you can put a fence around a regular lot. The length of the lot must be at least 60ft. The cost of the fence along the length is $1.50 per foot. And the width is 2$ per foot. The total cost can not excede $360.

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- anonymous

use two variables to write a system of inequalities that models the problem
what is the maximum width if the length is 60 feet?

- anonymous

@Hero alittle help?

- anonymous

ok, so x is the length. y is the width.
x = 60ft
(60*1.5)+(y*2) <= 360

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## More answers

- anonymous

ok thanx you know how the first one is done?

- tkhunny

\(x \ge 60\;ft\)
\((2x(1.5))+((2y)2)) \le360\)
Also, this is a "regular" lot. This might mean \(x = y\). It's not as clear as I'd like it to be.

- anonymous

ok there is one more so instead of the length having a limit what if the width had a limit of 36 ft? same kind you just did tomo

- tkhunny

If x = y, it is the same sort of problem. Notice, however that in tomo's setup, there is only one length and one width. In mine, there are two of each. In my opinion, the unclear problem statement makes it difficult to tell which is the correct interpretation.

- anonymous

you can put a fence around a regular lot. The length of the lot must be at least 60ft. The cost of the fence along the length is $1.50 per foot. And the width is 2$ per foot. The total cost can not excede $360
what is the maximum length if the width is 36ft?

- tkhunny

I get it. Still just as unclear. Is a "regular" lot "square"? Is there a side or two missing? Perhaps there was a drawing?

- anonymous

no drawing itz rectangle

- tkhunny

Well, there you go. \(y \le 36\;ft\)

- anonymous

you don't need to know whether it's a square or a rectangle the maximum length could make a square, it could maybe make a rectangle. All it asks is to maximize the length given the restriction and an equation. Tomo put it correctly. The max length would be obtained by using the most money therefore you can make it an equality. The restriction is simply a guideline it seems

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