anonymous
  • anonymous
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,,
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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radar
  • radar
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?
anonymous
  • anonymous
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...
radar
  • radar
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?

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anonymous
  • anonymous
yes i agree
radar
  • radar
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?
anonymous
  • anonymous
agreed..
radar
  • radar
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.
radar
  • radar
Did you get the same?
radar
  • radar
Do you have any questions?
anonymous
  • anonymous
i see where i went wrong... i kept the negative sign on the -5 when i divided it...
radar
  • radar
O.K, those signs can be very tricky. good luck in your studies
anonymous
  • anonymous
thank you a lot ..

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