Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
2. Standard Form of a Quadratic Function
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Exercise 38 Page 207

Notice that one side of the playground does not require fencing.

1250ft^2

Practice makes perfect

Let's define a variable for each dimension of the rectangle. Let w be the width and l be the length.

The perimeter of a rectangle is P=2l + 2w, but one side does not require fencing. This means the perimeter of the playground is P=2l + w. Since there is 100ft of fence, then 2l+w = 100, which means that w = 100 - 2l.

Now, we can write the area of the rectangular playground. A=l* w ⇔ A = l(100-2l) This is a quadratic function that models the area of the playground. Let's rewrite it in standard form. ccc A(l) = l(100-2l) ⇕ A(l) = -2l^2 + 100l + 0 Let's identify the coefficients of the above quadratic function. As we can see a = - 2, b = 100, and c = 0. Since a is a negative number the parabola opens downward, which implies that its maximum point corresponds to the vertex. Let's find the x-coordinate of the vertex. l = -b/2 a ⇔ l = -100/2( -2) Let's simplify above quotient to calculate l.
l = - 100/2(- 2)
l = - 100/- 4
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Simplify right-hand side
l = - (- 100/4)
l = 100/4
l = 25
The length that maximizes the area of the playground is l=25ft. To know what is the greatest area, we must find A(25).
A(l) = -2l^2 + 100l
A( 25) = -2( 25)^2 + 100( 25)
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Simplify right-hand side
A(25) = -2(625) + 100(25)
A(25) = -1250 + 2500
A(25) = 1250
Consequently, the greatest area we can fence is 1250ft^2.