Question

A poster is to have an area of 240 in² with 1-inch margins at the bottom and sides and a 2-inch margin at the top.

Answer 1

Answer:

$4\sqrt{10}$ inches and $6\sqrt{10}$ inches.

Step-by-step explanation:

Let x and y represent sides of the poster.

We have been given that a poster has an area of 240 in². We can represent this information in an equation as:

$x\cdot y=240$

$y=\frac{240}{x}$

The side of new poster with 1-inch margin on sides would be $(x-2)$ and 2 inch margin on top and 1 inch margin at the bottom would be $y-3$.

The area of new poster would be $A=(x-2)(y-3)$.

Upon substituting $y=\frac{240}{x}$  in area equation, we will get:

$A=(x-2)(\frac{240}{x}-3)$

$A=240-3x-\frac{480}{x}+6$  

$A=246-3x-\frac{480}{x}$

$A=246-3x-480x^{-1}$

Since, we need to maximize the area, so we will find critical points.

First of all, we will find derivative of area function as:

$A'=\frac{d}{dx}(246)-\frac{d}{dx}(3x)-\frac{d}{dx}(480x^{-1})$

$A'=0-3+480x^{-2}$

Now, we will equate derivative with 0 to find critical values:

$-3+480x^{-2}=0$

$480x^{-2}=3$

$\frac{480}{x^2}=3$

$3x^2=480$

$x^2=\frac{480}{3}$

$x^2=160$

$x=\pm\sqrt{16*10}$

$x=\pm4\sqrt{10}$

Since length cannot be negative, therefore, side length would be $4\sqrt{10}$.

$y=\frac{240}{4\sqrt{10}}$

$y=\frac{240\sqrt{10}}{4*10}=\frac{240\sqrt{10}}{40}=6\sqrt{10}$

Therefore, side length of $4\sqrt{10}$ inches and $6\sqrt{10}$ inches will give the largest printed area.