one below this point $(x,y)\text{.}$ As with the equation of a circle, we can resolve this equation into two functions, $\ds f_1(x,y)=\sqrt{4-x^2-y^2}$ and $\ds f_2(x,y)=-\sqrt{4-x^2-y^2}\text{,}$ representing the upper and lower hemispheres, respectively. Each of these is an example of a function with a restricted domain: only certain values of $x$ and $y$ make sense (namely, those for which $x^2+y^2\le 4$) and the graphs of these functions are limited to a small region of the plane.

Interactive Demonstration. Click to move the graphic below investigating the surface described by $x^2+y^2+z^2=4\text{.}$

###### Example4.5. Square Root.

Describe and graph the function $f(x,y)=\sqrt x+\sqrt y\text{.}$

Solution

This function is defined only when both $x$ and $y$ are non-negative. When $y=0$ we get $f(x,y)=\sqrt x\text{,}$ the familiar square root function in the $x$-$z$-plane, and when $x=0$ we get the same curve in the $y$-$z$-plane. Generally speaking, we see that starting from $f(0,0)=0$ this function gets larger in every direction in roughly the same way that the square root function gets larger. For example, if we restrict attention to the line $x=y\text{,}$ we get $f(x,y)=2\sqrt x$ and along the line $y=2x$ we have $f(x,y)=\sqrt x+\sqrt{2x}=(1+\sqrt2)\sqrt x$ (see graph below).

Interactive Demonstration. Click to move the graphic below investigating the surface described by $f(x,y)=\sqrt x+\sqrt y\text{.}$

###### Example4.6. Elliptic Paraboloid.

Describe and graph the function $f(x,y)=x^2+y^2\text{.}$

Solution

When $x=0$ this becomes $f(y)=y^2\text{,}$ a parabola in the $y$-$z$-plane; when $y=0$ we get the “same” parabola $f(x)=x^2$ in the $x$-$z$-plane.

Finally, picking a value $z=k\text{,}$ at what points does $f(x,y)=k\text{?}$ This means $x^2+y^2=k\text{,}$ which we recognize as the equation of a circle of radius $\sqrt k\text{.}$ So the graph of $f(x,y)$ has parabolic cross-sections, and the same height everywhere on concentric circles with centre at the origin. This fits with what we have already discovered.

Interactive Demonstration. Click to move the graphic below investigating the surface $z=f(x,y)=x^2+y^2$ and its level curves.

As in this example, the points $(x,y)$ such that $f(x,y)=k$ form a curve, called a level curve of the function. A graph of some level curves can give a good idea of the shape of the surface; it looks much like a topographic map of the surface. By drawing the level curves corresponding to several admissible values of $k\text{,}$ we obtain a contour map. In the graph of Example 4.6 both the surface and its associated level curves are shown. Note that, as with a topographic map, the heights corresponding to the level curves are evenly spaced, so that where curves are closer together the surface is steeper.

###### Example4.7. Level Curves.

Sketch the level curves of the function $z=f(x,y)=4x^2-y$ corresponding to $z=-2,-1,0,1,2\text{.}$

Solution

We find the level curves of $f$ by setting $f(x,y)=z$ to be a constant. For $z=-2,-1,0,1,2\text{,}$ we find the equations

\begin{equation*} \begin{array}{c} y=4x^2+2 \\[1ex] y=4x^2+1 \\[1ex] y=4x^2 \\[1ex] y=4x^2-1 \\[1ex] y=4x^2-2 \end{array} \end{equation*}

and so each level curve is a parabola in the $x$-$y$-plane, where only the $y$-intercept changes. The surface and its level curves are illustrated below.

Interactive Demonstration. Click to move the graphic below investigating the surface $z=f(x,y)=4x^2-y$ and its level curves.

###### Example4.8. Level Curves.

Sketch the level curves of the function $z=f(x,y)=\sqrt{x^2+y^2}$ corresponding to $z=0,1,2,3\text{.}$

Solution

For each level curve corresponding to $z=c\text{,}$ we have

\begin{equation*} \sqrt{x^2+y^2} = c \implies x^2+y^2 = c^2\text{,} \end{equation*}

which is the equation of a circle centred at $(0,0)$ of radius $c\text{,}$ as shown below.

