mathematics question (urgent)

[匿名]

2014-03-02 11:43 pm

A manufacturer can produce at most 120 units of a certain product each year. The
price function for the product is
p = q2 - 100q + 3200
and the manufacturer’s average-cost function is
c=2/3q^2-40q+10000/q
where q is the number of units, and both p and ¯c are expressed in dollars per unit.
(a) Determine the level of output at which profit is maximized by using the first
derivative test.
(b) Determine the price at which maximum profit occurs.

Please show me all the steps. Really urgent. Thank you for answering.

回答 (1)

用戶10012

2014-03-03 4:43 am

✔ 最佳答案

Revenue = pq = q^3 - 100q^2 + 3200q.
Total cost = cq = 2q^3/3 - 40q^2 + 10000
So profit, P = pq - cq = q^3 - 100q^2 + 3200q - 2q^3/3 + 40q^2 - 10000
dP/dq = 3q^2 - 200q + 3200 - 2q^2 + 80q = q^2 - 120q + 3200
Put dP/dq = 0,
q^2 - 120q + 3200 = 0
(q - 80)(q - 40) = 0
q = 80 or 40.
When q = 35, dP/dq is +ve
When q = 45, dP/dq is - ve.
So q = 40 is a max. point.
When q = 75, dP/dq is -ve.
When q = 85, dP/dq is +ve.
So q = 80 is a min. point.
So profit is maximum when q = 40.
(b) For q = 40, price, p = (40)^2 - 100(40) + 3200 = 800.

2014-03-02 20:53:03 補充：
Remark :
For q = 40, profit = 43,333. For q = 120, profit = 86,000, which is higher than q = 40. So profit is at a max. when q = 120. ( q = 40 is a local maximum, q = 120 is a global maximum). Price at q = 120 is (120)^2 - 100(120) + 3200 = 5600.

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