Answer :
Answer:
[tex]4\sqrt{10}[/tex] inches and [tex]6\sqrt{10}[/tex] 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:
[tex]x\cdot y=240[/tex]
[tex]y=\frac{240}{x}[/tex]
The side of new poster with 1-inch margin on sides would be [tex](x-2)[/tex] and 2 inch margin on top and 1 inch margin at the bottom would be [tex]y-3[/tex].
The area of new poster would be [tex]A=(x-2)(y-3)[/tex].
Upon substituting [tex]y=\frac{240}{x}[/tex] in area equation, we will get:
[tex]A=(x-2)(\frac{240}{x}-3)[/tex]
[tex]A=240-3x-\frac{480}{x}+6[/tex]
[tex]A=246-3x-\frac{480}{x}[/tex]
[tex]A=246-3x-480x^{-1}[/tex]
Since, we need to maximize the area, so we will find critical points.
First of all, we will find derivative of area function as:
[tex]A'=\frac{d}{dx}(246)-\frac{d}{dx}(3x)-\frac{d}{dx}(480x^{-1})[/tex]
[tex]A'=0-3+480x^{-2}[/tex]
Now, we will equate derivative with 0 to find critical values:
[tex]-3+480x^{-2}=0[/tex]
[tex]480x^{-2}=3[/tex]
[tex]\frac{480}{x^2}=3[/tex]
[tex]3x^2=480[/tex]
[tex]x^2=\frac{480}{3}[/tex]
[tex]x^2=160[/tex]
[tex]x=\pm\sqrt{16*10}[/tex]
[tex]x=\pm4\sqrt{10}[/tex]
Since length cannot be negative, therefore, side length would be [tex]4\sqrt{10}[/tex].
[tex]y=\frac{240}{4\sqrt{10}}[/tex]
[tex]y=\frac{240\sqrt{10}}{4*10}=\frac{240\sqrt{10}}{40}=6\sqrt{10}[/tex]
Therefore, side length of [tex]4\sqrt{10}[/tex] inches and [tex]6\sqrt{10}[/tex] inches will give the largest printed area.