Question

A sports memorabilia collector needs to create a display case to show her baseball cards

Answer 1

Answer:

For a box of length L, width W, and height H, the volume is:

V = H*L*W

in this case, we have a piece of cardboard of dimensions 40in by 60 in.

If we cut squares of side length X in the corners and we fold the corners, we will have a height of X.

And we can define:

Length = 60in - X

Width = 40 in - X

Then the volume of the box will be:

V = X*(60in - X)*(40in  - X)

V = X^3 - 100in*X^2 + 2400in^2*X

We want to maximize this, then we need to find the zero of the first derivative.

We have that:

V' = 3*X^2 - 2*100in*X + 2400in^2

V' = 3*X^2 - 200in*X + 2400in^2

This is the equation we will use to maximize the volume of the box, but let's not end here, let's complete the problem and find the value of X and the maximum volume.

We want to solve:

V' = 0  = 3*X^2 - 200in*X + 2400in^2

For X.

Then we can use Bhaskara's formula, we will get that the solutions are:

$X = \frac{-(-200in) +-\sqrt{(-200in)^2 - 4*3*(2400in)} }{2*3} = \frac{200in +- 105.8in}{6}$

Then we have two possible solutions for X.

X = (200in + 105.8in)/6 = 50.97 in

But this is larger than the side of 40 in, so we can not cut this much, we can discard this option.

The other one is:

X = (200in - 105.8in)/3 = 15.7in

This seems correct.

Then the maximum volume of the box is:

V = 15.7in*(60in - 15.7in)*(40in - 15.7in) = 16,900.893 in^3