Damon, Help :)
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?
You answered:
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
Now I have a stupid question... what happened to the y in the c= equations?
x y = 1
so
y = 1/x
In the equations, the variable "y" represents one of the sides of the rectangular region, which will not be adjacent to the road. Since the rancher wants to use a better quality wood for the side adjacent to the road, we can assume that the other three sides (represented by "x" and "y") will be made of the cheaper wood.
Therefore, in the equation "c = 3x + x + 2y," the first term "3x" represents the cost of the side along the road (which is three times more expensive), the second term "x" represents one of the other sides made of the cheaper wood, and the third term "2y" represents the sum of the other two sides (both made of the cheaper wood).
To find the dimensions that minimize the cost of the fence, we differentiate the cost equation with respect to "x" and set it equal to zero to find the critical point:
dc/dx = 4 - 2/x^2 = 0
Solving this equation, we find that x = 1/sqrt(2) = sqrt(2)/2, which represents the width of the region along the road.
Since the area of the rectangle is given as 1 square kilometer, we can find the length of the region (represented by "y") by dividing the area by the width:
y = 1/(sqrt(2)/2) = sqrt(2)
Therefore, the dimensions of the rectangle that minimize the cost of the fence are x = sqrt(2)/2 along the road and y = sqrt(2).