A rancher wants to build a rectangular pen with an area of 150 m2. Let W be the width of the pen and L be the length of the pen.
a. Find an equation for the perimeter P in terms of W and L .
b. Use the given area to write an equation that relates W and L .
c. Find the pen dimensions that require the minimum amount of fencing.
Width =
Length =
a. The perimeter P of a rectangular pen is the sum of all the sides, which is twice the length plus twice the width: P = 2L + 2W.
b. The area A of a rectangular pen is given by A = L * W. In this case, the area is 150 m^2, so the equation relating W and L is L * W = 150.
c. To find the dimensions that require the minimum amount of fencing, we can use the fact that for a given area, the perimeter is minimized when the dimensions are equal. This means that W = L.
Substituting W = L into the equation relating area and dimensions, we get L * L = 150, which simplifies to L^2 = 150. Taking the square root of both sides gives L = sqrt(150).
So, the length is approximately L ≈ 12.25 m, and since W = L, the width is also approximately W ≈ 12.25 m.
a. The perimeter of a rectangle is given by the equation P = 2W + 2L.
b. The area of a rectangle is given by the equation A = WL. Since we are given that the area is 150 m2, the equation relating W and L is 150 = WL.
c. To find the dimensions that require the minimum amount of fencing, we can use calculus optimization. Let's solve the equation 150 = WL for one variable, substitute it into the perimeter equation, and then find the minimum perimeter.
From the area equation, we can solve for W in terms of L:
W = 150/L
Substituting this value of W into the perimeter equation gives:
P = 2(150/L) + 2L
Simplifying this equation further:
P = 300/L + 2L
To find the minimum perimeter, we take the derivative of P with respect to L and set it equal to 0:
dP/dL = -300/L^2 + 2
-300/L^2 + 2 = 0
-300/L^2 = -2
L^2 = 150
Taking the square root of both sides:
L = √150
L ≈ 12.25 m
Substituting this value of L back into the equation W = 150/L gives:
W = 150/(√150)
W ≈ 10.95 m
So the pen dimensions that require the minimum amount of fencing are approximately Width = 10.95 m and Length = 12.25 m.
P = 2L + 2W
150 = LW
150/L = W
P = 2L + 300/L
Can you finish from here?