Armon specializes in framed art. for a specific design of framed art, the width of the frame is 10 inches longer than the width of the art inside the frame, and the length of the frame is twice as long as the frame's width. if x represents the length of the art inside the frame, what function could be used to model the area of the entire design, frame and art included?

The answer is: f(x) = 2x^2 + 40x + 200

Explanation:

Given: x is the width of the Artwork

Let W be the Width of the Frame

Let L be the Length of the Frame

Width = x + 10

Length = Width * 2, so by substitution, Length = 2(x + 10)

Area = Width * Length

By substitution:

Area = (x + 10) * 2(x + 10)

f(x) = (x + 10) * (2x + 20)

f(x) = 2x^2 + 20x + 20x + 200

f(x) = 2x^2 + 40x + 200

To test this, if a picture 8" wide is put in this frame, the width would be 8+10 or 18 inches. The length would be twice the width, or 36 inches. The Area is 18 x 36 = 648.

Using the formula:

f(8) = (2 * 8^2) + (40 * 8) + 200

f(8) = (2 * 64) + 320 + 200

f(8) = 128 + 320 + 200

f(8) = 648

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