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?

Question

anonymous · Answer

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.

anonymous · Answer

thanx , you are right its a calculus optimization problem

anonymous · Answer

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.

anonymous · Answer

Above I made a typo.  When we take the derivative, we take it with respect to L, not W.

anonymous · Answer

ahaa got it ,, tanx

anonymous · Answer

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!

radar · Answer

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.

anonymous · Answer

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.

radar · Answer

By gosh thats right.  I was thinking (simply)LOL, that calc wouldn't be necessary.  Wrong!!