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Plan to enclose a rectangular garden that includes a fence dividing the interior into two separate pieces. The fencing on the outside will cost $5 per linear foot, but the fencing inside will only cost $2 per linear foot. I have budgeted a total of $300 for the fencing. Find the dimensions of the largest garden I can enclose.

$300=5(2x+2y)+2y$

$300=10x+12y$

$y=\frac{300-10x}{12}$

How do they get

$A(x)=\frac{x(300-10x)}{12}$

$A(x)=\frac{300x-10x^2}{12}$

They ask to find the critical numbers and to test the critical numbers

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    I suppose the fence inside is somehow _required_ to be parallel to one of the sides of the rectangle, or else that the subdivided pieces are required to be equal (or at least not too unequal). Otherwise I will enclose a triangle of area 1 square angstrom in one corner of the field, and use (almost) the entire budget to fence a square garden of almost 15 feet on each side. – 2017-02-08

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You should write down what the variables mean. Here it looks like $x$ is the length of the garden, $y$ is the width and the divider fence runs along the width. The first equation expresses the cost constraint-do you see how? Then the area is $A=xy$ and they have substituted in the third equation for $y$. Since you want the maximum area, you should take $\frac {dA}{dx}$, set to zero, solve for $x$.