If x represents the side length of the square cut from each corner, then the dimensions of the base of the box are 26-2x by 43-2x. The volume of the box is
... V = x(26-2x)(43-2x)
where x is restricted to the domain 0 ≤ x ≤ 13.
Plotting this equation using a graphing calculator shows the maximum volume to be 2644.7 in³.
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The analytical solution can be found by setting the derivative of V to zero.
... V = 4x³ - 138x² +1118x
... dV/dx = 0 = 12x² -276x +1118
This has solutions given by the quadratic formula, where a=12, b=-276, c=1118:
... x = (-b±√(b²-4ac))/(2a)
The plus sign will give a solution that is not in the allowed domain of x, so the only viable solution is the one where the radical is subtracted.
... x = (276 - √(76176--53664))/24 = (23/2) - √(469/12) ≈ 5.2483 . . . inches
Then the corresponding maximum volume is found from the equation for V.
... V ≈ (5.2483)(26 - 2·5.2483)(43 - 2·5.2483) ≈ 2644.6877 in³ . . . . as above