# An open-topped box is being made from a piece of cardboard measuring 12 in. by 30 in. The sides of the box are formed when four congruent squares are cut from the corners, as shown in the diagram. The base of the box has an area of 208 sq. in. Create an equation that represents the surface area of the base of the box, then determine the side length, x, of the square cut from each comer.

An equation describing the area of the base of the box is (12-2x) (30-2x)=208, and the cut was 2 inches.

We are given a rectangular piece of material measuring 12 inches by 30 inches. Identical square cuts are made at the corners, and the resulting flaps are folded up to make an open-top box. The base of the box has an area of 208 square inches.

(See attachment.)

The width of the bottom of the box is 12-2x inches. Notice that we removed x inches from both ends. Similarly, the length is 30-2x inches. The area is found by taking the product of the length and the width, so `A=(12-2x)(30-2x)`.

We know the value of A, since we are given the area of the base to be 208 square inches. Substituting, we get

`(12-2x)(30-2x)=208`.

Multiplying the binomials on the left using the distributive property, we get

`360-24x-60x+4x^2=208`.

Combining like terms and moving all terms to the left side, we get

`4x^2-84x+152=0`.

Factor out the common factor of 4:

`4(x^2-21x-38)=0`.

By the zero-product property, the quadratic must be zero. We can solve the quadratic by factoring (if it factors over the rationals) or using completing the square or the quadratic formula.

`4(x-19)(x-2)=0` ; it factors. Again applying the zero-product property (if ab=0, then a=0, b=0, or both a and b are zero), we see that either x=19 or x=2.

`x ne 19`, since for this application, we would be removing more material than we started with, so x=2.

Checking: 12-2(2)=8, 30-2(2)=26, and 8(26)=208, as required.

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