Answer to Question #156056 in Trigonometry for anne

Question

A poster is to have an area of 180 in with 1-inch margins at the bottom and sides and a 2-inch margin at the top. What dimensions will give the largest printed area?

Answer 1

Let the poster is of width $a$ and length $b$. Then $b=\frac{180}{a}$. Let us sketch the picture:

It follows that the printed part of poster is of width $a-2$ and length $b-3$. The printed area is equal

$S(a)=(a-2)(\frac{180}{a}-3)=186-\frac{360}{a}-3a$.

Then $S'(a)=\frac{360}{a^2}-3$. If $S'(a)=0$ , then $3a^2=360$, and thus $a=\sqrt{120}=2\sqrt{30}$ (since $a>0$). It follows that $b=\frac{180}{2\sqrt{30}}=3\sqrt{30}$. Since $S'(a)=-\frac{720}{a^3}<0$ for $a>0$, we conclude that $a=2\sqrt{30}$ is a point of maximum of area function $S(a)$. Therefore, the largest printed area for poster of width $2\sqrt{30}$ and length $3\sqrt{30}.$