Question 298862
First write the equation for the volume of the box.
If you cut squares from each corner where the side of the square is x inches, then the volume of the resulting box when the flaps are folded up can be expressed as:
{{{V = (15-2x)(7-2x)X}}} which simplifies to:
{{{V = 4x^3-44x^2+105x}}} To find the value of x which yields the maximum of volume (V), take the first derivative of this equation and set it equal to zero.
{{{dV/dx = 12x^2-88x+105}}} Set this equal to zero and solve for x;
{{{12x^2-88x+105 = 0}}} Solve by the quadratic formula:{{{x = (-b+-sqrt(b^2-4ac))/2a}}} where a = 12, b = -88, and c = 105. You can do this and you should get:
{{{x = 5.833}}} or {{{x = 1.5}}}
Discard {{{x = 5.833}}} because it would result in a negative value for the length of the bottom of the box: {{{7-2x = 7-11.66}}}={{{-4.66}}}, so...
{{{x = 1.5}}}inches.
The dimensions of the box are:
Length, {{{L = 15-2x}}}={{{15-2(1.5) = 15-3}}}={{{12}}}inches.
Width, {{{W = 7-2x}}}+{{{7-2(1.5) = 7-3}}}={{{4}}}inches.
Height, {{{h = x}}}={{{1.5}}}inches.
The maximum volume of the box would be:
{{{V[max] = L*W*h}}}
{{{V[max] = 12*4*1.5}}}
{{{V[max] = 72}}}cubic inches.