i need to maximize the volume of this box.?

Question

i start out with a 26 X 16 piece of carboard. equal squares are cut off each corner and the sides are turned up to form an open rectangular box. 

i took calculus a while ago, and i pretty much forgot everything and i wanted to know how to go about this problem.

Scythian1950 · Accepted Answer

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".

morningfoxnorth · Answer

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.

Modus Operandi · Answer

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.

Elwood Blues · Answer

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.

i need to maximize the volume of this box.?

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