Answer to Question #155011 in Calculus for Phyroe

Question

Find the most economical dimensions for a closed cylindrical can containing a quart.

Answer 1

It sounds like "most economical dimensions" means you want to use the least amount of material for the can so that it still contains a quart. In other words, you're minimizing the surface area of the can.

Here is not given the volume $V$ of cylinder is $V=\pi \ r^2 \ h=\frac{1}{4} \ \pi \ d^2 \ h=1 quart$ . I applied differentiation like this:

$\frac{1}{4}\pi [d^2\frac{dh}{dd}+2dh]=0;$

$\frac{dh}{dd}=-\frac{2h}{d};$

Total area (closed both ends):

$A_T=2(\frac{1}{4}\pi d^2)+\pi dh$ ;

$A_T=\frac{1}{2}\pi d^2+\pi dh;$

$\frac{dA_T}{dd}=πd+π[d\frac{dh}{dd}+h]=0;$

$dd\frac{dh}{dd}+h=0;$

$d+d(\frac{−2h}{d})+h=0;$

$d=h;$

Result: The diameter of the base of the can is equal to the height of this can.