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

Solution;

$Volume=πr^2h$

Cost of material;

$C=3(2πr^2)+2(2πrh)=6πr^2+4πrh$

$But ;$

$h=\frac{V}{πr^2}$

Substituting into the cost;

$C=6πr^2+4πr(\frac{V}{πr^2})$

$C=6πr^2+4\frac Vr$

Differentiate cost w.r.t r;

$\frac{dC}{dr}=12πr-\frac{4V}{r^2}=0$

We have ;

$r^3=\frac{4V}{12π}=\frac{V}{3π}$

$r=(\frac{V}{3π})^{\frac13}$

While;

$h=\frac{V}{πr^2}=\frac{V}{π(\frac{V}{3π})^{\frac13}}$

Hence;

$h=2.08(\frac{V}π)^{\frac13}=1.42V^{\frac13}$