Question

A cylindrical can, open at the top, is to hold 840 cm3 of liquid. Find the height and radius that minimize the amount of material needed to manufacture the can. Enter your answer with rational exponents, and use pi to represent π.

Answer 1

Answer:

the radius of the base is equal to $x=2\sqrt[3]{\frac{105}{\pi } }$

And the height is equal to:

$y=\frac{840}{\pi(2\sqrt[3]{\frac{105}{\pi}})^{2} }$

Step-by-step explanation:

We write the volume function

$f(x,y)=840=\pi x^{2} y$ where x is the radius of the base and y is the height of the cylinder

$y=\frac{840}{\pi x^{2} }$

The surface of a cylinder is given by

$S(x)=\pi x^{2} +2\pi xy=\pi x^{2} +\frac{1680}{x}$ on the interval from 0 to infinity

We now determine the critical values by differentiating and making the equation equal to 0

$f'(x)=2\pi*x- \frac{1680}{x^{2}} =\frac{2\pi x^{3}-1680}{x^{2} } =0$

This equation have 2 complex solutions and one real solution

$x=2\sqrt[3]{\frac{105}{\pi } }$

For x=0 and infinity the equation goes to infinity also so the radius of the base is equal to $x=2\sqrt[3]{\frac{105}{\pi } }$

And the height is equal to:

$y=\frac{840}{\pi (2\sqrt[3]{\frac{105}{\pi } })^{2} }$

$y=\frac{840}{\pi(2\sqrt[3]{\frac{105}{\pi}})^{2} }$