An open box is to be made from a square piece of cardboard whose sides are 8 inches long, by cutting squares of equal size from the corners and bending up the sides. determine the size of the square that is to be cut out so that the volume may be a maximum.

The squares that are to be cut out must have dimensions of 4/3in x 4/3in to maximize the volume of the box.

Step-by-step explanation:

In order to solve this problem we must first draw a diagram that will represent the situation. (Diagram attached)

As you may see, the box should have a width of (8-2x) a length of (8-2x) and a height of x.

I determined the width and the length to be of 8-2x because notice that for each side of the cardboard I must subtract two little squares and each little square will have a side of x.

So now that we determined these dimensions, we can now find an equation for the volume of the box:

V=(8-2x)(8-2x)x

so now we can simplify it:

Once we got our equation, we can now take it's derivative so we can optimize the problem:

We can now set it equal to zero to find the value of x we need.

This can be simplified if we factored a common 4, so we get:

So we can solve it by factoring, when factoring the equation we get:

this equation yields two answers:

x=4 and x=4/3

the first answer cannot be our answer because if x was 4, that means that the length and width of the box would both be zero, which gives us the smallest possible box.

So the answer is x=4/3

If we cut squares whose sides are 4/3inx4/3in, our box will have a maximum volumne of

the angle that is vertical to ABI is EBH