A school is fencing in a rectangular area for a playground. It plans to enclose the playground using fencing on three sides (the fourth is a wall) . The school has budgeted enough money for 75 ft of fencing material and would like to make a playground with an area of 600 ft^2. Solve for the width of the fence by "completing the square".

let the width be x

let the length be y, along the wall

so 2x + y = 75
y = 75-2x

area = xy = 600
x(75 - 2x) = 600
-2x^2 + 75x = 600
x^2 - 37.5x = -300
x^2 - 37.5x + 351.5625 = -300 + 351.5625
(x - 18.75)^2 = 51.5625
x - 18.75 = appr 7.18

x = appr 25.93
y = 23.139

thnks man you helped me big time

To solve for the width of the fence using completing the square, let's assume the length of the rectangular area is L and the width is W.

We know that the perimeter of the rectangular area is given by:
Perimeter = 2L + W

In this case, since the fourth side is a wall, we only need to fence three sides, which means the perimeter is equal to the given fencing material:
Perimeter = 75 ft

Also, the area of the rectangular area is given by:
Area = L × W = 600 ft²

By rearranging the perimeter equation, we can express L in terms of W:
L = (75 - W)/2

Substituting this value of L into the equation for the area, we get:
(75 - W)/2 × W = 600

Now, let's simplify and solve for W using completing the square.

Multiply both sides of the equation by 2 to eliminate the fraction:
(75 - W) × W = 1200

Expand and rearrange the equation:
75W - W² = 1200
W² - 75W + 1200 = 0

To complete the square, add and subtract half of the coefficient of W (√(75/2)² = 37.5) squared to both sides:
W² - 75W + (37.5)² - (37.5)² + 1200 = 0
(W - 37.5)² - 1406.25 + 1200 = 0
(W - 37.5)² - 206.25 = 0

Now we have a perfect square trinomial, so we can write the equation as:
(W - 37.5)² = 206.25

Take the square root of both sides to solve for W:
W - 37.5 = ± √206.25
W - 37.5 = ± 14.36

Solve for W by adding and subtracting the constants:
W = 37.5 ± 14.36

The two possible values for W are:
W₁ = 37.5 + 14.36 ≈ 51.86 ft
W₂ = 37.5 - 14.36 ≈ 23.14 ft

Therefore, the width of the fence is approximately 51.86 feet or 23.14 feet, depending on the specific requirements and constraints of the playground area.

To solve for the width of the fence by "completing the square," we'll set up a quadratic equation based on the given information.

Let's assume the length of the rectangular playground is L, and the width of the rectangular playground (which is also the width of the fence) is W. According to the problem, we need fencing on three sides: two sides with a length of L and one side with a length of W. The equation representing the amount of fencing required is:

2L + W = 75 ----------(Equation 1)

We also know that the area of the rectangular playground is equal to 600 square feet:

L * W = 600 ----------(Equation 2)

Now, let's solve equation 2 for L:

L = 600 / W

Substitute this value of L into Equation 1:

2(600 / W) + W = 75

Simplify the equation:

1200 / W + W = 75

Multiply through by W to remove the denominator:

1200 + W^2 = 75W

Rearrange the equation to get all terms on one side:

W^2 - 75W + 1200 = 0

To complete the square, we need to move the constant term (1200) to the other side of the equation:

W^2 - 75W = -1200

Now, take half of the coefficient of W (-75/2) and square it ((-75/2)^2 = 5625/4):

W^2 - 75W + 5625/4 = -1200 + 5625/4

Simplify the right side of the equation:

W^2 - 75W + 5625/4 = -4800/4 + 5625/4

Combine the fractions:

W^2 - 75W + 5625/4 = 825/4

Now we have a perfect square trinomial on the left side. Rewrite the left side as a perfect square:

(W - 75/2)^2 = 825/4

Take the square root of both sides:

W - 75/2 = ±√(825/4)

Simplify the right side:

W - 75/2 = ±√825/2

Now, solve for W:

W = 75/2 ±√825/2

This gives us two possible solutions. However, since the width of a fence cannot be negative, we'll only consider the positive solution:

W = 75/2 + √825/2

This is the width of the fence, which can be approximately calculated.