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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

Fiona Lu
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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. – David K Feb 08 '17 at 04:57

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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 $x$.

Ross Millikan
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