Interactive Demonstration. Click to move the graphic below investigating the level curves of $z=f(x,y)=\sqrt{x^2+y^2}\text{.}$

###### Example4.9. Home Mortgage Payments.

The monthly payment for a condo that amortizes a loan of $A$ dollars in $t$ years when the interest rate is $r$ per year is given by

\begin{equation*} P=f(A,r,t)=\dfrac{Ar}{12\left[1-\left(1+\frac{r}{12}\right)^{-12t}\right]} \end{equation*}

Find the monthly payment for a home mortgage of $240,000 to be amortized over 25 years when the interest rate is 4% per year. Solution $P$ is a function of three variables: $A\text{,}$ $r$ and $t\text{.}$ To find the required monthly payments, we evaluate $P$ at the given values, \begin{equation*} \begin{split} P\amp =f(240000,0.04,25)\\ \amp =\dfrac{240000(0.04)}{12\left[1-\left(1+\frac{0.04}{12}\right)^{-12(25)}\right]}\\ \amp \approx 1266.81. \end{split} \end{equation*} Thus, the monthly payments would be about$1266.81.

Functions $f\colon\mathbb{R}^n\to\mathbb{R}$ for $n > 2$ behave much like functions of two variables. The principal difficulty with such functions is visualizing them, as they do not “fit” in the three dimensions we are familiar with. For $n=3$ variables there are various ways to interpret functions that make them easier to understand. For example, $f(x,y,z)$ could represent the temperature at the point $(x,y,z)\text{,}$ or the strength of a magnetic field. Similar to considering level curves with two-variable functions, it is useful to consider those points at which $f(x_1,x_2,\dots,x_n) = k\text{,}$ where $k$ is some constant for $n$-variable functions. This collection of points is called a level set. For three variables, a level set is a surface, called a level surface.

###### Example4.10. Level Surfaces.

Suppose the temperature at $(x,y,z)$ is $T(x,y,z)=e^{-(x^2+y^2+z^2)}\text{.}$ Describe the level surfaces.

Solution

This function has a maximum value of 1 at the origin, and tends to 0 in all directions. If $k$ is positive and at most 1, the set of points for which $T(x,y,z)=k$ is those points satisfying $x^2+y^2+z^2=-\ln k\text{,}$ a sphere centred at the origin. The level surfaces are the concentric spheres centred at the origin.

##### Exercises for Section 4.1.

Find the equations and describe the shapes of the cross-sections when $x=0\text{,}$ $y=0$ and $x=y\text{.}$ Plot the surface with a three-dimensional graphing tool.

1. $f(x,y)=(x-y)^2$

$z=y^2$ , $z=x^2 \text{,}$ $z=0\text{.}$
Solution

Let $z=f(x,y)=(x-y)^2\text{,}$ and consider the following cross-sections. Taking a slice along the $y$-axis, we have

\begin{equation*} z = f(0,y) = (0-y)^2 = (-y)^2 = y^2\text{.} \end{equation*}

This cross-section will then look like:

Taking a slice along the $x$-axis, we have

\begin{equation*} z = f(x,0) = x^2\text{,} \end{equation*}

and graphically,

Finally, along the line $x=y\text{,}$

\begin{equation*} z=f(x,y) = (0)^2 = 0\text{,} \end{equation*}

and so $f$ is constant (and zero) along this cross-section. The surface $z=f(x,y)$ is plotted below for $(x,y) \in [-5,5]\times[-5,5]\text{.}$

2. $f(x,y)=|x|+|y|$

$z= |y|\text{,}$ $z= |x| \text{,}$ $z=2|y|\text{.}$
Solution

Let $z=f(x,y)=|x|+|y|\text{,}$ and consider the following cross-sections. Taking a slice along the $y$-axis, we have

\begin{equation*} z = f(0,y) = |y|\text{.} \end{equation*}

This cross-section will then look like:

