Asked by Jackie
A farmer uses 1034 meters of fencing to enclose a rectangular region and also to subdivide the region into three smaller rectangular regions by placing a fence parallel to one of the sides. Find the demensions that produce the greatest enclosed area.
Answers
Answered by
Reiny
After making your sketch and creating the 3 rectangles
let the length of each rectangle be x, let the width of each rectangle be y
So the area of the whole enclosed area = 3xy
also : 4x + 6y = 1034
6y = 1034 - 4x
y = (1034 - 4x)/6
area = 3xy - 3x(1034-4x)/6
= 517x - 2x^2
If you know Calculus ....
d(area)/dx = 517 - 4x
= 0 for a max of area
x = 517/4 =129.25
y = 86.167 m
(maximum area = 3xy = 33411.125
Check by taking a value of x slightly smaller and slightly larger and finding their areas
e.g
x =129 , then y = 86.3333
area = 3xy = 33411 , just a bit smaller
x = 130, then y = 85.6666..
area = 3xy = 33410 , again a bit smaller
my answer is plausible.
let the length of each rectangle be x, let the width of each rectangle be y
So the area of the whole enclosed area = 3xy
also : 4x + 6y = 1034
6y = 1034 - 4x
y = (1034 - 4x)/6
area = 3xy - 3x(1034-4x)/6
= 517x - 2x^2
If you know Calculus ....
d(area)/dx = 517 - 4x
= 0 for a max of area
x = 517/4 =129.25
y = 86.167 m
(maximum area = 3xy = 33411.125
Check by taking a value of x slightly smaller and slightly larger and finding their areas
e.g
x =129 , then y = 86.3333
area = 3xy = 33411 , just a bit smaller
x = 130, then y = 85.6666..
area = 3xy = 33410 , again a bit smaller
my answer is plausible.
Answered by
arlene
What is the answer for 104.19 + 4.9
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