Question

A cylindrical can is to be constructed to hold a fixed volume of liquid. The cost of the

material used for the top and bottom of the can is 3 cents per square inch, and the cost of the

material used for the curved side is 2 cents per square inch. Use calculus to derive a simple

relationship between the radius and height of the can that is the least costly to construct.

Answer 1

Let r be radius and h, height of the container.

Then;

$V = \pi ×r ^ 2 ×h$

and

$S(r) =(\pi ×r ^ 2) + (2\pi × r × h)$

Then the cost is;

$C(r) = 0.03πr² +0.04 rh = C$

hence;

$h = (C - 0.03\pi ×r ^ 2)/(0.04\pi × r)$

and;

$V = (r(C - 0.03\pi × r ^ 2))/0.04$

The goal is to maximize V, since

$V= 25(C – 0.09²)$

Then,

$V' = 0 \\ r = (C/0.09π)^{1/2}\\ V" (r) = -4.5πr$

Hence,

V has maximum at $r =(\dfrac{C}{0.09 ×\pi })^ {1/2}$

Since;

$h/r = \dfrac{(C - 0.03\pi r ^ 2)}{(0.04\pi r ^ 2)}\\ r = 2/3h$