Damon help again :) (last one I promise)
why is X along the road sqrt2/2?
] A rancher plans to set aside a rectangular region of one square kilometer for cattle and wishes to build a wooden fence to enclose the region. Since one side of the region will run along the road, the rancher decides to use a better quality wood for that side which costs three times as much as the wood for the other sides. What dimensions will minimize the cost of the fence?
x y = 1
c = 3 x + x + 2 y = 4 x + 2 y
c = 4 x + 2/x
dc/dx = 4 - 2/x^2
= 0 for minimum
4 = 2/x^2
x^2 = 1/2
x = 1/sqrt 2 = sqrt 2/2 along road
y = sqrt 2
This question has already been answered by Damon
Yes, I know, but I i had question about the answer... I'm math stupid :)
The smaller of x or y will be the expensive fence along the road.
To find the dimensions that minimize the cost of the fence, we can start by setting up an equation to represent the cost of the fence.
Let's assume the length of the rectangular region along the road is represented by 'x' and the width is represented by 'y'. The area of the rectangular region is given as one square kilometer, so we can write the equation: x * y = 1.
The cost of the fence is determined by the quality of the wood used. Since the rancher decides to use a better quality wood for the side along the road, the cost of that side will be three times the cost of the other sides. Therefore, the cost equation can be written as: c = 3x + x + 2y = 4x + 2y.
Now, we need to express the cost equation in terms of a single variable, either 'x' or 'y', so that we can find the minimum point. Let's express it in terms of 'x': c = 4x + 2/y.
To find the value of 'x' that minimizes the cost of the fence, we can take the derivative of the cost equation with respect to 'x' and set it equal to zero.
dc/dx = 4 - 2/y^2 = 0 for minimum.
Simplifying, we have: 4 = 2/y^2.
Now, solve for 'y^2': y^2 = 2.
Taking the square root of both sides, we have: y = sqrt(2).
Since we know that x * y = 1, we can substitute the value of 'y' into this equation to find 'x': x * sqrt(2) = 1.
Solving for 'x', we get: x = 1/sqrt(2) = sqrt(2)/2.
Therefore, the dimensions that will minimize the cost of the fence are: x = sqrt(2)/2 along the road, and y = sqrt(2) perpendicular to the road.