Let the side length of the squares that are cut out from the corners be x, then the length of the base of the open box formed is 28 - 2x while the width of the box is 20 - 2x and the height is x.
The volume of a rectangular box is given by length x width x height
Thus,
[tex]Volume = x(28 - 2x)(20 - 2x) \\ \\ =x(560 - 96x + 4x^2)=4x^3-96x^2+560x[/tex]
For maximum volume, the differentiation of the volume with respect to x is 0 and the second derivative test yeilds a negative number,
Thus
[tex] \frac{dV}{dx} =0 \\ \\ \Rightarrow12x^2-192x+560=0 \\ \\ \Rightarrow x\approx12.16\ or\ x\approx3.84[/tex]
[tex]\left. \frac{d^2V}{dx^2} \right|_{x=12.16}=[24x-192]_{x=12.16} \\ \\ =24(12.16)-192=291.84-192 \\ \\ =99.84 \\ \\ \left. \frac{d^2V}{dx^2} \right|_{x=3.84}=[24x-192]_{x=3.84} \\ \\ =24(3.84)-192=92.16-192 \\ \\ =-99.84[/tex]
Since the second derivative is negative when x = 3.84, thus the value x and hence the size of the square which gives the box of largest volume is 3.84 inches.
