# Find maximum of a system of equations

You have 300 meters of fencing with which to build two enclosures. One will be a square, and the other will be a rectangle where the length of the base is exactly twice the length of the height.

(a) Give the dimensions of the square and rectangle that minimize the combined area. [SOLVED]

My equations:

\begin{align}A &= W^2 + 2L^2 ... 300 = 4w + 6L\\ w &= 75-1.5L\\ A &= (75-1.5L)^2 + 2L^2\\ A &= 4.25L^2 - 225L + 5675\\ -\frac{b}{2a} &= \frac{225}{8.5} \text{(minimum)}\end{align}

Using substitution I solved this correctly.

(b) What is the maximum combined area? [HELP]

• Well if you can find a quadratic in one variable with positive coefficient, it's vertex is the minimum. – Alfred Yerger Jun 3 '14 at 6:22
• Alfred that is how i solved part (a), not sure how to do (b) – Scherf Jun 3 '14 at 6:29

The first thing that should come to mind to you in a question like this is that any usage of my 300m of fence can be described by a single parameter. For instance, if I give you one of the edge lengths of the square, $s$, then you have all the information you need to work out what the edge lengths of the rectangle are and in turn what the total fenced area is.

So now we create a function $A$ whose value is the area fenced. We parameterize this function using the variable $s$-- the side length of the square. Given $s$, we know that we have $300-4s$ meters of fence left to use on the rectangular enclosure. The perimeter of this fence is equal to $8$ times the short edge length which I wil call $l$. $$8l=300-4s$$ The area of the rectangular enclosure is $2l^2$. $$2l^2=2\cdot\left(\frac{300-4s}{8}\right)^2$$ The area of the square enclosure is of course $s^2$. So now we can quantify the total area $A$ with the single parameter being $s$.

$$A\left(s\right)=s^2 + 2\left(\frac{300-4s}{8}\right)^2$$ $$A\left(s\right)=s^2 + \frac{1}{2}\left(75-s\right)^2$$

Now you can solve for the maximum and minimum of this function like you would for any function of a single variable. Note that $0 \leq s \leq 75$ because we are not permitting $s$ or $l$ to be negative. Find when the first derivative of this function is zero in this domain and also check the boundaries $\left(s=0 \ \rm{and} \ s=75\right)$. These are your local extrema and the minimum area will be the smallest of these and the maximum will be the greatest of these.

The essence of a question like this is recognizing the minimum number of parameters you need to describe the system and using standard techniques to solve for the extrema.

Use the same substitution you used for calculating the minimum combined area. Write down what you get (by editing the question), and write down the substitution as well. That way, we will be able to help you further.

Edit: You calculated that the total area is $A=4.25 L^2 - 225 L + 5675$. Indeed, the minimum of this function is $\frac{225}{8.5}$. Now you must calculate the maximum of the function. Of course, if I take $L$ large enough, the value of $A$ will become arbitrarily large, but you must realize that $L$ is bounded as it must be between $0$ and $50$. Therefore, you must find the maximum of $A$ on $[0,50]$.

• OK, I added the steps – Scherf Jun 3 '14 at 6:29
• Hint: $0\leq L\leq 50$ – poolpt Jun 3 '14 at 6:42
• @Scherf I edited my answer. – 5xum Jun 3 '14 at 6:47