SOLUTION: This week, I'm studying the axis of symmetry and vertexes in my math class. I'm having trouble with the word problems. Here's one of them: A gardener has enough money to build 260

Question 735118: This week, I'm studying the axis of symmetry and vertexes in my math class. I'm having trouble with the word problems. Here's one of them:
A gardener has enough money to build 260 feet of garden wall. He plans to use an existing wall for one side of the enclosure. What is the largest rectangular area he can enclose?
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
A gardener has enough money to build 260 feet of garden wall. He plans to use an existing wall for one side of the enclosure. What is the largest rectangular area he can enclose?
.
Let x = width
and y = length
.
perimeter:
260 = 2x+y (equation 1)
.
area:
area = xy (equation 2)
.
Solve equation 1 for y:
260 = 2x+y
260-2x = y
.
Substitute above into:
area = xy
area = x(260-2x)
area = 260x-2x^2
area = -2x^2 + 260x
.
This is a parabola that opens downwards (because the leading coefficient is negative). So, the vertex must be the maximum.
.
x-value at vertex is defined by the "axis of symmetry":
x = -b/(2a)
x = -260/(2(-2))
x = -260/(-4)
x = 65 feet
.
to find y, we substitute above into:
260 = 2x+y
260 = 130)+y
130 feet = y
.
Largest rectangular area is then:
65*130 = 8450 square feet

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