A company is constructing an open-top, square-based, rectangular metal tank that will have a volume of 71.5 ft3 What dimensions yield the minimum surface area? Round to the nearest tenth

Question

A company is constructing an open-top, square-based, rectangular metal tank that will have a volume of 71.5 ft³ What dimensions yield the minimum surface area? Round to the nearest tenth, if necessary.

12 ft × 12 ft × 0.5 ft

6 ft × 6 ft × 2 ft

4.2 ft × 4.2 ft × 4.2 ft

5.2 ft × 5.2 ft × 2.6 ft

Answer 1

The hardest part with this one (as is often the case) is the set-up. Once we get our two equations constructed, the rest of the problem (the part that actually requires calculus) is fairly simple. With that in mind...

We'll let x = the side length of the base and y = the container's height.

Surface area equation:

bottom + sides = area

x^2 + 4xy = A

Volume equation:

l * w * h = V

x^2*y = 71.5

And since we are concerned with minimizing surface area, it's going to be useful to get that in terms of one variable. So let's use the volume equation to do that.

y = 71.5/x^2

Substituting into the surface area equation:

x^2 + 4x*71.5/x^2 = A

Simplifying:

x^2 + 286/x = A

Great! Now for the calculus part. Since a max or min occurs when the derivative equals zero, let's first differentiate our equation:

2x - 286/x^2 = A'

Then we set this equal to zero, and solve.

2x - 286/x^2 = 0

(2x^3 - 286)/x^2 = 0

2x^3 - 286 = 0

x^3 = 143

x = 5.229

And from there we substitute this value for x in our volume equation, which gives us y.

5.229^2*y = 71.5

y = 2.615

Which, after rounding, gives us:

5.2 ft × 5.2 ft × 2.6 ft.

I hope that helped!