a college is planning to construct a new parking lot. the parking lot must be rectangular and enclose 6000 square meters of land. A fence will surround the parking lot, and another fence parallel to one of the sides will divide the parking lot into two sections, what are the dimension in meters of the rectangular lot that uses the least amount of fencing?
been there, done that
To find the dimensions of the rectangular lot that uses the least amount of fencing, we need to consider the concept of optimization. In this case, we want to minimize the amount of fencing required.
Let's assume the length of the rectangular lot is L meters, and the width is W meters. Since the area of the lot is given as 6000 square meters, we have the equation:
L * W = 6000
Now we need to express the amount of fencing required in terms of L and W.
There are two sides of the parking lot that require fencing, each equal to L meters. The other two sides are divided by a fence, so each side requires 2W meters of fencing.
Therefore, the total amount of fencing required is:
2L + 2W = 2(L + W)
To minimize the fencing, we need to find the dimensions that minimize the total fencing, which is represented by 2(L + W).
Since we have an equation for the area of the lot (L * W = 6000), we can solve for one variable in terms of the other and substitute it in the equation for the fencing.
Let's solve for L in terms of W:
L = 6000 / W
Now, substitute L in the equation for the fencing:
2(6000 / W + W) = 2(6000/W + W)
Next, simplify the equation:
2(6000/W + W) = 2(6000/W + W) = (12000 / W) + 2W
To find the dimensions that minimize the fencing, we need to find the derivative of this expression with respect to W and set it equal to zero:
d(12000 / W) / dW + d(2W) / dW = 0
To find the derivative of 12000/W with respect to W, we can use the quotient rule, which states:
d(u/v) / dW = (v * du/dW - u * dv/dW) / v^2
Let's find the derivative of 12000/W with respect to W using the quotient rule:
d(12000 / W) / dW = (W * 0 - 12000 * 1) / W^2 = -12000 / W^2
Now, let's find the derivative of 2W with respect to W:
d(2W) / dW = 2
Now, set the derivative equal to zero and solve for W:
-12000 / W^2 + 2 = 0
Multiply both sides by W^2:
-12000 + 2W^2 = 0
Rearrange the equation:
2W^2 = 12000
Divide both sides by 2:
W^2 = 6000
Take the square root of both sides:
W = sqrt(6000) ≈ 77.46
Now substitute the value of W to find L:
L = 6000 / W = 6000 / 77.46 ≈ 77.46
Therefore, the dimensions of the rectangular lot that uses the least amount of fencing are approximately 77.46 meters by 77.46 meters.