Ann wants to build an enclosed area behind her house. One wall of the enclosed area will be the back of the house. She needs the total to be 120 sq feet. She wants to minimize the cost of fence materials. For the sides (W) fence materials cost $3/ft, for the length (L) they cost $5/ft. What dimensions should her fence be?
c = 2*w*3 + 5 L = 6 w + 5 L
L = 120/w
so
c = 6 w + 600/w
dc/dw = 6 - 600/w^2 = 0 for min or max
w^2 = 100
w = 10
L = 12
Well, Ann certainly wants to maximize her savings, doesn't she? Let's put on our thinking cap and calculate the dimensions that will minimize her cost.
Let's assume the width of the enclosed area is W, and the length is L. We know that one side of the enclosed area will be the back of the house, which means we only need to build fences on the remaining three sides.
The equation we need to solve is: 2W + L = 120 (since there are two sides of width and one side of length, and the total area should be 120 sq ft).
Now, let's determine the cost of the fence materials. The sides (W) cost $3/ft, so the cost for both sides would be 2W * $3 = 6W dollars. The length (L) costs $5/ft, so the cost for the length would be L * $5 dollars.
Since we want to minimize the cost, we can express it as a function: Cost = 6W + 5L.
To minimize the cost, we can substitute the value of L from our equation into the cost function. So, we have: Cost = 6W + 5(120 - 2W).
Now, let's simplify this equation: Cost = 6W + 600 - 10W = -4W + 600.
To minimize the cost, we need to find the value of W that makes this equation as small as possible. And since we're all about minimizing costs, we'll aim for the smallest value.
Since we're dealing with a straight line, the smallest value of W that satisfies 2W + L = 120 is when W = 0. However, that doesn't make sense in this context.
So, let's say W = 10. That means L = 120 - 2W = 100. Therefore, the dimensions that would minimize Ann's cost are 10 feet width and 100 feet length.
Keep in mind that this answer assumes we are only looking for the cheapest way to enclose an area of 120 sq ft. If there are other factors, such as aesthetics or functionality, Ann might want to consider those as well.
To minimize the cost of fence materials, Ann should determine the dimensions of the fence that would enclose an area of 120 sq feet while using the least amount of expensive fencing materials.
Let's assume the width of the enclosed area is W, and the length is L.
Given that one wall of the enclosed area will be the back of the house, we can say the length (L) will be the same as the width of the house.
The area of the enclosed space can be calculated by multiplying the width and length: A = W * L.
We know that A = 120 sq feet, so we can substitute that value into the equation:
120 = W * L
Since L = W, we can rewrite the equation as:
120 = W * W
Simplifying further:
120 = W^2
To find the value of W (width), we'll take the square root of both sides:
√(120) = √(W^2)
Approximately, √(120) = 10.95 (rounded to two decimal places)
So, the width of the enclosed area should be around 10.95 feet.
Since the length (L) is the same as the width (W), the dimensions of the fence would be approximately 10.95 feet by 10.95 feet.
To minimize the cost of fence materials, Ann should determine the dimensions of the enclosure that result in the least amount of fencing required.
Let's assume the width (W) of the enclosure is x feet. As mentioned, one wall of the enclosure will be the back of Ann's house. Therefore, the length (L) of the enclosure will be determined based on the area required (120 sq feet) and the width (W).
The formula for the area of a rectangle is A = L × W, where A is the area, L is the length, and W is the width.
In this case, we know that A = 120 square feet. Thus, the equation becomes:
120 = L × W
To minimize the cost of the fence materials, we need to minimize the amount of fencing used. The amount of fencing required will depend on the length (L) and the width (W).
The amount of fencing required for the sides (W) will be the perimeter of the rectangle, which is 2W. Since fence materials for the sides cost $3/ft, the cost for the sides will be 3 × 2W = 6W dollars.
The amount of fencing required for the length (L) will be the perimeter minus the width (W), which is 2(L - W). Since fence materials for the length cost $5/ft, the cost for the length will be 5 × 2(L - W) = 10(L - W) dollars.
To minimize the cost, we need to find the values of L and W that give the minimum value for the total cost (6W + 10(L - W)).
To solve this optimization problem, we can substitute the value of L from the earlier equation (120 = L × W) into the cost formula to obtain a formula in terms of only one variable, W.
Let's now substitute the value of L from 120 = L × W into the cost formula:
Cost = 6W + 10(L - W) = 6W + 10(120/W - W) = 6W + 1200/W - 10W
Now, we can simplify the formula:
Cost = 6W + 1200/W - 10W = -4W + 1200/W
To find the minimum cost, we need to find the value of W that minimizes the Cost formula. This can be done by taking the derivative of the Cost equation with respect to W and setting it equal to zero:
dCost/dW = -4 + (-1200/W^2) = 0
Simplifying:
-4 - 1200/W^2 = 0
-1200/W^2 = 4
W^2 = -1200/4
W^2 = -300
Since we cannot have a negative width, it is clear that there is no real solution to this equation. Therefore, there are no real dimensions for the fence that will minimize the cost of fence materials while enclosing an area of 120 square feet.