Asked by Alex
A rectangular field is to be fenced in on four sides with a fifth piece of fencing placed Parallel to one of the shorter sides, so that the field is split in two parts. If 1600 m of fencing is available, find the largest possible area for this enclosure. What dimensions give this maximum area?
Answers
Answered by
Reiny
let the length be y m
and each of the shorter pieces (the width) be x
we know 2y + 3x = 1600
y = 800 - 3x/2
area = xy
= x(800-3x/2)
= 800x - 3x^2 /2
find the first derivative, set it equal to zero and find x
then find y, and the area xy
and each of the shorter pieces (the width) be x
we know 2y + 3x = 1600
y = 800 - 3x/2
area = xy
= x(800-3x/2)
= 800x - 3x^2 /2
find the first derivative, set it equal to zero and find x
then find y, and the area xy
Answered by
scott
xmax is on the axis of symmetry of the parabola
-b / 2a = -800 / (2 * -3/2) = 800 / 3
-b / 2a = -800 / (2 * -3/2) = 800 / 3
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