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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
which can then be substituted into the first equation.
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.
and a maximum or minimum occurs at its roots.
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
Therefore, the maximum enclosable area by a wall of length 1050ft is 68906.25 square feet.
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