A rectangular rose garden will be surrounded by a brick wall on three sides and by a fence on the fourth side. The area of the garden will be 1000m^2. The cost of the brick wall is $192/m. The cost of the fencing is $48/m. Find the dimensions of the garden so that the cost of the materials will as low as possible.
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To find the dimensions of the garden that will minimize the cost of the materials, we can start by assigning variables to the dimensions.
Let's say that the length of the garden is L and the width is W.
Since there are three sides surrounded by a brick wall, the total cost of the brick wall will be 3 times the perimeter of the garden multiplied by the cost of the brick per meter. So the cost of the brick wall is given by: 3 * (2L + W) * $192.
The remaining side is enclosed by a fence, so the cost of the fence will be the length of that side multiplied by the cost of the fence per meter. Thus, the cost of the fence is given by: W * $48.
The total cost of the materials is the sum of the cost of the brick wall and the cost of the fence. So, the total cost is: 3 * (2L + W) * $192 + W * $48.
The area of the garden is given as 1000 m^2, so L * W = 1000.
To minimize the cost, we need to find the dimensions that minimize the total cost. To do this, we can express the total cost as a function of one variable, either L or W, by substituting the area equation into the cost equation.
So, the total cost can be expressed as: C = 3 * (2(1000/W) + W) * $192 + W * $48.
To find the minimum cost, we can take the derivative of this expression with respect to W, set it equal to zero, and solve for W.
dC/dW = 0
Simplifying the equation and solving for W will give us the value of W that minimizes the cost. Substituting this value of W back into the area equation will allow us to find the corresponding value of L.
Let's solve this step-by-step.
To find the dimensions of the garden that will result in the lowest cost of materials, we can use optimization techniques. Let's start by introducing some variables:
Let's call the length of the garden L and the width of the garden W.
The garden is surrounded by a brick wall on three sides, so the sum of the lengths of those three sides (perimeter) will be L + 2W.
The remaining side is surrounded by a fence with length W.
We know that the area of the garden is 1000m^2, so we can write the equation:
L * W = 1000 (equation 1)
The cost of the brick wall is $192/m, and the cost of the fence is $48/m. Therefore, the total cost of the materials can be expressed as:
Cost = (perimeter of brick wall) * $192 + (length of the fence) * $48
We can substitute the expressions for the perimeter and length of the fence into the cost equation:
Cost = (L + 2W) * $192 + W * $48
Cost = $192L + $384W + $48W
Cost = $192L + $432W (equation 2)
Now, we have two equations (equation 1 and equation 2) that relate the dimensions of the garden (L and W) and the cost of the materials (Cost).
To find the dimensions that minimize the cost, we need to find the values of L and W that minimize the Cost. We can achieve this by finding the critical points of Cost with respect to L and W.
To do this, we take partial derivatives of Cost with respect to L and W separately and set them equal to zero:
∂Cost/∂L = $192 + 0 = 0
∂Cost/∂W = $432 + 0 = 0
Solving these equations will give us the values of L and W that minimize the cost:
$192 + 0 = 0 => L = 0 (not possible)
$432 + 0 = 0 => W = 0 (not possible)
Since we can't have zero dimensions, the minimum cost occurs at the boundaries of the feasible region, which means we need to consider the endpoints for optimization.
Given that the garden area is 1000m^2 (L * W = 1000), we can substitute this into equation 1 and solve for L:
L * W = 1000
L = 1000/W
Substituting this value for L into equation 2, we get:
Cost = $192L + $432W
Cost = $192(1000/W) + $432W
Cost = $192000/W + $432W
To find the minimum cost, we can find the value of W that minimizes this Cost function.
One way to do this is by finding the critical points, which occur when the derivative of Cost with respect to W is zero. Let's calculate:
dCost/dW = - $192000/W^2 + $432
Setting this derivative equal to zero and solving for W:
- $192000/W^2 + $432 = 0
$432 = $192000/W^2
W^2 = $192000/$432
W^2 = 444.44
W = √444.44
W ≈ 21.08
Now, we can substitute the value of W back into equation 1 to find the corresponding value of L:
L * W = 1000
L * 21.08 = 1000
L ≈ 47.44
Therefore, the dimensions that minimize the cost of materials are approximately L = 47.44m and W = 21.08m.
If you are the same "anonymous" , give yourself a nickname, and try one of these yourself.
I have helped you and done two of these for you now, I think it is time you show me what you have done so far.
This one is the easiest of the three.