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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.

This leads to

$$-2y^2 + 600y = -x^2 + 300x $$ 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 $(300,0)$.

enter image description here

Question from HW

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

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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.

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