Question

A closed-top cylindrical container is to have a volume of 250 in2. 250 , in squared , . What dimensions (radius and height) will minimize the surface area?

Answer 1

Answer:

radius r = 3.414 in

height h = 6.8275 in

Step-by-step explanation:

From the information given:

The volume V of a closed cylindrical container with its surface area can be expressed as follows:

$V = \pi r^2 h$

$S = 2 \pi rh + 2 \pi r^2$

Given that Volume V = 250 in²

Then;

$\pi r^2h = 250 \\ \\ h = \dfrac{250}{\pi r^2}$

We also know that the cylinder contains top and bottom circle and the area is equal to πr²,

Hence, if we incorporate these areas in the total area of the cylinder.

Then;

$S = 2\pi r h + 2 \pi r ^2$

$S = 2\pi r (\dfrac{250}{\pi r^2}) + 2 \pi r ^2$

$S = \dfrac{500}{r} + 2 \pi r ^2$

To find the minimum by determining the radius at which the surface by using the first-order derivative.

$S' = 0$

$- \dfrac{500}{r^2} + 4 \pi r = 0$

$r^3 = \dfrac{500 }{4 \pi}$

$r^3 = 39.789$

$r =\sqrt[3]{39.789}$

r = 3.414 in

Using the second-order derivative of S to determine the area is maximum or minimum at the radius, we have:

$S'' = - \dfrac{500(-2)}{r^3}+ 4 \pi$

$S'' = \dfrac{1000}{r^3}+ 4 \pi$

Thus, the minimum surface area will be used because the second-derivative shows that the area function is higher than zero.

Thus, from $h = \dfrac{250}{\pi r^2}$

$h = \dfrac{250}{\pi (3.414) ^2}$

h = 6.8275 in