3.10 Optimization (continued)

Optimization on a Restricted Domain

Consider a heated circular plate of radius 1 meter with location on the plate described relative to the center. The domain $R=\{(x,y):x^2+y^2 \leq 1\}$ of our temperature function is a closed and bounded region, as shown in Figure.

Suppose the temperature $T$ at each point on the circular plate $x^2+y^2 \leq 1$ is given by

. We would like to find the absolute maximum and absolute minimum temperature of the circular plate.

Because the domain of $T$ is closed and bounded, the Extreme Value Theorem guarantees that $T$ has an absolute maximum and minimum on the plate. The graph of $T$ over its domain $R$ is shown in Figure. We will find the hottest and coldest points on the plate.

If the absolute maximum or absolute minimum occurs inside the disk, then the absolute extreme point will be at a critical point. We begin by looking for critical points inside the disk. To do this, notice that critical points are given by the conditions $T_x= 4x=0$ and $T_y=2y - 1=0$. This means that there is one critical point of the function at the point $(x_0,y_0) =(0,1/2)$, which does lie inside the disk.

We now find the hottest and coldest points on the boundary of the disk, which is the circle of radius 1. To turn this into a single-variable calculus problem, we use the familiar parameterization of the unit circle as

where $0\leq t \leq 2\pi$. The temperature at a point on the circle is then described by

.

To find the hottest and coldest points on the boundary, we look for the critical points of this single-variable function on the interval $0\leq t\leq 2\pi$. We have

Therefore, we need to identify the points where $\cos(t) (-2\sin(t) - 1) =0$. These happen when $\cos(t) = 0$ or $\sin(t) = -1/2$ on the interval $0\leq t\leq 2\pi$. This occurs when $t=\pi/2$, $3\pi/2$, $7\pi/6$, and $11\pi/6$. Since $x(t) = \cos(t)$ and $y(t) = \sin(t)$, the corresponding points are -$(x,y) = (0,1)$ when $t = \frac{\pi}{2}$-$(x,y) = \left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$ when $t = \frac{11\pi}{6}$-$(x,y) = (0,-1)$ when $t = \frac{3\pi}{2}$-$(x,y) = \left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$ when $t = \frac{7\pi}{6}$ These are the critical points of $T$ on the boundary which means that this collection of points includes the hottest and coldest points on the boundary.

We now have a list of candidates for the hottest and coldest points: the critical point in the interior of the disk and the critical points on the boundary. We find the hottest and coldest points by evaluating the temperature at each of these points, and find that -$T\left(0,\frac{1}{2}\right) = -\frac{1}{4}$-$T\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right) = \frac{9}{4}$-$T\left(0,1\right) = 0$-$T\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right) = \frac{9}{4}$-$T\left(0,-1\right) = 2$ Hence, the maximum temperature on the disk $x^2+y^2\leq 1$ is $\frac{9}{4}$, which occurs at the two points $\left(\pm\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$ on the boundary, and the minimum value of $T$ on the disk is $-\frac{1}{4}$ which occurs at the critical point $\left(0,\frac{1}{2}\right)$ in the interior of $R$.

Find all of the critical points of $f$ in $R$.

Parameterize the edge of $R$ that is on the $x$-axis and find the critical points of $f$ on that edge.

Parameterize the edge of $R$ that is on the $y$-axis and find the critical points of $f$ on that edge.

Parameterize the diagonal edge of $R$ and find the critical points of $f$ on that edge.

Find the absolute maximum and absolute minimum values of $f$ on $R$. Write a couple of sentences to describe how the surface plot of $f$ shown below illustrates your results.

If a continuous function $f$ of a single variable has two critical numbers $c_1$ and $c_2$ at which $f$ has relative maximum values, then $f$ must have another critical number $c_3$, because “it is impossible to have two mountains without some sort of valley in between. The other critical point can be a saddle point (a pass between the mountains) or a local minimum (a true valley).” (From Calculus in Vector Spaces by Lawrence J. Corwin and Robert H. Szczarb.) Consider the function $f$ defined by $f(x,y) = 4x^2e^y -2x^4 -e^{4y}$. (From Ira Rosenholz in the Problems Section of the Mathematics Magazine, Vol. 60 NO. 1, February 1987.) Show that $f$ has exactly two critical points, and that $f$ has relative maximum values at each of these critical points. Explain how this function $f$ illustrates that it really is possible to have two mountains without some sort of valley in between. Use appropriate technology to draw the surface defined by $f$ to see graphically how this happens.

If a continuous function $f$ of a single variable has exactly one critical number with a relative maximum at that critical point, then the value of $f$ at that critical point is an absolute maximum. In this exercise we see that the same is not always true for functions of two variables. Let $f(x,y) = 3xe^y-x^3-e^{3y}$ (from “The Only Critical Point in Town” Test by Ira Rosenholz and Lowell Smylie in the Mathematics Magazine, VOL 58 NO 3 May 1985.). Show that $f$ has exactly one critical point, has a relative maximum value at that critical point, but that $f$ has no absolute maximum value. Use appropriate technology to draw the surface defined by $f$ to see graphically how this happens.