Answer:
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:
[tex]V=(64-16x-16x+4x^{2} )x\\\\V=(64-32x+4x^{2} )x\\\\V=64x-32x^{2} +4x^{3}[/tex]
Once we got our equation, we can now take it's derivative so we can optimize the problem:
[tex]V=4x^{3}-32x^{2} +64 x\\\\V'=12x^{2} -64x+64[/tex]
We can now set it equal to zero to find the value of x we need.
[tex]12x^{2} -64x+64=0[/tex]
This can be simplified if we factored a common 4, so we get:
[tex]4(3x^{2} -16x+16)=0\\\\3x^{2} -16x+16=0[/tex]
So we can solve it by factoring, when factoring the equation we get:
[tex](x-4)(3x-4)=0[/tex]
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 [tex]37.93in^{3}[/tex]
