# Area question

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• Nov 21st 2010, 12:20 PM
Dudealadude
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
• Nov 21st 2010, 01:34 PM
Archie Meade
Quote:

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.
• Nov 21st 2010, 02:26 PM
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!
• Nov 21st 2010, 02:56 PM
Archie Meade
Quote:

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.