A rectangular box of given volume is to be "a" times long as it is wide. find the dimension for which

Question

it has the least total surface area ?
(calcuus)
optimization problem
plz give me the exxact answer in detail

Answer 1

I sure hope I'm not just doing your homework for you. Well anyway, here goes the solution.

Let v be the volume of the box.
Let s be the width of the box.

Then:
Width = s
Length = as
Height = v / (a s^2)

To find the area, you need to find the area of each of the six sides of the box, and add all those areas together. The sides of the box are in identical pairs. (Front has same area as back. Top has same area as bottom. etc.)

Total Area = 2as^2 + 2v/s + 2v/(as)

Now we need to find the value of s that will minimize the area of the box. To do that, we take the derivative of the area expression, and then find the value of s that makes the derivative equal to 0. That value of s is either a minimum, or a maximum, or a point of inflection. In this case, it will be a minimum. I will leave it to you to understand why that is so. (Try drawing a graph. Also try examining the second derivative of the function.)

The derivative of the area is:
4as - 2v/(s^2) - 2v/(as^2)

So form this equation:
0 = 4as - 2v/(s^2) - 2v/(as^2)

Now solve that equation for s. You will get:
s = cube_root( v/a + v/(a^2) )

That's the width of the box with least surface area. (You can easily find the height and depth from this.)

The answer is given as a formula involving v and s, because the problem didn't give us a specific value for v and s.