Question

A rectangular area is to be fenced in with 300 feet of chicken wire. find the maximum area that can be enclosed.

Answer 1

Let the lengths of the sides of the rectangle be x and y. Then A(Area) = xy and 2(x+y)=300. You can use substitution to make one equation that gives A in terms of either x or y instead of both.

2(x+y) = 300
x+y = 150
y = 150-x

A=x(150-x) <--(substitution)

The resulting equation is a quadratic equation that is concave down, so it has an absolute maximum. The x value of this maximum is going to be halfway between the zeroes of the function. The zeroes of the function can be found by setting A equal to 0:
0=x(150-x)
x=0, 150

So halfway between the zeroes is 75. Plug this into the quadratic equation to find the maximum area.

A=75(150-75)
A=75*75
A=5625

So the maximum area that can be enclosed is 5625 square feet.

Answer 2

The maximum area that can be enclosed is 5625 feet²

A rectangle is a quadrilateral in which opposite sides are equal and parallel to each other.

Let x represent the length and y represent the width. Hence:

perimeter = 2(length + width)

300 = 2(x + y)

150 = x + y

y = 150 - x

The area of a rectangle (A) = length * width

A = x * (150 - x)

A = 150x - x²

The maximum area is at dA/dx = 0, hence:

dA/dx = 150 - 2x

150 - 2x = 0

2x = 150

x = 75 feet

y = 150 - x = 150 - 75 = 75 feet

The maximum area = x * y = 75 * 75 = 5625 feet²

The maximum area that can be enclosed is 5625 feet²

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