A farmer wishes to enclose a rectangular pen with area 100 square feet next to a road. The fence along the road is to be reinforced and costs $34 per foot. Fencing that coast $16 per foot can be used for the other three sides. What dimensions for the pen will minimize the cost to the farmer. What is the minimum cost?
1X100
let the length be x, let the width be y
xy = 100
y = 100/x
cost = 34x + 16x + 16(2y) = 34x + 48y
= 34x + 48(100/x)
d(cost)/dx = 34 - 4800/x^2 = 0 for a minimum cost
34 = 4800/x^2
x^2 = 4800/34
x = appr. 11.88 ft
pen is 8.42 by 11.88 ft, with the 11.8 ft along the road
minimum cost = 34x + 4800/x = 807.96
check:
take x = 12 , cost = 808
take x = 11 , cost = 810.36
take x = 1 (Steve's answer) , cost = 4834
take x=100 , cost = 3448
The answer of 11.88 by 8.42 is correct for a min cost of $807.96
I was joking with the 1x100. Also, I notice that your solution is incorrect, because 16(2y) is not 48y.
The correct solution has appeared elsewhere as 8x12.5
To find the dimensions of the pen that minimize the cost to the farmer, we can express the cost function in terms of one variable and then find its minimum using calculus.
Let's denote the length of the pen along the road as x and the width as y. Since the area of the pen is given as 100 square feet, we have the equation xy = 100.
To find the cost function, we need to calculate the cost for each side of the pen. The fence along the road costs $34 per foot, so the cost for that side is 34x. The other three sides cost $16 per foot, resulting in a combined cost of 16(2x + y).
The total cost function is then given by:
C(x, y) = 34x + 16(2x + y)
To minimize the cost, we need to find the critical points of the cost function, which occur when its partial derivatives with respect to x and y are equal to zero.
Taking the partial derivative with respect to x, we get:
∂C/∂x = 34 + 16(2) = 34 + 32 = 66
Setting the partial derivative equal to zero, we have:
66 = 0
Since this equation has no solution, we move on to the next partial derivative.
Taking the partial derivative with respect to y, we get:
∂C/∂y = 16
Setting the partial derivative equal to zero, we have:
16 = 0
Similarly, this equation has no solution.
Since neither partial derivative is equal to zero, we conclude that there are no critical points for the cost function.
However, the cost function must have a minimum when considering the practical constraints of the problem. Since the cost cannot be negative, we know that the cost function will reach its minimum at the boundaries of the feasible region.
By substituting xy = 100 into the cost function, we can express it in terms of a single variable:
C(x) = 34x + 16(2x + 100/x)
Simplifying this expression, we get:
C(x) = 34x + 32x + 1600/x
To find the minimum, we can take the derivative of C(x) and set it equal to zero.
Taking the derivative, we get:
C'(x) = 34 + 32 - 1600/x²
Setting C'(x) equal to zero, we have:
34 + 32 - 1600/x² = 0
Combine like terms:
66 - 1600/x² = 0
Move 66 to the other side:
1600/x² = 66
Cross-multiply:
x² = 1600/66
Simplify:
x² = 24.24
Take the square root of both sides:
x ≈ 4.924
Since the length of the pen cannot be negative, we take the positive solution.
The approximate length of the pen along the road is x ≈ 4.924 feet.
To find the width, we can substitute the value of x into the equation xy = 100:
4.924y = 100
Divide both sides by 4.924:
y ≈ 20.33
The approximate width of the pen is y ≈ 20.33 feet.
Now, to find the minimum cost, substitute the values of x and y into the cost function:
C(min) = 34(4.924) + 16(2(4.924) + 20.33)
Calculate the value of C(min):
C(min) ≈ $293.783
Therefore, the dimensions that will minimize the cost to the farmer are approximately 4.924 feet by 20.33 feet, and the minimum cost is approximately $293.783.