A school wants to build a fence to enclose a rectangular 4-sided playground of 144m^2. What is the min cost for a fence if it costs $5.70 per meter of fencing?
2. The principal checks the budget an sees that only $228 is available for the purchase of fencing. What s the largest rectangular 4-sided area that she can enclose?
I am stuck on question 2 but I know how to do question 1. The two questions are linked!
228/5.7 = 40 meters
40 / 4 = 10 meters = 100 square meters
To solve question 1, we need to find the minimum cost for a fence to enclose a rectangular playground of 144m². We are given that the cost per meter of fencing is $5.70.
To find the perimeter of the rectangular playground, we need to calculate the sum of all sides since the fence will enclose all four sides. Let's assume the length of the rectangular playground is L meters, and the width is W meters.
The perimeter of the rectangular playground can be calculated as: P = 2L + 2W
Given that the area of the playground is 144m², we have the equation: L * W = 144
Now, we can express one of the variables in terms of the other, substituting it into the perimeter equation:
L = 144 / W
Substituting this into the perimeter equation, we get:
P = 2(144 / W) + 2W
Simplifying further, we can rewrite this as:
P = 288/W + 2W
To find the minimum cost of the fence, we need to find the value of P that minimizes the cost. Since the cost is given by the perimeter multiplied by the cost per meter of fencing ($5.70), we can express the cost equation as:
Cost = 5.70 * P
Substituting the expression for P, we get:
Cost = 5.70(288/W + 2W)
To find the minimum cost, we can take the derivative of the cost equation with respect to W, set it equal to zero, and solve for W. However, solving this equation analytically can be complicated.
Therefore, an alternative approach is to graph the cost equation and find the value of W that corresponds to the minimum point on the graph. We can use tools like graphing calculators or software to plot the graph.
Now, moving on to question 2, we are given that the budget for the purchase of fencing is $228. We need to determine the largest rectangular 4-sided area that can be enclosed within that budget.
Using the same variables and equations as in question 1, we substitute the given budget into the cost equation and solve for W:
5.70(288/W + 2W) = 228
After simplifying, we obtain a quadratic equation in the form:
1.35W^2 - 228W + 1599.60 = 0
Solving this quadratic equation will give us the values of W. However, since we are looking for the largest rectangular area, we select the value of W that maximizes the area.
The area of the rectangle can be calculated as A = L * W. So, once we have the value of W from solving the quadratic equation, we can substitute it back into our area equation to find the corresponding value of L.
Thus, by solving the quadratic equation and finding the width and length values that maximize the area, we can determine the largest rectangular area that can be enclosed within the given budget.