An aluminum can is to be constructed to contain 1200 𝑐𝑚3 of liquid. Let 𝑟 and ℎ be the radius of the base and the height of the can respectively.

Question

 An aluminum can is to be constructed to contain 1200  of liquid. Let  and  be the radius of the base and the height of the can respectively.

a) Express h in terms of r. (If needed you can enter  as pi.)

b)Express the surface area of the can in terms of 𝑟

c)Approximate the value of r that will minimize the amount of required material (i.e. the value of r that will minimize the surface area). What is the corresponding value of h?

Answer 1

a) Let's start with the volume of a cylinder:

V = πr²h ,

where r is the radius and h is the height. To express h in terms of r, we simply solve for h. So

h = V/(πr²) ,

and since we know the volume is 1200 cm³, we have

h = 1200/(πr²) .

b) The surface area of a (closed) cylinder is the area of the two end caps (the top and bottom) plus the area of the lateral surface, or

A_surface = A_ends + A_lateral .

The two end caps are circles, so that part of the area is

A_ends = 2πr²

and the lateral surface is found by unraveling the surface into a rectangle whose base is the circumference of the end cap, and whose height is the height of the cylinder, or h. Thus,

A_lateral = 2πr*h , so

A_surface = 2πr² + 2πr*h .

Now, since we found h in terms of r for part a), we can say:

A_surface = 2πr² + 2πr*1200/(πr²)

=πr² + 2400/r²

=πr² + 2400r^-1

c) To minimize our values, we take the derivative of the surface area and set it equal to zero.

A_surface' = 4πr - 2400/r² = 0

And now solve for r. We get

4πr = 2400/r²

4πr³ = 2400

r³ = 2400/4π

r=(2400/4π)^1/3

r = 5.76 cm .

To find h, just plug in this value to the formula we derived for h:

h = 1200/(πr²)

= 1200/(π(5.76)²)

= 11.52 cm .

If you plug those values into our original formula for volume, you'll get 1200 cm³. Hope this helps!

Answer 2

Please see video for explanation.