High School

A rancher has 10,000 linear feet of fencing and wants to enclose a rectangular field and then divide it into two equal pastures with an internal fence parallel to one of the rectangular sides. What is the maximum area of each pasture? Round to the nearest square foot.

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.

For more information about area of rectangle, refer:

https://brainly.com/question/26290174

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