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anonymous

  • 5 years ago

a rectangular animal enclosure is to be constructed having one side along an existing long wall and the other 3 sides fenced, if 100m of fence are available , whats the largest possible area for the enclosure?

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  1. anonymous
    • 5 years ago
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    I remember from geometry that the largest possible area of an area with a definite perimeter (100 m) is going to be a square with that perimeter. This would mean each side of the square would be 25 meters long. Area of a square with s = 25 is 25^2 = 625. Wasn't sure if this was a calculus optimization problem. If it is, I can do that too... answer's still the same, though.

  2. anonymous
    • 5 years ago
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    thanx , you are right its a calculus optimization problem

  3. anonymous
    • 5 years ago
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    Just so you can see it, I'll put the actual optimization stuff up. We write an equation for the perimeter and an equation for the area. I use the letter Q to represent the area that were are going to maximize so I remember to maximize it. \[2l+2w=100\] \[lw = Q\] Solve for L or W in the perimeter equation. \[2w=100-2l\] \[w=50-l\] Substitute w into area equation and take the derivative of Q with respect to w. \[Q=l(50-l)=50l-l^2\] \[Q'(l)=50-2l\] The derivative is equal to zero when there is either a maximum or a minimum at that point. 0 = 50 - 2L L = 25 In this case, we can assume that this is a maximum. However, for other problems in your problem sets and problems you undertake the real world, you'll want to check whether this L = 25 is a maximum or a minimum. Substitute L = 25 into perimeter equation to get W = 25.

  4. anonymous
    • 5 years ago
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    Above I made a typo. When we take the derivative, we take it with respect to L, not W.

  5. anonymous
    • 5 years ago
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    ahaa got it ,, tanx

  6. anonymous
    • 5 years ago
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    Just one problem... ;) One side is a wall that doesnt need a fence. So you have to distribute the 100m of fence to three sides instead of four. Think about it!

  7. radar
    • 5 years ago
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    Nightie points out that the fencing makes up only three of sides. Wouldn't the sides then only be 1/3 of 100M. Area then would be 100/3*100/3 or 10000/9 giving a value of 1111.111 sq. meters rather than the previous solution of 625 where length was computed as 25.

  8. anonymous
    • 5 years ago
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    You would think that's the logical solution, but if you do the calculations you will find out that it's not true. You'll find the answer by: 2y+x=100 xy=Area(A) 100y-2y^2=A Deriving to find the critical point: A´=100-4y 100-4y=0 100=4y 25=y Which gives the solution of two sides being 25m each and opposite the wall it's 50m, total area of 1250 sq. meters.

  9. radar
    • 5 years ago
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    By gosh thats right. I was thinking (simply)LOL, that calc wouldn't be necessary. Wrong!!

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