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?

  • $\begingroup$ 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

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 $A=160$.

  • $\begingroup$ Oh I see; I got confused by my own variables lol $\endgroup$ – suomynonA Jan 18 '17 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.