The total cost, C (in thousand of dollars) , of producing x (in hundred) units of some commodity is given by C= 0.1x^(3) + 9 where x is related to time, t (in months), by the equation x = 2t + 5
a) Find the rate of change of total cost with respect to number of unit produced.
b) Find the rate of change of total cost with respect to time after 4 months.
I have the unit transform problems in doing the question, would you give me hint? Thanks!
One easy differentiation
2013-11-04 3:00 am
回答 (3)
2013-11-07 3:39 am
✔ 最佳答案
Answer for part a) is 3^2 thousands dollars per hundred units
2013-11-06 20:47:30 補充：
Answer for part b) is
$101,400 per month
2013-11-06 20:48:41 補充：
Just differentiate both sides,
then be careful of the unit,
then no problem
參考: myself
2013-11-05 4:06 am
but the first answer is not correct when compare to my textbook answer
2013-11-04 2:31 pm
a) Rate of change of total cost w. r. t no. of unit produced = dC/dx
= (0.1)(3)x^2 = 0.3x^2.
b)
Rate of change of total cost w. r. t. time, t = dC/dt = (dC/dx)(dx/dt).
dx/dt = 2, so dC/dt = (0.3x^2)(2) = 0.6x^2 = 0.6(2t + 5)^2.
So rate of change when t = 4 is 0.6(8 + 5)^2 = 101.40 = $101,400 per month.
2013-11-05 21:39:32 補充：
Remark : Please tell me what is the answer in the textbook, I think the difference is in the unit since C is in thousands of dollars and x is in hundreds.
= (0.1)(3)x^2 = 0.3x^2.
b)
Rate of change of total cost w. r. t. time, t = dC/dt = (dC/dx)(dx/dt).
dx/dt = 2, so dC/dt = (0.3x^2)(2) = 0.6x^2 = 0.6(2t + 5)^2.
So rate of change when t = 4 is 0.6(8 + 5)^2 = 101.40 = $101,400 per month.
2013-11-05 21:39:32 補充：
Remark : Please tell me what is the answer in the textbook, I think the difference is in the unit since C is in thousands of dollars and x is in hundreds.
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