1. Finding maximum volume: We need to design a cylindrical can with radius r and

Question

1. Finding maximum volume: We need to design a cylindrical can with radius r and height h. The

top and bottom must be made of copper, which will cost 2 cents per square inch. The curved

side is to be made of aluminum, which will cost 1 cent per square inch. We seek the dimensions

that will maximize the volume of the can. The only constraint is that the total cost of the can is to

be 300? cents.

2. Finding minimum cost: The cost of fuel to propel a boat through the water (in dollars per hour)

is proportional to the cube of the speed. A certain ferry boat uses $100 worth of fuel per hour

when cruising at 10 miles per hour. Apart from fuel, the cost of running this ferry (labor,

maintenance and so on) is $675 per hour. At what speed should it travel so as to minimize the

cost per mile traveled?

Explanation / Answer

1.

cost of top and bottom Ce = (r^2 * pi * 2) * 2 ..........(1)
cost of side Cs = h * 2*r * pi * 1 ...........................(2)

Total cost 300pi = Ce + Cs ................................(3)

Volume V = r^2 * pi * h .......................................(4)

Take 1 & 2 and substitute into 3, then solve for h.
300pi = (r^2 * 4 + h * 2*r) * pi
150 = r^2 * 2 + h * r
150 - r^2 * 2 = h * r
150/r - r*2 = h ..........................................(5)

Substitute 5 into 4
V = r ^ 2 * pi * (150/r - r*2)
V = (150 * r - r^3 * 2) * pi

Now maximize V with respect to r.
dV/dr = (150 - 3r^2 * 2) * pi

Find when dV/dr = 0 using quadradic equation.
a = -6pi
b = 0
c = 150pi
r = +/- sqrt(-4* (-6pi) * (150*pi)) / (2 * (-6pi))
r = +/- sqrt(3600*pi^2) / (-12pi)
r = +/- 60*pi / (-12pi)
r = -/+ 5

r = -5 is kind of a problem, so scratch that and say r = +5.

Use (5) to solve for h
h = 150/5 - 5*2
h = 20

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