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Anyone willing to help me with an optimization problem?

Here is the problem:

A box has a square base of side x and height y. Find the dimensions for which the surface area is 20 and the volume is as large as possible.

When I solved the problem, I came up with one of the sides equaling the square root of 10/3, but once I tried it in the first derivative test, it fails. Can someone help me out, my book is bad and I don't know where else but here to look.

Public Comments

1. If the box has a top your answer is correct. It would be in the shape of a cube (you will have to derive this). And you would have:
   
   
   
   Surface area = 6x² = 20
   
   x² = 20/6 = 10/3
   
   x = √(10/3)
   
   
   
   However, if the box does NOT have a top you will get a different answer. And from reading what you wrote I suspect this is the case. You would have:
   
   
   
   x = length base
   
   x = width base
   
   y = height
   
   S = surface area
   
   V = volume
   
   
   
   V = x²y
   
   S = x² + 4xy = 20
   
   
   
   Try working with this and see what you get.
2. Vol = x²y
   
   Area = 2x² + 4xy = 20
   
   
   
   Solve for y in the area equation. Substitute it in the first equation, then do dV/dx = 0.
   
   
   
   If you do it correctly, you will get x = 1.8257
   
   
   
   Plug it into the area equation to get y, which should also equal 1.8257, because the minimum area for a maximum volume is a cube, as you would expect.