Exercise 37 Page 230

Let's begin by plotting the painting with the frame. Let x be the width of the frame. Our mission is finding the value of x.

From this picture, we see that the area of the painting with the frame is given by the expression below. A_(Total) = (24+2x)(16+2x)To find the area of just the frame, we subtract the area of the painting from the expression above. A_(Frame) &= A_(Total) - A_(Painting) &= (24+2x)(16+2x) - 24* 16 Let's simplify the right-hand side of the expression above.

Now, remember we are told that we have a total of 276in^2 to frame the painting. A_(Frame) = 276 ⇒ 4x^2+80x = 276 By solving the equation above for x, we will find the width of the frame.

To continue solving the equation we use the Quadratic Formula.

From the final equation, we obtain two solutions: one using the positive sign, and another with the negative sign.

Since x represents the width's frame it cannot be negative, so we disregard the negative option. In conclusion, the width of the frame is 3in.