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A farmer is building four connected pens for his livestock. He has $160$ feet of fencing. What is the maximum area of one sector?

So here's a (crude) drawing I made; the red lines represent congruent lines.

enter image description here

My attempted solution:

$8$ fences of length $y$ and $5$ fences of length $x$ add up to $160$ feet of fence, so I get the equation $8y+5x=160$, which is equivalent to $x=-\dfrac {8}5y+32$

The area of one sector is the product of $x$ and $y$, so I get the equation $xy=A$

substituting, I get $-\dfrac {8}5y^2+32y=A$

Using the formula $x=-\dfrac{b}{2a}$, I get $x=20$.

Substituting this into the first equation, I get $y=7.5$, and substituting both values into the second equation, I get $A=150$.

The correct answer is $160$. What did I do wrong?

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  • $\begingroup$ I deleted my comment.. because I'm getting 1600/9 for the answer which is kinda odd... but I'm posting a solution for your 1x4 arragement (which is probably what the problem assumes). $\endgroup$ – pie314271 Jan 18 '17 at 3:45
  • $\begingroup$ @pie314271 Yes; I just did the question again and got 40/3 for x... $\endgroup$ – suomynonA Jan 18 '17 at 3:46
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Your $-\dfrac b{2a}$ is incorrect; you have tried to find $x$ with the formula but you have a quadratic in $y$ so you will find $y$.

$y=\dfrac{-b}{2a}=\dfrac{-32}{-\frac{16}{5}}=10$ so $x=16$ and the answer is $A=160$.

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  • $\begingroup$ Oh I see; I got confused by my own variables lol $\endgroup$ – suomynonA Jan 18 '17 at 3:49
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The thing is we have to maximise the area, $xy $ with the total perimeter $8y+5x=160$ as constant. So, substituting, we have to maximise the expression $$f (y)=-\frac {8}{5}y^2+32y $$ Now letting $f'(y)=0$ and solving for $y $ gives us $y=10$ and thus $x=16$.

The area is thus $xy=160$ square units . Hope it helps.

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  • $\begingroup$ So basically as pie said, I solved for y and called it x $\endgroup$ – suomynonA Jan 18 '17 at 3:49
  • $\begingroup$ @suomynonA Yes, there lies the mistake. $\endgroup$ – Rohan Jan 18 '17 at 3:51

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