Taking a slice along the $x$-axis, we have

\begin{equation*} z = f(x,0) = |x|\text{,} \end{equation*}

and graphically,

Finally, along the line $x=y\text{,}$

\begin{equation*} z=f(x,y) = |y|+|y| = 2|y|\text{,} \end{equation*}

and graphically,

The surface $z=f(x,y)$ is plotted below for $(x,y) \in [-5,5]\times[-5,5]\text{.}$

3. $f(x,y)=e^{-(x^2+y^2)}\sin(x^2+y^2)$

$z=e^{-(y^2)}\sin(y^2) \text{,}$ $z= e^{-(x^2)}\sin(x^2) \text{,}$ $z= e^{-(2y^2)}\sin(2y^2)\text{.}$
Solution

Let $z=f(x,y)=e^{-(x^2+y^2)}\sin(x^2+y^2)\text{,}$ and consider the following cross-sections. Taking a slice along the $y$-axis, we have

\begin{equation*} z = f(0,y) = e^{-(y^2)}\sin(y^2)\text{.} \end{equation*}

This cross-section will then look like:

Taking a slice along the $x$-axis, we have

\begin{equation*} z = f(x,0) = e^{-(x^2)}\sin(x^2)\text{,} \end{equation*}

and graphically,

Finally, along the line $x=y\text{,}$

\begin{equation*} z=f(x,y) = e^{-(y^2+y^2)}\sin(y^2+y^2) = e^{-(2y^2)}\sin(2y^2)\text{,} \end{equation*}

and graphically,

The surface $z=f(x,y)$ is plotted below for $(x,y) \in [-5,5]\times[-5,5]\text{.}$

4. $f(x,y)=\sin(x-y)$

$z=-\sin(y) \text{,}$ $z= \sin(x)\text{,}$ $z=0 \text{.}$
Solution

Let $z=f(x,y)=\sin(x-y)\text{,}$ and consider the following cross-sections. Taking a slice along the $y$-axis, we have

\begin{equation*} z = f(0,y) = \sin(-y) = -\sin(y)\text{,} \end{equation*}

where we used the fact that sine is an even function. This cross-section will then look like:

Taking a slice along the $x$-axis, we have

\begin{equation*} z = f(x,0) = \sin(x)\text{,} \end{equation*}

and graphically,

Finally, along the line $x=y\text{,}$

\begin{equation*} z=f(x,y) = \sin(0) = 0\text{,} \end{equation*}

and so $f$ is constant (and zero) along this cross-section. The surface $z=f(x,y)$ is plotted below for $(x,y) \in [-2\pi,2\pi]\times[-2\pi,2\pi]\text{.}$

5. $f(x,y)=(x^2-y^2)^2$

$z =y^4 \text{,}$ $z =x^4 \text{,}$ $z =0 \text{.}$
Solution

Let $z=f(x,y)=(x^2-y^2)^2\text{,}$ and consider the following cross-sections. Taking a slice along the $y$-axis, we have

\begin{equation*} z = f(0,y) = (0-y^2)^2 = (-y^2)^2=y^4\text{.} \end{equation*}

This cross-section will then look like:

Taking a slice along the $x$-axis, we have

\begin{equation*} z = f(x,0) = x^4\text{,} \end{equation*}

and graphically,

Finally, along the line $x=y\text{,}$

\begin{equation*} z=f(x,y) = (0)^2 = 0\text{,} \end{equation*}

and so $f$ is constant (and zero) along this cross-section. The surface $z=f(x,y)$ is plotted below for $(x,y) \in [-5,5]\times[-5,5]\text{.}$

Find the domain of each of the following functions:

1. $f(x,y) = \ds\sqrt{9-x^2}+\sqrt{y^2-4}$

$\{(x,y)\mid |x|\le3\ \hbox{and}\ |y|\ge2\}$
Solution

The domain of $f(x,y)=\sqrt{9-x^2}+\sqrt{y^2-4}$ is given by

\begin{equation*} 9-x^2 \geq 0 \ \ \text{ and } \ \ y^2-4 \geq 0\text{.} \end{equation*}

This will be the area of the $x$-$y$-plane such that $x \in [-3,3]$ and $y \in (-\infty,-2]\cup[2,\infty)\text{.}$ This area is given by the two infinite rectangles graphed below.

