# Thread: Area to cost based on parimiter

1. ## Area to cost based on parimiter

I have an area A = 2500ft2 of a rectangular fence
the area is found base * height
the base is X and height is H

Each fence post is $25; meaning$100 for all 4 of them
Only 3 of the sides are being fenced, X X and H
it costs $5 per ft on sides of X and$10 on H
C = cost
I am lost as to how to make a formula to tie all this information together. To make a formula in terms of C(x)=

H is something i made up to hold the place of Height the rest are given.

2. ## Re: Area to cost based on parimiter

cost ...

C = 100 + 5x + 5x + 10h = 10x + 10h = 10(x+h)

xh = 2500

solve for h in the second equation, substitute the result in for h in the second equation to get the cost as a function of x

3. ## Re: Area to cost based on parimiter

Originally Posted by blondedude092
I have an area A = 2500ft2 of a rectangular fence
the area is found base * height
the base is X and height is H

Each fence post is $25; meaning$100 for all 4 of them
Only 3 of the sides are being fenced, X X and H
it costs $5 per ft on sides of X and$10 on H
C = cost
I am lost as to how to make a formula to tie all this information together. To make a formula in terms of C(x)=

H is something i made up to hold the place of Height the rest are given.
You can get it all in terms of x using the fact that the area is \displaystyle \begin{align*} 2500\,\textrm{ft}^2 \end{align*}

\displaystyle \begin{align*} x\,h &= 2500 \\ h &= \frac{2500}{h} \end{align*}

4. ## Re: Area to cost based on parimiter

So i am getting the equation of C = 100+10(x+(2500/x))

thanks!

5. ## Re: Area to cost based on parimiter

Right. Better to show this way: C = 100 + 10(x^2 + 2500) / x

Cheapest when x = h : $1100 ; do you understand why? 6. ## Re: Area to cost based on parimiter Originally Posted by Wilmer Right. Better to show this way: C = 100 + 10(x^2 + 2500) / x Cheapest when x = h :$1100 ; do you understand why?
Or even \displaystyle \begin{align*} C = \frac{10x^2 + 100x + 25\,000}{x} \end{align*}

7. ## Re: Area to cost based on parimiter

Originally Posted by Prove It
You can get it all in terms of x using the fact that the area is \displaystyle \begin{align*} 2500\,\textrm{ft}^2 \end{align*}

\displaystyle \begin{align*} x\,h &= 2500 \\ h &= \frac{2500}{h} \end{align*}
Typo. You mean, of course,
\displaystyle \begin{align*} x\,h &= 2500 \\ h &= \frac{2500}{x} \end{align*}
or
\displaystyle \begin{align*} x\,h &= 2500 \\ x &= \frac{2500}{h} \end{align*}

8. ## Re: Area to cost based on parimiter

Originally Posted by HallsofIvy
Typo. You mean, of course,
\displaystyle \begin{align*} x\,h &= 2500 \\ h &= \frac{2500}{x} \end{align*}
or
\displaystyle \begin{align*} x\,h &= 2500 \\ x &= \frac{2500}{h} \end{align*}
Of course, I meant h in terms of x