optimization: a page of a book...

[zebra]

A page of a book is to have an area of 90in(squared), with 1-inch margins at the bottom and sides and a 1/2-inch margin at the top. Find the dimensions of the page that will allow the largest printed area.

i let the sides of the page =y and the top and bottom of the page equal x.

from that i got that the Area of the whole page equals x times y. since the Area equals 90, i got the connector:
A=xy

then i drew the pic and got that the Area of the printed part of the page equals x minus the margins and y minus the margins and i got this equation:
A(of printed part)= (x-2)(y-1.5)

then i used the connector to solve for y and got the equation:
A(of printed page)=[x-2][(90/x)-1.5]

i distributed it out and then found that the deriv is 178.5
dA/dx=178.5

and i dont kno where to go 4m there...

 

[soroban]

Hello, zebra!

You did something mysterious at the end . . .

A page of a book is to have an area of 90 in²
with 1-inch margins at the bottom and sides and a 1/2-inch margin at the top.
Find the dimensions of the page that will allow the largest printed area.

i let the sides of the page = \(\displaystyle y\) and the top and bottom of the page = \(\displaystyle x\).

from that i got that the Area of the whole page equals x times y.
since the Area equals 90, i got the connector: \(\displaystyle xy \:= \:90\;\) . . . right!

then i drew the pic and got that the Area of the printed part of the page
. . equals \(\displaystyle x\) minus the margins times \(\displaystyle y\) minus the margins
and i got this equation: \(\displaystyle \:A(\text{print}) \:= \: (x\,-\,2)(y\,-\,1.5)\l\) . . . yes!

then i used the connector to solve for \(\displaystyle y\) and got the equation:
. . \(\displaystyle A(\text{print})\:=\x\,-\,2}\left(\frac{90}{x}\,-\,1.5\right)\;\) . . . correct!

i distributed it out and then found that the deriv is 178.5 . ??

We have: \(\displaystyle \:A \:=\x - 2)\left(\frac{90}{x}\,-\,1.5\right) \;=\;90 \,-\,15x\,-\,\frac{180}{x}\,+\,3\)
. . Then: \(\displaystyle \:A \:=\:93\,-\,15x\,-\,180x^{-1}\)

Differentiate: \(\displaystyle \:A' \:=\:-15\,+\,180x^{-2} \:=\:0\)
Multiply by -\(\displaystyle x^2:\;\;15x^2\,-\,180\:=\:0\;\;\Rightarrow\;\;x^2\:=\:12\;\;\Rightarrow\;\;\fbox{x\,=\,2\sqrt{3}}\)

Substitute into the connector: \(\displaystyle \:y \:=\:\frac{90}{2\sqrt{3}}\;\;\Rightarrow\;\;\fbox{y \,=\,15\sqrt{3}}\)

 

[zebra]

thx again...wat i did was when i took the deriv, i accidentally wrote that the deriv of -180/x was just 180 b/c i forgot that the x was to the -1. next time ill be sure to right out all my work. thank you lots!!!