2. $h(u,v)=\sqrt{2-u^2-v^2}$

$\{(u,v)\mid u^2+v^2 \leq 2\}$
Solution

The domain of $h(u,v)=\sqrt{2-u^2-v^2}$ is given by

\begin{equation*} u^2+v^2 \leq 2\text{.} \end{equation*}

This will be the area of the circle of radius $\sqrt{2}\text{,}$ including the boundary.

3. $f(x,y) = \ds\sqrt{16-x^2-4y^2}$

$\{(x,y)\mid x^2+4y^2\le16\}$
Solution

The domain of $f(x,y)=\sqrt{16-x^2-4y^2}$ is given by

\begin{equation*} \dfrac{x^2}{4^2}+\dfrac{y^2}{2^2} \leq 1\text{.} \end{equation*}

This will be the area of the ellipse with major radius equal to $4$ and minor radius equal to $2\text{,}$ including the boundary.

The domain of the function $f(x,y)=\sqrt{16-x^2-y^2}$ is all points $(x,y)$ such that $x^2+y^2 \leq 16\text{.}$ This gives a disk in the $x$-$y$-plane:

4. $f(x,y)= x+4y$

All real $x$ and $y$
Solution

The domain of $f(x,y)=x+4y$ is all $(x,y) \in \R \times \R\text{.}$

5. $g(x,y,z)=x^2+y^2+z^2$

All real $x\text{,}$$y\text{,}$ and $z$
Solution

The domain of $g(x,y,z)=x^2+y^2+z^2$ is all $(x,y,z) \in \R \times \R \times \R\text{.}$

6. $f(x,y) = \arcsin(x^2+y^2-2)$

$\{(x,y)\mid 1\le x^2+y^2\le3\}$
Solution

The domain of $f(x,y)=\arcsin(x^2+y^2-2)$ is given by

\begin{equation*} -1 \leq x^2+y^2-2 \leq 1 \implies 1 \leq x^2 + y^2 \leq 3\text{.} \end{equation*}

This is the area between the two circles $x^2+y^2 =1$ and $x^2+y^2=3$ (inclusive of the boundaries):

7. $h(u,v)= \dfrac{uv}{u-v}$

$\{(u,v)\mid u \neq v\}$
Solution

The domain of $h(u,v) = \dfrac{uv}{u-v}$ is the set of all points $(u,v) \in \mathbb{R} \times \mathbb{R}$ such that $u \neq v\text{:}$

8. $f(s,t)=\sqrt{s^2+t^2}$

All real $s$ and $t$
Solution

The domain of $f(s,t) = \sqrt{s^2+t^2}$ is all $(s,t) \in \mathbb{R} \times \mathbb{R}\text{.}$

9. $g(r,s)=\sqrt{rs}$

$\{(r,s)\mid rs > 0\}$
Solution

The domain of $g(r,s)=\sqrt{rs}$ is given by

\begin{equation*} rs \geq 0\text{.} \end{equation*}

This is given by

\begin{equation*} r \geq 0 \ \text{ and } \ s \geq 0 \end{equation*}

or

\begin{equation*} r \leq 0 \ \text{ and } \ s \leq 0 \end{equation*}
10. $f(x,y)=e^{-xy}$

All real $x$ and $y$
Solution

The domain of $f(x,y)=e^{-xy}$ is all $(x,y) \in \R \times \R\text{.}$

11. $h(x,y)=\ln(x+y-7)$

$\{(x,y)\mid x+y > 7 \}$
Solution

The domain of $h(x,y)=\ln(x+y-7)$ is given by

\begin{equation*} x+y-7 > 0 \ \text{ or } \ y > 7-x\text{.} \end{equation*}

Below are two sets of level curves at a sequence of equally-spaced heights $z\text{.}$ One is for a cone, one is for a paraboloid. Which is which? Explain.

