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Spent WAY too long trying to figure this out and I just don't know what I am doing wrong.

"A rectangular region is to be fenced using 5100 feet of fencing. If the rectangular region is to be separated into 4 regions by running 3 lines of fence parallel to two opposite sides, determine the dimensions of the region which maximizes the area of the region.

Give the numerical values of the length and the width (in feet) of the entire enclosed region."

_______y_________
|x  |x  |x  |x |x
_______y_________

Drawn really badly, but the above is basically what I am dealing with right? $$A = x\cdot y$$

$$5100 = 5x + 2y$$

$$120 - \frac25y = x$$

$$A = y(120 - \frac25y)$$

$$A = 120y - \frac25y^2$$

$$A' = 120 - \frac45y$$

$$y = 150$$

and so: $$x = 960$$

150 is the absolute max because if you plug any number higher into the derivative of the area, you get a negative number

What is wrong with my strategy?

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  • $\begingroup$ $x=-\frac{2}{5}y + 1020$ $\endgroup$ – Thoth Dec 16 '15 at 1:04
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You have made an error when dividing by 5 on your second line.

$$5100\div5=1020$$

So your answer would become:

$$5100 = 5x + 2y$$

$$1020 - \frac25y = x$$

$$A = y(1020 - \frac25y)$$

$$A = 1020y - \frac25y^2$$

$$A' = 1020 - \frac45y$$

$$y = 1275$$

and so: $$x = 510$$

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