A fence is to be built to enclose a rectangular area of 200 square feet. The fence along three sides is to be made of material that costs 3 dollars per foot, and the material for the fourth side costs 13 dollars per foot.
Find the length L and the width W (with W less than or equal to L) of the enclosure that is most economical to construct.
To find the most economical dimensions for the enclosure, we need to consider the cost of the fence for different dimensions.
Let's assume the length of the rectangular enclosure is L and the width is W (with W ≤ L).
The area of a rectangle is given as A = L * W, and we know that the area is 200 square feet, so we can write the equation: L * W = 200.
Now, we need to express the cost of the fence in terms of L and W.
The fence along three sides will cost $3 per foot, and the fourth side will cost $13 per foot. The perimeter of the rectangle is P = 2L + W because two sides will have length L and one side will have length W. The total cost of the fence is then:
C = 3(2L + W) + 13W.
Now, we want to minimize the cost function C with respect to the dimensions L and W. In other words, we want to find the values of L and W that minimize C.
To do this, we can take the derivative of C with respect to L and W separately and set them equal to zero to find critical points:
∂C/∂L = 0 and ∂C/∂W = 0.
Differentiating C with respect to L:
∂C/∂L = 6 + 13(∂W/∂L).
Differentiating C with respect to W:
∂C/∂W = 3 + 13.
Setting both derivatives equal to zero and solving:
6 + 13(∂W/∂L) = 0
∂W/∂L = -6/13.
Since W ≤ L, the value of ∂W/∂L should be negative. However, -6/13 is positive, which means there are no critical points within the given constraints. Therefore, we need to examine the endpoints of the possible range for L and W.
Since L * W = 200, W = 200/L.
Now, let's consider some reasonable values of L and calculate the corresponding cost C:
When L = 1, W = 200.
C = 3(2(1) + 200) + 13(200) = 6600.
When L = 2, W = 100.
C = 3(2(2) + 100) + 13(100) = 803.
When L = 4, W = 50.
C = 3(2(4) + 50) + 13(50) = 545.
When L = 5, W = 40.
C = 3(2(5) + 40) + 13(40) = 515.
Thus, the smallest cost is $515, which corresponds to the dimensions L = 5 and W = 40.
To find the length and width of the enclosure that is most economical to construct, we need to minimize the cost of the fence.
Let's assume the width of the enclosure is W (in feet) and the length is L (in feet), where W ≤ L.
We are given that the area of the enclosure is 200 square feet, so we can write the equation:
L * W = 200
We also know that the cost of the fencing material for three sides (two sides of length L and one side of width W) is $3 per foot, and the cost for the fourth side (of length W) is $13 per foot.
The cost of the fencing material for the three sides is:
3 * (2L + W) dollars
The cost of the fencing material for the fourth side is:
13 * W dollars
Therefore, the total cost, C, of the fence is:
C = 3 * (2L + W) + 13 * W
Now, we need to express C in terms of a single variable, either L or W, so we can find its minimum value.
Using the equation L * W = 200, we can write W in terms of L:
W = 200 / L
Substituting this into the expression for C:
C = 3 * (2L + 200/L) + 13 * (200 / L)
Simplifying:
C = 6L + 600/L + 2600/L
Now, we need to take the derivative of C with respect to L and set it equal to zero to find the minimum value of C.
dC/dL = 6 - 600/L^2 - 2600/L^2
Setting dC/dL equal to zero:
6 - 600/L^2 - 2600/L^2 = 0
Combining like terms:
6L^2 - 600 - 2600 = 0
Simplifying:
6L^2 - 3200 = 0
Dividing by 6:
L^2 = 3200/6 = 533.33
Taking the square root:
L = √533.33 ≈ 23.09
Since W ≤ L, we can substitute L = 23.09 into the equation W = 200/L to find the width:
W = 200/23.09 ≈ 8.66
So, the length L of the enclosure that is most economical to construct is approximately 23.09 feet, and the width W is approximately 8.66 feet.