# Basic Optimization Problem

I sat for an exam a few days ago. I managed to answer every question except for question $1$c in the calculus paper. Provided that I got question $2$d correct (my answer was $m=0.5$), the absence of an answer for question $1$c should not in any way affect my overall grade. Out of interest, however, I am wondering if you could start me off on how to approach problem $1$c. Both the questions can be found below.

Thank you very much.

Question $1$c:

An enclosure at a reserve is divided into two rectangular cages.

One side of the enclosure is a solid wall.

There is a fence around the rest of the enclosure and another between the male and female cages.

The total length of the fences is $275$ m.

$5$ male birds are kept in the smaller cage and $8$ female birds in the larger cage.

The males have an average area of $250$ sq.m each in their cage.

The area of the female cage is to be maximised.

Find the average area for each of the female birds in their cage.

Question $2$d:

The diagram below shows part of the graph of the function $y=x^2, x>0$.

The shaded region between the curve and the $X$-axis, and between $x = m$ and $x = m + 2$, has an area of $5 \frac16$ sq.units. Find the value of $m$.

Explain the choice of the solution.

Here is the link to the entire exam.

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The answer $m=\frac{1}{2}$ is correct. –  André Nicolas Nov 22 '11 at 3:27
Welcome to math.stackexchange.com! I hope you don't mind, I've taken the content of the question from the external source and inputted it here to make it easier for people to answer. –  process91 Nov 22 '11 at 3:43

From the picture above, the total length of the fence is $3a+b+c$. We are given that this is $275$. Hence, we the first constraint

$$3a+b+c = 275.$$

We are further given that the $5$ males have an average area of $250$ in their cage. This gives us that

$$\frac{ac}{5} = 250 \implies ac = 1250.$$

We now want to maximize the average area for each of the female birds in their cage i.e. we want to maximize $\displaystyle \frac{ab}{5}$ subject to these constraints. Maximizing $\displaystyle \frac{ab}{5}$ is equivalent to maximizing $ab$. Hence, the optimization problem is

$$\text{Maximize } ab \text{ subject to } 3a+b+c = 275 \text{ and } ac = 1250.$$

Eliminating $b$ using the first constraint gives us $b = 275 - 3a - c$. Hence the function we want to maximize becomes $$275a - 3a^2 - ac$$ subject to $ac = 1250$. Plugging in the value of $ac$, we get that we want to maximize

$$275a - 3a^2 - 1250.$$

The maximum for this function can be found easily by completing the squares (or) by differentiation and the maximum occurs at $$a = \frac{275}{6}$$ giving a maximum of $$\frac{60625}{12}.$$

Hence, the average area for each of the female birds in their cage is $\frac{60625}{96}$.

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I hate to be a nuisance but I get a little lost after you state "Maximize ab subject to 3a+b+c=275 and ac=1250." Can you please further explain how you derive the expression "275a−3a2−ac"? Is this explanation making use of the Lagrange multiplier method that @process91 referred to earlier? I am not familiar with this method, but am about to read up on this now. Thank you all very much for your help. –  clookid Nov 22 '11 at 8:19
Ah, I see what you have done now. "ab = 275a - 3a^2 - 1250." That's a clever approach. Thanks, once again. –  clookid Nov 22 '11 at 8:44

Using the following labeling:

The total length of the fences is 275m

Therefore $3x+y+z=275$.

5 male birds, average area of 250m$^2$

Therefore $\frac {xz} 5 = 250$. You are to maximize the area of the female cage, which is given by $A(x,y) = xy$.

You can use substitution (with regard to the constraints) to write a function for the area of the cage which is only dependent on one variable, and then maximize it using the normal approach of taking the derivative and setting it to zero. Alternatively, you can use the Lagrange multiplier method.

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Thank you very much for your help. Can you please check my working (see: i43.tinypic.com/s1jbma.jpg)? –  clookid Nov 22 '11 at 4:00
@CharlesSalmon Yes, that is correct. Your answer also agrees with the answer that Siviram gave in his answer of $60625/96$, the difference in approach is that you chose to maximize the average area for the female birds, while Siviram maximized the area of the female cage. You are lucky, in this respect, because maximizing the average area also maximized the area of the cage and the answer turned out to be correct, however a question could be chosen in a tricky way where this is not the case so make sure you are maximizing what has been asked. –  process91 Nov 22 '11 at 4:15