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Calculus IChapter 4: Application of DerivativesSection 4-9

More Optimization — Question 8

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Question 8

A poster is to have an area of 192 in2^2. It must have 1-inch margins on the sides, a 2-inch margin at the top, and a 3-inch margin at the bottom. What dimensions of the poster will allow the largest printed area?

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Question 8 - Solution

Let xx be poster width and yy its height. Then xy=192xy=192, and positive printed dimensions require 2<x<192/52<x<192/5.

A(x)=(x−2)(192/x−5)=202−5x−384/x.A(x)=(x-2)(192/x-5)=202-5x-384/x.

A′(x)=−5+384/x2=0⇒x=384/5.A'(x)=-5+384/x^2=0\quad\Longrightarrow\quad x=\sqrt{384/5}.

Since A″(x)=−768/x3<0A''(x)=-768/x^3<0, this is the unique maximum.

x=384/5≈8.76 in,y=480≈21.91 in.\boxed{x=\sqrt{384/5}\approx8.76\text{ in},\qquad y=\sqrt{480}\approx21.91\text{ in}.}

The exact width, rather than its rounded value, is used to compute the height.

Original worksheet page 2: question and worked solution for 4-9-008

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