Hi, I need help.
The organisers of a sheepdog competition have 100 metres of fencing available to fence an enclosure for some sheep. They wish to make the area rectangular and as large as possible. What dimensions should the enclosure have to maximise the area if
a) the 100m of fencing is to be used for all four sides,
b) an existing wall forms one side and the fencing is used for the other three.
I've attempted A) by drawing the diagram and using 4x=100 and I got 25m as my answer, I want to know if there is another way.
As for b) I have tried everything and still can't figure it out, I would appreciate help.
October 4th 2010, 08:30 PM
a) Yes, there is another way of solving it:
x * y = maximum area
2x + 2y = 100 meters. Rearrange equation to get: y = -x + 50
Substitute the second equation into the first to get:
x * (-x + 50) = maximum area
y = x * (-x + 50)
Now expand it and then differentiate it and set dy/dx = 0 to get the length of x.
This is also how you solve b. But instead of having 2x + 2y = 100, this time it is x + 2y = 100. Can you solve it now?