A manufacturer finds that he can sell x refrigerators per month at $y each
where 0≤x≤2000 and y=-0.038x+800. The cost $C of producing x refrigerators
per month is given by C=580+620x-0.003x^2. Find the value of x for which
the profit is the greatest.
I calculated that x is larger than 2000. Please help!!
F.4Maths M1 differentiation
2014-05-31 1:07 am
回答 (2)
2014-05-31 3:04 am
✔ 最佳答案
Answer: $4259420profit
= selling price - cost
= xy - C
= x(-0.038x + 800) - (580 + 620x - 0.003x^2)
= -0.035x^2 + 220x - 580
d(profit) / dx = -0.07x + 220
set d(profit) / dx = 0, => x = 3142.857143
therefore, for 0 ≤ x ≤ 2000, there are no turning points.
and the maximum profit should appear at two ends.
for x = 0, profit = -0.035(0)^2 + 220(0) - 580 = -580
for x = 2000, profit = -0.035(2000)^2 + 220(2000) - 580 = 4259420
So, the greatest profit is $4259420
2014-05-30 19:05:14 補充：
Sorry, the answer should be 2000
參考: knowledge
2014-05-31 11:46 pm
Re: YTC
Within 0 ≤ x ≤ 2000, there is no point such that dy/dx = 0.
If y is differentiable within the range 0 ≤ x ≤ 2000, then either dy/dx > 0 in the whole range or dy/dx < 0 in the whole range.
That is, y is always increasing or always decreasing.
2014-05-31 15:46:32 補充：
Therefore, the maximum and minimum occur at the end points.
Within 0 ≤ x ≤ 2000, there is no point such that dy/dx = 0.
If y is differentiable within the range 0 ≤ x ≤ 2000, then either dy/dx > 0 in the whole range or dy/dx < 0 in the whole range.
That is, y is always increasing or always decreasing.
2014-05-31 15:46:32 補充：
Therefore, the maximum and minimum occur at the end points.
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