A rectangular area is to be fenced off with a wall 1050 ft in length. Find the dimensions of the fence for it to be enclosed. What is the maximum area the wall can enclose?

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labrat256 | (Level 2) Adjunct Educator

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A rectangular wall encloses an area of `A=lxxw`  where l is the length of the rectangle and w is the width. The perimeter of this wall is `P=2(l+w)=1050ft` .

This can be rearranged as 

`w=1050/2-l`

which can then be substituted into the first equation.

`A=lxx(525-l)`

which gives

`A=-l^2+525l`

which needs to be maximised. The maximum of the area occurs where the gradient is zero. To find this, you need to differentiate the equation.

`(dA)/(dl)=-2l+525`

and a maximum or minimum occurs at its roots. 

`-2l+525=0`

`2l=525`

l=262.5

As you might see, this is a quarter of 1050, so w=l. This isn't suprising as a rectangle can only enclose a lesser area than a square of the same perimeter. 

As we have now established that l=w=262.5 for the maximum area, we can calculate that 

`A=262.5^2=68906.25ft^2`

Therefore, the maximum enclosable area by a wall of length 1050ft is 68906.25 square feet.

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