Posted by **Melanie** on Friday, July 15, 2011 at 4:51pm.

A farmer has 100 yeards of fencing with which to enclose two adjacent rectangular pens both bordering a river. The farmer does not need to fence the side with the river. What should the dimensions of the two pens together (a rectangle) be in order to yield the largest possible area?

- College Algebra -
**bobpursley**, Friday, July 15, 2011 at 5:51pm
Perimeter = 3w+4L=100

area= 2LW

but W=(100-4L)/3

area=2L (100-4L)/3

take the derivative with respect to L, set to zero, and solve for L, then go back and find W.

- College Algebra -
**bobpursley**, Friday, July 15, 2011 at 5:54pm
Ooops. You are not a calculus student yet.

area=2L(100-4L)/3= 200L/3 -8L^2/3

This is a parabola, with zeroes at L=25, and L=0. So max must be halfway between, L=25/2

if L = 25/2 then W=(100-50)/3 and that is it.

- College Algebra -
**Melanie**, Friday, July 15, 2011 at 6:41pm
Thank you bobpursley!

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