Left: paraboloid. Right: cone.
Solution

Both sets of level curves are concentric circles, however the left plot must be for the paraboloid, and the right plot must be for the cone. For the left set of level curves, the difference in radii is decreasing as we move away from the origin, indicating that the slope of the surface is getting steeper and steeper. This is indicative of a paraboloid. In the right set of level curves, the radii are equally spaced, which means that the slope of the surface is constant in all directions away from the origin. This is consistent with a cone.

Sketch the level curves of the function corresponding to the given values of $z\text{.}$

1. $f(x,y)=3x+2y\text{,}$ $z=-2,-1,0,1,2$

Solution

The level curves of $z=f(x,y)=3x+2y$ are

\begin{equation*} k=3x+2y \implies y=\frac{1}{2}(k-3x) \end{equation*}

for some constant $k\text{.}$ We sketch below the level curves corresponding to $k=-2,-1,0,1,2\text{.}$

2. $f(x,y)=x^2-y\text{,}$ $z=-2,-1,0,1,2$

Solution

The level curves of $z=f(x,y)=x^2-y$ are

\begin{equation*} k=x^2-y \implies y=x^2-k \end{equation*}

for some constant $k\text{.}$ We sketch below the level curves corresponding to $k=-2,-1,0,1,2\text{.}$

3. $f(x,y)=2y+x^2\text{,}$ $z=-2,-1,0,1,2$

Solution

The level curves of $z=f(x,y)=2y+x^2$ are

\begin{equation*} k=2y+x^2 \implies y=\frac{1}{2}(k-x^2) \end{equation*}

for some constant $k\text{.}$ We sketch below the level curves corresponding to $k=-2,-1,0,1,2\text{.}$

4. $f(x,y)=2xy\text{,}$ $z=-4,-2,2,4$

Solution

The level curves of $z=f(x,y)=2xy$ are

\begin{equation*} k=2xy \implies y=\dfrac{k}{2x} \end{equation*}

for some constant $k\text{.}$ We sketch below the level curves corresponding to $k=-4,-2,2,4\text{.}$

5. $f(x,y)=\sqrt{25-x^2-y^2}\text{,}$ $z=0,1,3,5$

Solution

The level curves of $z=f(x,y)=\sqrt{25-x^2-y^2}$ are

\begin{equation*} k=\sqrt{25-x^2-y^2} \implies x^2+y^2=25-k^2 \end{equation*}

for some constant $k\text{.}$ We sketch below the level curves corresponding to $k=0,1,3,5\text{.}$

6. $f(x,y)=e^y-x^2\text{,}$ $z=1,2,3,4$

Solution

The level curves of $z=f(x,y)=e^y-x^2$ are

\begin{equation*} k=e^y-x^2 \implies y=\ln(x^2+k) \end{equation*}

for some constant $k\text{.}$ For $k=1,2,3,4\text{,}$ we therefore have

A clothing retailer sells casual and business jackets. Let $q_c$ denote the quantity of casual jackets demanded monthly and $q_b$ denote the quantity of business jackets demanded monthly. It is determined that the unit price of casual jackets is approximately

\begin{equation*} p_c=100-\frac{1}{4}q_c-\frac{1}{7}q_b\text{,} \end{equation*}

and that the unit price of business jackets is approximately

\begin{equation*} p_b=150-\frac{1}{10}q_c-\frac{1}{3}q_b \end{equation*}

dollars, respectively.

1. Determine the monthly revenue, $R(q_c,q_b)\text{.}$

$R(q_c,q_b) = \frac{1}{420}\left(-105q_c^2-102q_cq_b+42000q_c-140q_b^2+63000q_b\right)$
Solution

The monthly revenue is a two-variable function given by the sum of the revenue from selling casual jackets and the revenue from selling the business jackets:

\begin{equation*} \begin{split} R(q_c,q_b) \amp = p_cq_c + p_bq_b\\ \amp = \bigl(100-\frac{1}{4}q_c - \frac{1}{7}q_b\bigr)q_c + \bigl(150-\frac{1}{10}q_c- \frac{1}{3}q_b\bigr)q_b\\ \amp = \frac{1}{420}\left(-105q_c^2-102q_cq_b+42000q_c-140q_b^2+63000q_b\right) \end{split} \end{equation*}
2. Determine the domain of $R\text{.}$

The region where all of $p_c,p_b,q_c,q_b \geq 0\text{.}$
Solution

The domain of $R$ is the region for which all of $q_c\text{,}$ $q_b\text{,}$ $p_c$ and $p_b$ are positive:

\begin{equation*} p_c \geq 0 \implies q_c \leq 400 - \frac{4}{7}q_b \end{equation*}
\begin{equation*} p_b \geq 0 \implies q_c \leq 1500 - \frac{10}{3}q_b \end{equation*}

The domain is the shaded region graphed below.

