Applications of Quadratic Equations

Question

A rectangular canvas picture measures 8 inches by 12 inches. The canvas is mounted inside a frame of uniform width, increasing the total area covered by both canvas and frame to 165 square inches. Find the uniform width of the frame.

Answer 1

The uniform width u of the frame is found by taking the original dimensions of length 8 and width 12 and increasing them on each side by the same distance u, giving us a complete dimension of width (8+2u) and length (12+2u). Once we multiply these two dimensions (8 + 2u) (12 + 2u), using distribution, we're left with 96 +24u +16u + 4u^2, which we can write in standard form 4u^2 +40u + 96. We can set that equal to 165 and then subtract 165 from both sides to get 4u^2 + 40u + 96 - 165 = 0, which simplifies to 4u^2 + 40u - 69 = 0. There is no easy factor, so we use the quadratic formula. Applying the quadratic formula, we get (-40 +/- sqrt(40)^2 - 4(4)(-69)))/(2(4)). We can then can simplify to (-40 +/- sqrt(1600 - (16)(-69)))/(8). Again, we can simplify to (-40 +/- sqrt(1600 + 1104))/(8). Still further, (-40 +/- sqrt(2704))/(8). And then (-40 +/- sqrt(2 * 1352))/(8). Still more, (-40 +/- sqrt(2 * 2 * 676))/8, and again (-40 +/- 2sqrt(676))/(8), and further (-40 +/- 2sqrt(2 * 338))/(8), and again (-40 +/- 2sqrt(2 * 2 * 169))/(8), and again (-40 +/- 4sqrt(169))/(8), and finally (-40 +/- 4(13))/(8) = (-40 +/ 52)/(8). Simplifying even more, we're left with (-5 +/- 6.5). This leaves us with u being equal either to -5 - 6.5 or -11.5, which is a negative number and irrelevant when looking at dimensions of a frame, or -5 + 6.5, which is 1.5 and the value of u. Put more simply, we thought about the width being the same around our picture, created a quadratic equation, applied the quadratic formula, and found the positive option. The above steps are thorough and can be done without a calculator. So, our frame is 1.5 inches wide:)

Answer 2

The area covered by the canvas alone is 8 inches by 12 inches, for a total of 96 in^2. With the addition of the frame, the total area covered by canvas and frame is 165 in^2. This means that the area covered by the frame alone is 165 in^2 - 96 in^2 = 69 in^2.

We will call the canvas's width w = 8 in and its length l = 12 in.

The frame's width is W = 8 + C in and its length is L = 12 + C in, where C is the constant width of the frame.

The area covered by the frame alone is thus

69 in^2 = WL - wl = (8+C)(12+C) - (8)(12)

69 in^2 = 96 + 20C + C^2 - 96

69 in^2 = 20C + C^2

0 = C^2 + 20C - 69

0 = (C - 3)(C + 23)

C = 3, C = -23

As the width of the frame must be positive, we thus have that C = 3. As C is how much we added in total to the canvas width, C/2 = 1.5 in is the frame width (as half was added to each side of the canvas.)

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