A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Five hundred and forty feet of fencing is used. Find the dimensions of the playground that maximize the total enclosed area. What is the maximum area?
To find the dimensions of the playground that maximize the total enclosed area, we need to understand the problem and apply calculus concepts, specifically optimization.
Let's define the dimensions of the playground:
- Length: denoted by L
- Width: denoted by W
According to the problem statement, the playground is divided into two equal halves by a fence parallel to one side. This means that the length of each half will be half of the total length.
Now, let's set up the equation for the total fencing used in terms of L and W:
Since there are two lengths and one width (the fence is parallel to one side), the total fencing used is given by:
2L + 3W = 540 feet. (Equation 1)
Next, we want to express the area of the playground in terms of L and W.
The total area is the sum of the areas of both halves. Each half has an area of LW, so the total area is given by:
2(LW) = 2LW. (Equation 2)
Now we have two equations: Equation 1 and Equation 2.
To solve the problem, we need to eliminate one of the variables (L or W) to express the total area, Equation 2, in terms of just one variable. We can do this by rearranging Equation 1 to solve for L and substituting it into Equation 2.
Step 1: Solve Equation 1 for L.
From Equation 1: 2L + 3W = 540
Rearrange the equation to solve for L:
2L = 540 - 3W
L = (540 - 3W)/2
Step 2: Substitute the expression for L from Step 1 into Equation 2.
Total Area (A) = 2LW = 2[(540 - 3W)/2]W = (540 - 3W)W.
Now we have the equation for the total area in terms of just one variable, W.
To find the dimensions that maximize the total enclosed area, we need to find the value of W that maximizes the area. We can use calculus to find the critical points.
Step 3: Differentiate the equation for the total area with respect to W.
dA/dW = 540W - 3W^2
Step 4: Set the derivative equal to zero and solve for W.
540W - 3W^2 = 0
3W(180 - W) = 0
Setting each factor equal to zero, we get:
3W = 0 or 180 - W = 0
W = 0 (not a valid solution for a rectangle)
180 - W = 0
W = 180
Step 5: Substitute the value of W = 180 into the equation for L from Step 1 to find L.
L = (540 - 3W)/2 = (540 - 3(180))/2 = 540 - 540/2 = 270
So, the dimensions of the playground that maximize the area are L = 270 feet and W = 180 feet.
Step 6: Find the maximum area by substituting the values of L and W into the equation for the total area, A = (540 - 3W)W:
A = (540 - 3(180))180 = 360 * 180 = 64,800 square feet.
Therefore, the dimensions of the playground that maximize the total enclosed area are L = 270 feet and W = 180 feet, and the maximum area is 64,800 square feet.
To find the dimensions of the playground that maximize the total enclosed area, we can start by assigning variables to the unknown dimensions. Let's call the length of the playground "L" and the width "W".
According to the given information, the total length of the fencing used is 540 feet.
Since the playground is divided into two equal parts by a fence parallel to one side, the length of fencing used for the sides is L + L = 2L, and for the ends is W + W = 2W. So, the total length of fencing used is:
2L + 2W = 540
Simplifying the equation, we have:
2(L + W) = 540
Dividing both sides by 2, we get:
L + W = 270
Now, we need to express one variable in terms of the other, so we can substitute it into the area formula. Let's solve the above equation for L:
L = 270 - W
Now, let's express the area of the playground in terms of L and W. The area of a rectangle is given by the formula:
Area = Length × Width
Substituting L = 270 - W, we have:
Area = (270 - W) × W
Expanding the equation, we get:
Area = 270W - W^2
To maximize the area, we need to find the maximum point of this equation. The maximum point of the area occurs at the vertex of the parabola, which is given by the formula:
W = -b / 2a
For the quadratic equation -W^2 + 270W, the coefficients are a = -1 (the coefficient of W^2) and b = 270 (the coefficient of W).
Substituting the values into the formula, we get:
W = -270 / (2 × -1)
Simplifying, we have:
W = 270 / 2
W = 135
Now, substituting the value of W back into the equation L + W = 270, we can solve for L:
L + 135 = 270
L = 270 - 135
L = 135
So, the dimensions of the playground that maximize the total enclosed area are 135 feet by 135 feet.
To find the maximum area, we can substitute the values of L and W into the area formula:
Area = (270 - W) × W
Area = (270 - 135) × 135
Area = 135 × 135
Area = 18225
Therefore, the maximum area of the playground is 18,225 square feet.
the fence must be divided equally among the two lengths and three widths. So, 540/2 = 270, so the playground is largest if it is
135x90
You can verify this by finding the vertex of the parabola
y = x(540-2x)/3