Mathematics 133 – Intensive Calculus for Science 1 Solutions for Discussion 6 November 29, 2001

A. Calling the length of the straight sides of the track x and the sides where the semicircles are attached y, we get

\begin{matrix}A & = xy + π(y/2)^{2} \\ 440 = 2x + 2πy \end{matrix}

We can solve the second equation for x as a function of y:

(1)  x = 220 − πy/2,

and substitute into A to get

A(y) = 220y − πy^{2}/2 + πy^{2}/4 = 220y − πy^{2}/4

Now we are ready to apply our optimization procedures. First compute the derivative:

A′(y) = 220 − πy/2

so there is one critical point:

\begin{aligned}0 & = 220 − πy/2 \\ πy/2 & = 220 \\ y & = 440/π \end{aligned}

By the second derivative test, A′′(y) = −π/2 < 0 for all y so this is a local maximum.

The problem asks for the dimensions of the track with maximum area, so we substitute back into (1) to find

x = 220 − (π/2) · (440/π) = 0

So the maximum area track is actually a circle (the straight sides have length zero(!)

B. Let the two numbers be x, y. We have x + y = 72 and we want to maximize P = xy. Solving for y in terms of x, y = 72 − x, so P (x) = x(72 − x) = 72x − x^{2}. Then P′(x) = 72 − 2x, so there is one critical point: 72 − 2x = 0 shows x = 36. This is a maximum by the Second Derivative Test, since P′′(x) = −2 < 0 for all x. Then y = 72 − 36 = 36 too, and the maximum product is 36 · 36 = 1296.

C. If you use x inches of the wire to make the circle and 10 − x inches to make the square, then the circle has circumference x, and the square has perimeter 10 − x. This shows that the radius of the circle is r = x/(2π), and the side of the square is (10 − x)/4. Hence the total area (circle plus square) is

A(x) = π · (x/(2π))^{2} + ((10 − x)/4)^{2} = \frac{x^{2}}{4π} + \frac{(10 − x)^{2}}{16}

Then

A′(x) = \frac{x}{2π} − \frac{10 − x}{8}

There is one critical point:

\begin{aligned}0 & = \frac{x}{2π} − \frac{10 − x}{8} \\ \frac{10 − x}{8} & = \frac{x}{2π} \\ 20π − 2πx & = 8x \\ 10π & = (π + 4)x \\ \frac{10π}{π + 4} & = x \end{aligned}

But the second derivative test shows that this critical point is actually a local minimum: A′′(x) = \frac{1}{2π} + \frac{1}{8} > 0. To find the global maximum, we need to consider the values of A(x) at the endpoints of the range of possible values for x: A(0) = 100/16 and A(10) = 100/(4π). We see that A(10) is larger, so the maximum area is obtained if we use all of the wire to make the circle(!).

D. The volume of the can is 30 = πr^{2}h. The surface area (top, bottom, and side wall) is

SA = 2πr^{2} + 2πrh

Solve for h from the volume and substitute:

SA(r) = 2πr^{2} + \frac{60}{r}

Then

SA′(r) = 4πr − \frac{60}{r^{2}}

There is one critical point r = ^{3} \sqrt{1}5/π ≐ 1.68. The corresponding value of h is h = 30/(π(1.68)^{2}) ≐ 3.38. This is a minimum by the second derivative test:

SA′′(r) = 4π + \frac{120}{r^{3}}

is positive at r ≐ 1.68 (positive, in fact, for all r > 0.)