My initial observations:

If the sides are 25 by 25 then the largest cut out could be no less than 12.

The length and width of the box is 25-2x if the size of the cut out is x.

The volume is v= (25-2x) (25-2x) (x)

I can solve this by graphing the function and finding the maximum volume.

I could also solve this by looking at a spreadsheet we can see that the maximum volume of approximately 1156 cubic units with a cut out of about 4 units.

AM-GM Inequality

First attempt gave me this mess:

[x(25-2x)(25-2x)]^(1/3) > (25-2x + 25-2x + x) /3

x(25-2x)(25-2x) >[ (50 -3x)/3]^3

4x^3 - 100x^2 +625x >-x^3+50x^2-(2500/3)x + 125000/27

5x^3-150x^2 + (4375/3)x- 12500/27> 0

[5(3x-40)(3x-25)^2]/27 >0

25/3, 40/3

My second attempt made much more sense:

The sum of the three expressions, x, 25-2x, 25-2x is 50-3x

I need to get three expressions to be equal, so I can use the form

V = k ( x ( ? - x) (? - x))

If k = 12.5 then the function can be expressed as

V = 12.5 [x(6.25 - .5x) (6.25 - .5x)]

So then x+ 6.25 - .5x + 6.25 - .5x = 12.5

By dividing by 12.5 by 3 I have the x value of 4.16666666.

Then V(x) = 1157.407046 cubic units

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