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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?
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanx , you are right its a calculus optimization problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just 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. \[2w=1002l\] \[w=50l\] Substitute w into area equation and take the derivative of Q with respect to w. \[Q=l(50l)=50ll^2\] \[Q'(l)=502l\] 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Above I made a typo. When we take the derivative, we take it with respect to L, not W.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just 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!

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Nightie 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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) 100y2y^2=A Deriving to find the critical point: A´=1004y 1004y=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.

radar
 5 years ago
Best ResponseYou've already chosen the best response.0By gosh thats right. I was thinking (simply)LOL, that calc wouldn't be necessary. Wrong!!
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