Thread: Area question

I have to find the maximum area of a rectangle that already has a 100m fence in the ground, and I am given 200m to add onto it. However, my answer doesn't make sense:

A=b x h
P=300=2b + 2h
300=2(100+x)+2h
300=200+2x+2h
h=50-x
A=b(50-x)
A=50b-xb
dA/db = 50-x
0=50-x
x=50

Originally Posted by Dudealadude

I have to find the maximum area of a rectangle that already has a 100m fence in the ground, and I am given 200m to add onto it. However, my answer doesn't make sense:

A=b x h
P=300=2b + 2h
300=2(100+x)+2h
300=200+2x+2h
h=50-x
A=b(50-x)

b=100+x

A=(100+x)(50-x)=5000-50x-x^2

dA/dx=-2x-50

x=-25

A=50b-xb
dA/db = 50-x
0=50-x
x=50

Hence, you must create a kink in the 100m side so that it is 75m long after.

I think the purpose of the question is that one side has to be atleast 100m, in your answer they are both 75. No kinks allowed. Thanks though!

Originally Posted by Dudealadude

I think the purpose of the question is that one side has to be atleast 100m, in your answer they are both 75. No kinks allowed. Thanks though!

In that case x=0

This is because the graph is an inverted U-shape with an absolute maximum at x=-25.
Hence for x non-negative, the area will be maximised at x=0,
since any positive x will give a lower value for area.

therefore, h=50.