# What (if any) system of equations would solve this problem?

I am helping my daughter with her high school pre-calc. We've both got stuck on this. Below is a copy of the exercise as presented.

We've come up with the following equations (none of which are in the answer set).

$$A = -y^2 + 600y$$ $$A = -x^2 + 300x$$

Since both areas are equal:

$$-y^2 + 600y = -x^2 + 300x$$ or $$y^2 - 600y = x^2 - 300x$$

Three trivial solutions (out of infinitely many): $$(0,0),(300,0)$$ and $$(0,600)$$ (curious ways to build a fence) can be seen in the graph below. Knowing the area would not make this problem seem any less strange.

Am I missing something embarrassingly obvious here?

Edit: I originally had the second equation as $$A = -2y^2 + 600y$$ and I "fixed" it. Working too late and losing efficiency.

$$-2y^2 + 600y = -x^2 + 300x$$ or $$2(y^2 - 300y) = (x^2 - 300x)$$

which produces a similar-looking graph and trivial (zero-area) results (among infinitely many): $$(0,0), (0,300)$$ and $$(300,0)$$.

• I think here the area $A$ is being treated as a known quantity. You have to select the particular $(x,y)$ that simultaneously satisfies $y^2 - 600y = x^2 - 300x,A = -x^2 + 300x,A = -y^2 + 600y$ – Shubham Johri Dec 20 '18 at 7:52
• You know what? I had -2y^2 and I "fixed" it. – Adam Hrankowski Dec 20 '18 at 8:05

The solution and that of the book doesn't make any sense because the variables $$x$$ and $$y$$ are not defined and the problem actually has four unknowns.

From the given data, using field dimensions $$w,h,w',h'$$, we have

$$\begin{cases}A=wh,\\2w+2h=600,\\A=w'h',\\w+2h'=600.\end{cases}$$

Eliminating $$h$$ and $$h'$$, we are left with

$$\begin{cases}A=300w-w^2,\\A=300w'-\dfrac{w'^2}2.\end{cases}$$

You can eliminate $$A$$ and end-up with a single equation in two unknowns, which is indeterminate.

Let the dimensions of the first field be $$x,y$$.

$$x+y=300\\A=xy=x(300-x)=300x-x^2$$

Let the dimensions of the second field be $$l,b$$, and say we have two sides of length $$b$$.

$$l+2b=600\\A=lb=b(600-2b)=600b-2b^2$$

So the systems $$(C),(D)$$ will provide us with the required dimensions.

As for your query, I believe the area $$A$$ is being treated as a known quantity. We have to select that $$(x,b)$$ which satisfies $$300x-x^2=A=600b-2b^2$$ simultaneously.