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Question

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3. A farmer with 1500 ft offence wants to enclose a rectangular area and he divides it into four equal pens with fencing parallel to one side of the rectangle. a) Give the equation for the amount of fencing needed. b) Give the equation for the area of the total enclosures. c) Eliminate one of the variables and, using calculus, find the dimensions of the largest possible total area of the pens?

Explanation / Answer

Solution:

If you drew it out you would have 2 long sides and 5 wide segments to break it up into 4 pens.

Perimeter = 2*L + 5*W = 1500

Solve for W

W = 300 - 2/5*L

Area of each pen = L/4*W

Substitute in W

Area = L/4*300 - L/4*2/5*L

Area = 75*L - 0.1*L2

The maximum Area occurs at the apex of the Area equation, or the derivative with respect to L = 0

0 = 75 - 0.2*L

L = 375 ft

Substitutethat back into the perimeter equation and solve for W

2*L + 5*W = 1500

2*375 + 5*W = 1500

5*W = 1500 - 750

W = 150 ft

Area for one pen = L/4*W = 375/4*150 = 14,062 ft2

Area for all the pens = L*W = 375*150 = 56,250 ft2

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