A certain manufacturer produces basic and enhanced versions of their product. Let $q_b$ denote the quantity of basic units daily demanded and $q_e$ denote the quantity of enhanced units demanded daily. It is determined that the unit price of the basic units is approximately

\begin{equation*} p_b=10-0.1q_b-0.5q_e\text{,} \end{equation*}

and that the unit price of the enhanced units is approximately

\begin{equation*} p_e=30-0.4q_b-q_e \end{equation*}

dollars, respectively.

1. Determine the daily revenue, $R(q_b,q_e)\text{.}$

$R = -0.1q_b^2-0.9q_bq_c+10q_b-q_c^2+30q_c$
Solution

The price $p_b$ per unit of $q_b$ is given by

\begin{equation*} p_b=10-0.1q_b-0.5q_e\text{,} \end{equation*}

and the price $p_e$ per unit of $q_e$ is given by

\begin{equation*} p_e=30-0.4q_b-q_e\text{.} \end{equation*}

Therefore, the revenue $R$ as a function of $q_b$ and $q_e$ is

\begin{equation*} \begin{split} R(q_b,q_e) \amp = p_bq_b + p_eq_e \\ \amp = \left(10-0.1q_b-0.5q_e\right)q_b+\left(30-0.4q_b-q_e\right)q_e\\ \amp = -0.1q_b^2-0.9q_bq_e+10q_b-q_e^2+30q_e \end{split} \end{equation*}
2. Determine the domain of $R\text{.}$

The region where all of $p_b,p_c,q_b,q_c \geq 0\text{.}$
Solution

The domain of $R$ is constrained by $p_b \geq 0$ and $p_e \geq 0\text{:}$

\begin{equation*} p_b=10-0.1q_b-0.5q_e\geq 0 \implies q_e \leq 20-0.2q_b \ \ \text{ and } \end{equation*}
\begin{equation*} p_e=30-0.4q_b-q_e \geq 0 \implies q_e \leq 30-0.4q_b\text{,} \end{equation*}

such that both $q_b,q_e \geq 0\text{.}$ The domain is the shaded region graphed below.

We can calculate the outstanding principal on a loan at the end of $i$ months by the formula

\begin{equation*} B(A,r,t,i)=A\left(\frac{\left(1+\frac{r}{12}\right)^i-1}{\left(1+\frac{r}{12}\right)^{12t}-1}\right) \ \ \ 0\leq i \leq 12t \end{equation*}

where $A$ is the principal loan, $r$ is the annual interest rate, and $t$ is the amortization period in years. Suppose the original amount borrowed is $100,000, and the interest rate charged by the bank is 3%. What is the amount owed to the bank after 2 years if the loan is to be repaid in equal instalments over 25 years? What is the amount owed after 20 years? Answer$50,148.53, $28,661.05 Solution We use the formula provided with $A=100,000\text{,}$ $r=0.03$ and $t=25\text{.}$ After 2 years ($i=24$ months), we find the amount owed to be approximately$50,148.53. After 20 years, the amount owed is approximiately $28,661.05. In economics, the given optimal quantity $Q$ of goods for a store to order is given by the Wilson lot-size formula: \begin{equation*} Q(C,N,H)=\sqrt{\frac{2CN}{H}} \end{equation*} where $C$ is the cost of placing an order, $N$ is the number of items the store sells per day, and $H$ is the daily holding cost for each item. What is the most economical quantity of winter tires to order if the store pays$25 for placing an order, pays \$2 for holding a tire per day, and expects to sell 60 tires a day?

We use the given formula with $C=25\text{,}$ $N=60$ and $H=2$ to find