Question

A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. the total volume of the solid is 2 cubic centimeters. find the radius of the cylinder that produces the minimum surface area. (round your answer to three decimal places.)

Answer 1

When you cut a sphere across its central axis, you create two equal hemispheres. When you place them to both ends of a right cylinder, it would look like that shown in the picture. The total volume is equal to the volume of the cylinder + the sphere. You equate this to 2 cm^3. On the other hand, the surface area of the solid is the sum of the surface area of the sphere and cylinder. The equations for the volume and surface area are

Volume = 4/3*π*r^3 + π*r^2*h = 2
Surface area = 4π*r^2 + 2πrh

First, you equate h in terms of r using the formula for volume:

$2= \frac{4}{3} \pi r^{3} + \pi r^{2} h$
$\pi r^{2} h=2- \frac{4}{3} \pi r^{3}$
$h= \frac{2- \frac{4}{3} \pi r^{2} }{ \pi r^{2} }$

Then, you substitute the expression for h in the formula for the surface area (SA):

$SA=4 \pi r^{2} +2 \pi r( \frac{2- \frac{4}{3} \pi r^{3} }{ \pi r^{2} } )$
$SA = 4 \pi r^{2} + \frac{4- \frac{8}{3} \pi r^{3} }{r}$

Then, you differentiate SA in terms of r and equate it to zero to find the value of r in order to get the maximum SA. Use the rules of differentiation in this part.

$\frac{d(SA)}{dr} =0=8 \pi r+ \frac{r(-8 \pi r^{2} -(4- \frac{8}{3} \pi r^{3}(1) }{ r^{2} }$

$8 \pi r + \frac{-8 \pi r^{3}-4+ \frac{8}{3} \pi r^{3} }{ r^{2} } =0$

This part involves iteration. That is a very tedious solution to be discussed here. Thus, I suggest you just use the scientific calculator to solve for r. From my calculations, r = 0.7816 cm. Then, you will know the value of h from previous derivation.

Thus, the answer is 0.7816 cm.