Thomas wants to make a box with no lid out of cardboard sheet to use as a food dish for his dog. He plans to make the box shape by cutting four congruent squares from all corners of the sheet.

1.) Thomas buys a cardboard sheet that is 12 by 8 inches. Create an equation for this situation, find the zeros, and sketch the function.

2.) What is the size of the cutout he needs to make so that he can fit the maximum amount of food in the box?

3.) If Thomas wants a volume of 12 cubic inches, what size does the cutout need to be? What would be the dimensions of this box?

4.) Using complete sentences, explain the connection between the cutout and the volume of the box.

5.) Design an equation that would work for any cardboard sheet length, q, and width, p.

1. If we let x as the side of the square cut-out, the formula for the capacity (volume) of the food dish is:

V = (12 - 2x) (8 - 2x) (x)

V = 96x - 40x^2 + 4x^3

To find the zeros, we equate the equation to 0, so, the values of x that would result to zero would be:

x = 0, 6, 4

2. To get the value of x to obtain the maximum capacity, we differentiate the equation, equate it to zero, and solve for x.

dV/dx = 96 - 80x + 12x^2 = 0

x = 5.10, 1.57

The value of x that would give the maximum capacity is x = 1.57

3. If the volume of the box is 12, then the value of x can be solved using:

12 = 96x - 40x^2 + 4x^3

x = 0.13, 6.22, 3.65

The permissible value of x is 0.13 and 3.65

4. Increasing the cutout of the box increases the volume until its dimension reaches 1.57. After that, the value of the volume decreases it reaches 4.

5. V = (q - 2x) (p - 2x) (x)