i need to maximize the volume of this box.?

Okay, let x = the side of those squares cut out at the corners, so that the cardboard can be folded up to the open top box. From this, we can figure out the length and width of the box as follows:

L = 26 - 2x

W = 16 - 2x

The volume is the product of all three, or:

V = x(26 - 2x)(16 - 2x) = 4 x³ - 84 x² + 416 x

To find the maximum, we set dV/dx = 0, or

12 x² -168 x + 416 = 0

The roots are

x = (1/3)(21 - √129) and (1/3)(21 + √129), or

x = 3.214 and 10.786

But only 3.214 is possible as a real solution, as 10.786 will result in "negative lengths".

Call the size of the squares, "A".

The corner cuts are A by A

The sides of the box are then

26 - 2A

A

16-2A

Multiply these 3 terms together, that's the volume.

To find the maximum volume, differentiate and set to zero.

Then solve the quadratic for A.

That will be the point were any change in A will make the volume smaller.

length = 26-x

width = 16-x

height = x

then Volume = (26-x)(16-x)x

multiply out.

Take the first derivative and set it = 0.

Should give you maxima/minima. Test the x values to see.

I don't have a calculator on me, but I believe the squares that you cut out should be (square root of 14) inches on a side, although I've been wrong before.