# Find the maximum area of one sector

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.

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?

• 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). Commented Jan 18, 2017 at 3:45
• @pie314271 Yes; I just did the question again and got 40/3 for x... Commented Jan 18, 2017 at 3:46

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 $A=160$.

• Oh I see; I got confused by my own variables lol Commented Jan 18, 2017 at 3:49

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.

• So basically as pie said, I solved for y and called it x Commented Jan 18, 2017 at 3:49
• @suomynonA Yes, there lies the mistake.
– user371838
Commented Jan 18, 2017 at 3:51