Squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring dimensions LxW, where L and W are fixed constants.The resulting piece of cardboard is then folded into a box without a lid. Find the dimensions of the box with the largest volume that can be formed in this way.
a.) what is the equation we are trying to optimize? what is the domain? draw a picture of the problem.
b.) optimize the function and answer the following questions:
a.)suppose l, the length of the cardboard piece, is 24" w is the same.what are the optimal dimensions(length, width, height) of the box with the maximum volume?
b.) what is the volume of this box?....
-----is the attached picture correct?---------

a) Consider reatangle with length L and width W. Squares with sides of length x are cut out of each corner of a rectangle. Then the length of inner rectangle (see attached diagram) is L-2x and the width is W-2x.

The resulting piece of cardboard is then folded into a box without a lid. The height of this box is x and the volume is

[tex]V=X\cdot (L-2x)\cdot (W-2x).[/tex]

Note that x must be positive [tex]<\dfrac{L}{2}\ and\ <\dfrac{W}{2}.[/tex]

b) If L=W=24 units, then

[tex]V(x)=X\cdot (24-2x)\cdot (24-2x)=2x(12-x)^2=4x(144-24x+x^2)=576x-96x^2+4x^3.[/tex]

Find the derivative:

[tex]V'(x)=576-96\cdot 2x+4\cdot 3x^2=576-192x+12x^2=12(48-16x+x^2).[/tex]

When [tex]V'(x)=0,[/tex] you get

[tex]x^2-16x+48=0,\\ \\D=(-16)^2-4\cdot 48=256-192=64,\\ \\x_{1,2}=\dfrac{16\pm 8}{2}=12,\ 4.[/tex]

Since [tex]12=\dfrac{24}{2}\ \left(\dfrac{L}{2}\right),[/tex] this solution is extra.

When x<4, V'(x)>0 (volume is increasing) and when x>4, then V'(x)<0 (volume is decreasing). This means that x=4 is point of maximum and maximal value for volume is

[tex]V(4)=576\cdot 4-96\cdot 4^2+4\cdot 4^3=2304-1536+256=1024\ un^3.[/tex]

Optimal dimensions of the box are 16 units, 16 units, 4 units.