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

Now, we can write the area of the rectangular playground.

A = ℓ \cdot w \Leftrightarrow A = ℓ (100 - 2 ℓ)

This is a quadratic function that models the area of the playground. Let's rewrite it in standard form.

A (ℓ) = ℓ (100 - 2 ℓ) ⇕ A (ℓ) = - 2 ℓ^{2} + 100 ℓ + 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 = - \frac{b}{2 a} \Leftrightarrow ℓ = - \frac{1 0 0}{2 ( - 2 )}

Let's simplify above quotient to calculate

ℓ .

ℓ = - \frac{1 0 0}{2 ( - 2 )}

MultPosNeg

$a (- b) = - a \cdot b$

ℓ = - \frac{1 0 0}{- 4}

▼

Simplify right-hand side

MoveNegDenomToFrac

Put minus sign in front of fraction

ℓ = - (- \frac{1 0 0}{4})

NegNeg

$- (- a) = a$

ℓ = \frac{1 0 0}{4}

CalcQuot

Calculate quotient

ℓ = 25

The length that maximizes the area of the playground is

ℓ = 25 ft .

To know what is the greatest area, we must find

A (25) .

A (ℓ) = - 2 ℓ^{2} + 100 ℓ

Substitute

$ℓ = 25$

A (25) = - 2 (25)^{2} + 100 (25)

▼

Simplify right-hand side

CalcPow

Calculate power

A (25) = - 2 (625) + 100 (25)

Multiply

A (25) = - 1250 + 2500

AddTerms

Add terms

A (25) = 1250

Consequently, the greatest area we can fence is

1250 ft^{2} .

Exercise