August 17th, 2014, 05:39 AM  #1 
Newbie Joined: Jul 2014 From: australia Posts: 16 Thanks: 0  quadratics
Question is attached: don't understand part c. Answer for part c is: 100m by 112.5m Solutions would be much appreciated! Last edited by skipjack; August 17th, 2014 at 06:23 AM. 
August 17th, 2014, 06:28 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
You need to assume that all of the available fencing is used. For part (c), they want you to calculate x and y.

August 17th, 2014, 08:57 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, jessjans11! Quote:
There are 9 lengths of fencing which are $x$ m each. There are 8 lengths of fencing which are $y$ m each. The total fencing is 1800 m: $\:9x + 8y \:=\:1800\;\;[1]$ Quote:
Solve equation [1] for $y$. $\quad y \;=\;\dfrac{18009x}{8} \quad\Rightarrow\quad y \;=\;225  \tfrac{9}{8}x\;\;[2]$ The area of a pen is: $\:A \;=\;xy \;=\;x\left(225\frac{9}{8}x\right)$ Therefore: $\:A \;=\;\frac{9}{8}x^2 + 225$ Quote:
We want to maximize $A$. The equation is a downopening parabola. Its maximum is at its vertex. The vertex is: $\:x \:=\:\dfrac{\text{}b}{2a} \:=\:\dfrac{\text{}225}{2(\text{}\frac{9}{8})} \:=\:100 $ Substitute into [2]: $\:y \:=\:225  \frac{9}{8}(100) \:=\: \dfrac{225}{2}$ Therefore, the dimensions are: $\:100\text{m} \times 112.5\text{m}$.  
August 18th, 2014, 06:19 AM  #4  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
And then when you substituted, you substituted into the wrong expression. You should have substituted the xvalue into A = 225x  (9/8)x^2.  

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