Asked by Melanie
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?
Answers
Answered by
bobpursley
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.
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.
Answered by
bobpursley
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.
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.
Answered by
Melanie
Thank you bobpursley!
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