I am having a lot of trouble with this word problem. Can anyone help?

There are 900 units of fencing available to enclose a rectangular plot of ground with a fence down the middle and parallel to two ends. What is the maximum area which can be enclosed?


I assume that you have made a sketch. That is the first step to solving the problem.

Imagine making the type of enclosure described. Let the sides parallel to the fence in the middle have length $x$, and let the other two sides of the rectangle each have length $y$. Write $x$ beside the two sides and the fence in the middle that have length $x$, and write $y$ beside the two sides that have length $y$.

From the picture, we see that the amount of fencing used is $3x+2y$. if we want to make the enclosure large, itt is clear that we are best off using all the available fencing. Thus we must have $3x+2y=900$. We want to maximize the area $xy$, given that $3x+2y=900$.

Now proceed as has been suggested for similar problems. We have $y=\frac{1}{2}(900-3x)$.

So we want to maximize $f(x)=\frac{1}{2}(x)(900-3x)$.

If we want to use calculus, note that $f(x)=450x -\frac{3}{2}x^2$. So $f'(x)=450-3x$. The derivative is $0$ at $x=150$.

The maximum area is therefore $f(150)$, which is not hard to compute. I would prefer to find $y$ from $3x+2y=900$. If $x=150$ then from $3x+2y=900$ we get $y=225$. Thus the maximum area that can be enclosed is $(150)(225)$.

Remark: If we want to be very fussy (and sometimes it can be important to be), we can note that $0\le x\le 300$ (where $x=0$ and $x=300$ do not give "real" fields). Our function $f$ attains a maximum in the interval $0\le x\le 300$. But obviously $f(0)=f(300)=0$, so the maximum is not attained at an endpoint. Thus the maximum is reached at a place where $f'(x)=0$. There is only one such place, namely $x=150$, so the maximum must be reached there. Thus our calculation does yield the maximum area.

  • $\begingroup$ So then what would be the answer? 3x+2y=900? $\endgroup$ – Amira Jun 11 '12 at 4:22
  • $\begingroup$ As André suggested, solve the maximization problem. This will give you $x$. Then use the $3x+2y=900$ formula to find $y$. Then compute $xy$ to find the maximum area. $\endgroup$ – copper.hat Jun 11 '12 at 4:25
  • $\begingroup$ I think it is very worthwhile pointing out the details you called 'fussy'. It shows that there is more to the problems than just churning through some formulae. And the details provide a good backdrop should the student move on to more complicated problems. $\endgroup$ – copper.hat Jun 11 '12 at 4:42

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.