A house is built in the shape of a rectangle, with $3$ rectangular interior sections separated by parallel walls, using fencing. The owner has $900$ feet of fencing, and he wants to enclose the largest possible area. What should the length, width, and area be?

Please help, I'm lost.


Let the two inside parallel walls each have length $x$. Let the sides of the rectangle perpendicular to these each have length $y$.

Then the total area enclosed is $xy$. The amount of fencing used is $4x+2y$. This is to be $900$, since it is clear that it is best to use up all the fencing.

So we want to maximize $xy$, under the constraint $4x+2y=900$.

Thus $y=450-2x$, and we want to maximize $x(450-2x)$.

Because of the physical situation, we need $x\ge 0$ and $y\ge 0$. This means $x\le 225$.

So mathematically, we want to minimize $f(x)=450x-2x^2$, where $0\le x\le 225$.

This can be done by standard tools, such as calculus or completing the square.

  • $\begingroup$ i got x= 225/2 and y=1 please help me. What are the correct answers $\endgroup$ – Michael Rametta Mar 27 '13 at 2:29
  • $\begingroup$ I agree with the $x=225/2$. The $y$ is then $225$. $\endgroup$ – André Nicolas Mar 27 '13 at 2:32
  • $\begingroup$ You are welcome. Sorry that it could not be accompanied by a picture. That is often a very important part of a solution. $\endgroup$ – André Nicolas Mar 27 '13 at 2:44

Here are more details than you wanted to know:

Suppose the house has length $l$ and width $w$. The length and with must be non-negative, of course. Then the area is $A(l,w) = lw$. The length of fence required is $2l+2w$ for the exterior wall and $2w$ for the interior separators, hence we have $\lambda(l,w) = 2(l+2w)$. So the problems is $\max\{ A(l,w) | \lambda(l,w) \le 900 , l \ge 0, w \ge 0\}$.

Inequality constraints are a little harder to work with in general, so we try to remove them.

Since the feasible set is compact, there is a maximum. If $(l,w)$ are non-negative, and $\lambda(l,w) < 900$, then we can increase $l$ and $w$ a little and increase $A(l,w)$. Hence the problem is equivalent to $\max\{ A(l,w) | \lambda(l,w) = 900 , l \ge 0, w \ge 0\}$.

The two non-negativity constraints are still present, so one way is to just ignore them and check if the resulting points are feasible.

Solving $\max\{ A(l,w) | \lambda(l,w) = 900 \}$ is straightforward since we can let $l = 450-2w$ (because $\lambda(l,w) = 900$), and then $A(450-2w,w) = (450-2w)w$. Note that this is a strictly concave quadratic, hence it has a maximum. Setting the derivative to zero and solving gives $\hat{w} = \frac{255}{2}$, and then we have $\hat{l}=255$. Since $\hat{w}, \hat{l} \ge 0$, this is a solution to the original problem.

Hence we have $\hat{w} = \frac{255}{2}$, $\hat{l}=255$, and $A(\hat{l},\hat{w}) = \frac{50625}{2}$.


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