Question

1. (#18 on exam 3 review) a production editor decided that a promotional flyer should have a 1 - in margin at the top and the bottom and a ½ - in margin on each side. the editor further stipulated that the flyer should have an area of 72 in². determine the dimensions of the flyer that will result in the maximum printed area (shaded area) on the flyer. your work must include the following: identify variables both on graph and in words; any equations or formulas; all mathematical work; and correct application of a derivative test. without correct application of a derivative test, no more than 50% credit for question.

(description: shaded rectangle within a rectangle. distance between shaded rectangle and outer rectangle is 1/2 in on vertical sides. distance between shaded rectangle and outer rectangle is 1 in on horizontal sides.)

Explanation:

Step1: Define variables

Let the width of the flyer be \(x\) inches and the height be \(y\) inches. The area of the flyer is \(A = xy=72\), so \(y=\frac{72}{x}\). The printed - area width \(w=x - 1\) (since \(0.5\) - inch margin on each side) and the printed - area height \(h=y - 2\) (since \(1\) - inch margin on top and bottom). The printed - area function \(P=(x - 1)(y - 2)\).
Substitute \(y=\frac{72}{x}\) into the printed - area formula: \(P=(x - 1)(\frac{72}{x}-2)\).
Expand the function:

$$
LATEXBLOCK0
$$

Step2: Find the derivative

Differentiate \(P(x)\) with respect to \(x\). Using the power rule \((x^n)^\prime=nx^{n - 1}\), we have \(P^\prime(x)=-2+\frac{72}{x^{2}}\).
Set \(P^\prime(x) = 0\) to find the critical points:

$$
LATEXBLOCK1
$$

Step3: Second - derivative test

Differentiate \(P^\prime(x)\) to get the second - derivative \(P^{\prime\prime}(x)=-\frac{144}{x^{3}}\).
When \(x = 6\), \(P^{\prime\prime}(6)=-\frac{144}{6^{3}}=-\frac{144}{216}=-\frac{2}{3}<0\). So the function \(P(x)\) has a maximum at \(x = 6\).
When \(x = 6\), \(y=\frac{72}{6}=12\).

Answer:

The dimensions of the flyer are \(x = 6\) inches (width) and \(y = 12\) inches (height).