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April 24, 2014

April 24, 2014

Posted by **JENNY** on Tuesday, March 30, 2010 at 7:25pm.

- CALCULUS -
**Reiny**, Tuesday, March 30, 2010 at 7:40pmlet the length be x feet, and the width be y feet

we know xy = 3000

so y = 3000/x

You don't say if the fence is along the length or along the width.

I will work it as if it the length has the fence.

If otherwise, just go through it again by changing the equation.

Cost = 10x + 10y + 25(2y)

= 10x + 60y

= 10x + 60(3000/x)

d(cost)/dx = 10- 180000/x^2 = 0 for a minimum Cost

10 = 180000/x^2

x^2 = 18000

x = 134.16

y = 3000/x = 22.36

minimum cost = 10(134.16) + 60(22.36) = 2683.28

- CALCULUS -
**JENNY**, Tuesday, March 30, 2010 at 8:59pmthank u sooo much

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