Asked by Karen
The ordering and transportation cost C for components used in a manufacturing process is approximated by the function below, where C is measured in thousands of dollars and x is the order size in hundreds.
C(x) = 12((1/x)+((x)/(x+3)))
(a) Verify that C(2) = C(15).
C(2) = 54/5
C(15) = 54/5
(b) According to Rolle's Theorem, the rate of change of the cost must by 0 for some order size in the interval (2, 15). Find that order size. (Round your answer to the nearest whole number.)
What is the answer to part b? I found the the derivative of C(x) and set it equal to zero and got approximately 4, but that isn't the right answer. Please help!!!
C(x) = 12((1/x)+((x)/(x+3)))
(a) Verify that C(2) = C(15).
C(2) = 54/5
C(15) = 54/5
(b) According to Rolle's Theorem, the rate of change of the cost must by 0 for some order size in the interval (2, 15). Find that order size. (Round your answer to the nearest whole number.)
What is the answer to part b? I found the the derivative of C(x) and set it equal to zero and got approximately 4, but that isn't the right answer. Please help!!!
Answers
Answered by
Steve
Rolle's Theorem says that since C(2)=C(15), there is some value 2<c<15 such that C'(c) = 0.
dC/dx = 12(2x^2-6x-9)/(x(x+3))^2
dC/dx=0 when 2x^2-6x-9=0, or
x = 3/2 (1±√3)
or, x = -1.1, 4.1
So, your answer of 4 is correct. The graph at
http://www.wolframalpha.com/input/?i=12%28%281%2Fx%29%2B%28%28x%29%2F%28x%2B3%29%29%29+
shows the minimum in the desired interval, where C' is zero.
dC/dx = 12(2x^2-6x-9)/(x(x+3))^2
dC/dx=0 when 2x^2-6x-9=0, or
x = 3/2 (1±√3)
or, x = -1.1, 4.1
So, your answer of 4 is correct. The graph at
http://www.wolframalpha.com/input/?i=12%28%281%2Fx%29%2B%28%28x%29%2F%28x%2B3%29%29%29+
shows the minimum in the desired interval, where C' is zero.
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