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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