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anonymous

  • 5 years ago

A pig rancher wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle (see the figure below). He has 410 feet of fencing available to complete the job. What is the largest possible total area of the four pens? if someone can help me find the constraint with this one i can do the rest... ??? im having trouble,,

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  1. radar
    • 5 years ago
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    Let x = the side that is parallel to the interior walls comprising the 4 pens in other words you will need 5x feet for this direction. Let y be the other dimension. So 2y + 5x = 410 ft Do you understand this line of reasoning?

  2. anonymous
    • 5 years ago
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    see i have this but for some reason ... its giving me a negative number for x or y when i solve for either of them, to be plugged back into the area formula...

  3. radar
    • 5 years ago
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    Good lets see why you get that. I would like to express y in terms of x. 2y=410-5x y=205-(5/2)(x) Do you agree with that?

  4. anonymous
    • 5 years ago
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    yes i agree

  5. radar
    • 5 years ago
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    Area will equal x*y or substituting for y A=(x)(205-(5/2)(x)) A=205x-(5/2) (x) (x) or\[A=(205x-(5/2)(x ^{2})\] Agree with that?

  6. anonymous
    • 5 years ago
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    agreed..

  7. radar
    • 5 years ago
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    dA/dx = 205-5x Let that = 0 for max/min -5x+205=0 5x=205 x=41 ft y=205-(5/2)41=102.5 A=4202.5 sq ft fence used is 5*41 + 2 * 102.5=205+205=410 ft.

  8. radar
    • 5 years ago
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    Did you get the same?

  9. radar
    • 5 years ago
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    Do you have any questions?

  10. anonymous
    • 5 years ago
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    i see where i went wrong... i kept the negative sign on the -5 when i divided it...

  11. radar
    • 5 years ago
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    O.K, those signs can be very tricky. good luck in your studies

  12. anonymous
    • 5 years ago
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    thank you a lot ..

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