A factory owner who employes m workers finds that they produce
q=14m(14m+43)^3/2
units of product per day.
The total revenue R in dollars is
R=540q/(293392+4q)^1/2
(a) From the fact that
revenue =(price per unit)(number of units)
it follows that
R=(price per unit)q
So when there are 15 workers, the price per unit is _______ dollars.
(b) When there are 15 workers, the marginal revenue is_______ dollars/(one unit of product).
(c) The marginal-revenue product is defined as the rate of change of revenue with respect to the number of employees. Therefore
marginal-revenue product=dR/dm
If q and R are given as above then, when m= 15, the marginal-revenue product is ______dollars/(one worker). This means that if employee number 16 is hired, revenue will increase by approximately _________dollars per day.
Marginal function derivative help plzz!?
2012-02-08 9:10 am
回答 (1)
2012-02-09 3:50 am
✔ 最佳答案
(a) R/q = 540/√(293392+4q), but we need to know q for m=15... q(15) = 14*15*(14*15+43)^(3/2) = 210(253^(3/2)) = 210*4024.211 = 845,084 (units)
.. R/q = 540/√3673728 ≈ 0.28173
The (average) price per unit is about $0.282
(b) dR/dq = 540/(293392+4q)^(1/2) - 540q(1/2)(4)/(293392+4q)^(3/2)
.. = 135 (146696 + q)/(73348 + q)^(3/2)
.. when q=845084, this is about 0.15212
The marginal revenue per unit is about $0.152
.. note that this is less than the average price. Apparently, after a large number are sold, the price per unit drops substantially.
(c) dR/dm = dR/dq*dq/dm
.. differentiating q, we get dq/dm = 14(14m+43)^(3/2) + 14m(3/2)(14m+43)^(1/2)(14)
.. evaluating at m=15, this is 7952√253 ≈ 126,484
.. and the revenue product is (0.15212)(126,484) ≈ 19,240 dollars/worker
One expects revenue to increase by about $19,240/day when worker 16 is hired. <=== Answer
(If you work through the values of q and r, you find that revenue increases by about $19,339/day when m=16.)
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