3. (#18 on exam 3 review) a production editor decided that a promotional flyer should have a 1 - in margin…

3. (#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.\n(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.)

3. (#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.\n(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.)

Answer

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: [ \begin{align*} P&=(x - 1)(\frac{72}{x}-2)\ &=x\cdot\frac{72}{x}-2x-\frac{72}{x}+2\ &=72-2x-\frac{72}{x}+2\ &=74-2x-\frac{72}{x} \end{align*} ]

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: [ \begin{align*} -2+\frac{72}{x^{2}}&=0\ \frac{72}{x^{2}}&=2\ x^{2}& = 36\ x&=6\quad(x>0) \end{align*} ]

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).