Answer :
For a 10,000 linear feet of fencing and wants to enclose a rectangular field, the maximum area of each pasture is equals to the 2,083,333.33 square feet .
A rancher wants to enclose about 10,000 linear feet of fencing in a rectangular field. There are two equal pastures. Let, W the Width and L the Length with 3 pieces. The total length will be equal to 2W + 3L. As we know, the Area of rectangle, A = L×W ---(1)
Perimeter of rectangle = 10,000 feet, so sum of all sides of rectangle, 2W + 3L = 10000
Solve for determining value of L, 3L
= 10000 - 2W
[tex]L = \frac{ 10000 - 2W }{3}[/tex]
Substitutes for L into the first equation, A = L×W
[tex]A = W(\frac{ 10000 - 2W }{3})[/tex]
For maximum area, set the 1st derivative = 0.
differentiating above Area equation w.r.t W,[tex] A = \frac{(10000 \: W - 2W^2)}{3}[/tex]
[tex] \frac{dA}{dW} = \frac{d(\frac{10000 \: W - 2W^2}{3})}{dW}[/tex]
=> [tex]\frac { 1}{3}(10000 - 4W) = 0 [/tex]
=> W = 2500 meters
Now, using above relation, 2W + 3L= 10000
=> 5000 + 3L = 10000
=> 3L = 5000
=> L = 1667 meters
Area = 2500× 5000/3 = 4,166,666.67 sq feet. Now, maximum area of each pasture
= 4,166,666.67/2 = 2,083,333.33333 sq. feet. Hence, required value is 2,083,333.33 sq. feet.
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