Mathematics 133 – Intensive Calculus for Science 1 Solutions for Discussion 7 November 17, 2005

A) Two non-negative numbers x, y add up to 72. What is the largest possible value of the product P = xy?

Solution. We have x + y = 72, so y = 72 − x. Then P (x) = x(72 − x) = 72x − x^{2}. To maximize, we first compute P′(x) and set equal to 0 to find critical points:

P′(x) = 72 − 2x = 0 ⇒ x = 36.

Then P′′(x) = −2 < 0, so the second derivative test tells us that x = 36 is a local maximum. It must also be the global maximum because this is the only critical point. (Alternate method: Note that 0 ≤ x ≤ 72 since it is given that x, y ≥ 0. Then P (0) = 0, P (36) = 36^{2} = 1296 and P (72) = 0. So we can see this way that x = 36 gives the global maximum.) The maximum value of the product is P (36) = 36^{2} = 1296.

B) A rectangular beam is cut from a cylindrical log of radius 30 cm. The strength of a beam of width w and height h is proportional to wh^{2}. Find the width and the height of the beam of maximum strength.

Solution. The cross-section of the log is a circle of radius 30, and the cross-section of the beam is a rectangle inscribed in this circle. So the Pythagorean Theorem shows w^{2} + h^{2} = 60^{2} = 3600. Then it is easiest to solve this for h^{2} in terms of w and to write the strength as a function of w: S = kw(3600 − w^{2}) = k(3600w − w^{3}), where k is the constant of proportionality. To maximize, we proceed as before:

S′ = k(3600 − 3w^{2}) = 0 ⇒ w^{2} = 1200,

so w = \sqrt{1200} = 20\sqrt{3} ≐ 34.64. This is a maximum for the strength because S′′(w) = −6kw. k is a positive constant and our value w ≐ 34.64 > 0, so S′′(34.64) < 0. Since this is the only critical point for w ≥ 0, we have a local and global maximum. To find the corresponding h, we substitute into the equation w^{2} + h^{2} = 60^{2} = 3600: 1200 + h^{2} = 3600, so h^{2} = 2400 and h = \sqrt{2400} = 20\sqrt{6} ≐ 48.99. The dimensions are

w ≐ 34.64  and  h ≐ 48.99.

C) Northern South Dakota State Agricultural and Veterinary Junior College is building a new running track for their prize-winning track team – the “Flying Farmers”. The track is to be the perimeter of a region obtained by putting two semicircles on opposite ends of a rectangle, and that perimeter should be ¡EM¿440 yards¡/EM¿ in length. Due to financial constraints, the administration has decided to grow sweet corn in the area enclosed by the track and sell it to the local grocery store to make some extra income. Determine the dimensions to build the track in order to maximize the area for growing corn.

Solution: Let x be the length of the straight segments of the track, and r be the radius of the semicircular portions. Then the length of the track is the (outside) perimeter:

440 = 2x + 2πr ⇒ 2x = 440 − 2πr.

The area inside the track is

A = x(2r) + πr^{2} = r(440 − 2πr) + πr^{2} = 440r − 2πr^{2} + πr^{2} = 440r − πr^{2}.

Then A′(r) = 440 − 2πr = 0 when r = 220/π. This is a local and global maximum for A(r) since A′′(r) = −2π < 0. The optimum track has r = 220/π and x = 220 − π(220/π) = 0 (!). In other words the optimum track is a circle.