Question 189837This question is from textbook
: A developer wants to enclose a rectangle. If the developer has 200 feet of fencing and does not enclose one side, what is the largest area that can be enclosed?
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Answer by solver91311(24713)   (Show Source):
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The answer is 5000 square feet. How you get there depends on whether you are studying Calculus and investigating the nature of local extrema using the first and second derivatives, or you are in Algebra investigating the properties of parabolas.
Common to both methods is the Area function. If two of the fenced sides of the rectangle are represented by x, and you have 200 feet of fence available, then the third side is 200 - 2x because you subtract the two pieces of fence that are x feet long from the available 200 feet of fencing to get the remainder which must be the length of the third side. The area is the length times the width, so:
 = (200 - 2x)x = 200x - 2x^2)
Algebra Solution
The function is a parabola opening downward. You know it opens downward because the coefficient on the high-order term is negative. Therefore the vertex is a maximum and the value of the function at the vertex will give us the maximum possible area.
The x-coordinate of the vertex of a parabola whose equation is in the form:
 = ax^2 + bx + c)
is given by:
For this problem:
} = 50)
So we achieve the maximum area when the value of x is 50.
To determine what that maximum area is, we need to evaluate:
 = 200(50) - 2(50)^2 = 10000 - 5000 = 5000)
Calculus Solution
) is a polynomial function, therefore it is continuous and differentiable across its entire domain. Therefore, a local extreme point will be found at any value a where  = 0)
So take the first derivative of the area function:
 = 200 - 4x)
Set it equal to zero, and then solve:
Now we need to investigate whether this local extreme is a minimum or a maximum. The first derivative is a polynomial, therefore continuous and differentiable, so we can take the second derivative:
)}{dx^2} = -4)
Which is negative for all x in the domain of A, therefore the extremum is a maximum.
Again, to determine what that maximum area is, we need to evaluate:
John
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