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a farmer with 10,000 meters of fencing wants to enclose a rectangular field and then divide it into two plots with a fence para...Asked by Joselito
A farmer with 10000 meters of fencing wants to enclose a rectangular field and divide it into two plots with a fence parallel to the sides. What is the largest area that can be enclosed?
Answers
Answered by
Reiny
He has 10 km of fence to fence in a field on a farm ??
Anyway ...
Let each of the lengths be y
let each of the shorter sides by x
so we need 2y + 3x
2y + 3x = 10000 ---> y = (10000-3x)/2
area = xy
= x(10000-3x)/2
= 5000x - (3/2)x^2
d(area)/dx = 5000-3x
= 0 for a max of area
3x = 5000
x = 5000/3 m or 1666 2/3 m
sub that into area
then area = 4,166,666.667 m^2
Anyway ...
Let each of the lengths be y
let each of the shorter sides by x
so we need 2y + 3x
2y + 3x = 10000 ---> y = (10000-3x)/2
area = xy
= x(10000-3x)/2
= 5000x - (3/2)x^2
d(area)/dx = 5000-3x
= 0 for a max of area
3x = 5000
x = 5000/3 m or 1666 2/3 m
sub that into area
then area = 4,166,666.667 m^2
Answered by
Anonymous
lika sum bodee
Answered by
Kameron
Will you do this same problem but the farmer has 400m of fencing.
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