A construction company has 840 feet of chain-link fence that is used to enclose storage areas for equipment and materials at construction sites. The supervisor wants to set up identical rectangular storage areas sharing a common fence.

Express the total area A(x) enclosed by both pens as a function of x. What is the maximum area?

  • $\begingroup$ Can you express the lengths of the sides of the two areas using a single variable given the information you have listed? Or perhaps as two variables, then solve for one as a function of the other? $\endgroup$ – abiessu Sep 12 '13 at 23:16
  • $\begingroup$ Ick. You might want to edit your title, since a function doesn't have an area. Perhaps "A function for the area"? $\endgroup$ – Rick Decker Sep 13 '13 at 0:28

HINT: You didn’t say, but I assume that $x$ is the length of the common side. Make a sketch. (That should almost always be your first step in a problem of this general type.)

           |       y      |      y      |  
           |              |             |  
           |              |             | x  
           |              |             |  

There are $3$ fences of length $x$. If $y$ is the other dimension of each of the rectangles, there are $4$ fences of length $y$ (or two of length $2y$, if you prefer). Clearly you should use all $840$ feet of the available fencing, so $3x+4y=840$, and you can express $y$ in terms of $x$. Equally clearly $A(x)=2xy$, and once you’ve expressed $y$ in terms of $x$, you can rewrite $A(x)$ as a function of $x$ alone.

After you’ve done that, go through the usual routine to find a maximum of $A(x)$: find the derivative $A'(x)$, set it to $0$, solve for $x$, and check that you’ve actually found a maximum. Then substitute that value of $x$ back into $A(x)$ to find the actual maximum area.


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