Maximizing Volume

Question

An open box is to be made out of a 8-inch by 20-inch piece of cardboard by cutting out squares of equal size from the four corners and bending up the sides. Find the dimensions of the resulting box that has the largest volume.

Answer 1

So firstly, we need to find a function for the volume.

So if we have a piece of cardboard of length l and width w (which are both constant), after cutting out open squares of sidelength x (greater than 0 and less than min(l/2, w/2)) from the boxes and bending it upward, what is the new length, width, and now height of the box?

--It should be l-2x for the length, w-2x for the width, and x for the height. So the volume as a function of x is V(x)=(l-2x)(w-2x)x. Now, how do we find the maximum of this function?

--If you recall, you can find a maximum value of the function V by finding the derivative V' and finding where it (1) V' is zero, (2) where there is a sign change V', and (3) where the sign change is from positive to negative. This should make sense because when the derivative is zero, the tangent slope is zero. If there is a sign change, it is a local maximum/minimum, and if the V' is positive before the point and after the point, this means V was increasing before the point and decreasing after the point, hence it being a maximum point.

So that should give you x, then you can use it to find the dimensions of the box. I hope this was helpful!