SOLUTION: A retangular field is to be enclosed by 500m of fence. What dimensions will give the maximum area? What is the maximum area? Be sure to use "let statements"
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Question 196947: A retangular field is to be enclosed by 500m of fence. What dimensions will give the maximum area? What is the maximum area? Be sure to use "let statements"
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Let's solve this one in general, that is for any given length of fence.
Let F represent the length of fence available.
Let w represent the width of the field.
Let l represent the length of the field.
The perimeter of a rectangle is:
So, in order to enclose the field,
So
The area of a rectangle is the length times the width so a function for the area in terms of the width is:
 = \left(\frac{F - 2w}{2}\right)w)
 = \frac{Fw - 2w^2}{2})
 = \frac{F}{2}w - w^2)
Algebra Solution:
The area function is a parabola, opening downward, with vertex at:
Since the parabola opens downward, the vertex represents a maximum value of the area function. The value of the width that gives this maximum value is one-fourth of the available fencing. Therefore, the shape must be a square, and the area is the width squared.
Calculus Solution:
The area function is continuous and twice differentiable across its domain, therefore there will be a local extrema wherever the first derivative is equal to zero and that extreme point will be a maximum if the second derivative is negative.
 = \frac{F}{2} - 2w)
Therefore the maximum area is obtained when
And that maximum area is:
 = \frac{F}{2}\left(\frac{F}{4}\right) - \left(\frac{F}{4}\right)^2 = \left(\frac{F}{4}\right)^2 = w^2)
John
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