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I have been working this problem for awhile and cannot seem to solve it even though its probably easier than I think...

There is a rectangular garden that needs fencing. For one side of the fencing wooden slats will be used costing 2 dollars per foot, the other three sides will use metal fencing costing 8 dollars per foot. If the total money to spend on fencing in $700, what is the largest area he can make.

I made the equation 2x+8x+2(8y)

(2x) for the one wooden side, (8x) for the metal side opposite the wooden side, and (2(8y)) for the two metal sides that face opposite each other.

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Let x represent the width, y the length of the garde.

We have:

$$A=xy (1)$$ $$700=2x+8(x+y+y) \implies 700=10x+16y \implies x=70-1.6y (2)$$

Plugging x into Equation 1, $$A(y)=70y-1.6y^2$$ $$A'(y)= 70 -3.2y$$ Critical Values where $A'(y)=0$: $$0=70-3.2y \implies y=21.875$$ Thus if $$y=21.875, \text{ from equation (2)}, x=35$$

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The "right" way to do this is with Lagrange multipliers. But a quick way is to take your first equation, which really ought to be $10x+16y = 700$, and supplement it with the equation for area $A=xy$. Into this equation, substitute from your first equation for either $x$ or $y$. Which one doesn't really matter. Let's do $x$ so you get $A=(700-16y)/10*y$. Now take the derivative of this w/ respect to $y$ to get where $y$ gives the maximum area. Plug this $y$ back into your first eqn to get the corresponding value of $x$. Finally, plug $x$ and $y$ into $A=xy$ to get the area.

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