Commercial Recording, Inc. is a manufacturer and distributor of reel-to-reel rec
ID: 2863294 • Letter: C
Question
Commercial Recording, Inc. is a manufacturer and distributor of reel-to-reel recording docks for commercial srudios. Revenue and cost relations are: R = $3000q - $0.5q^2 C = $ 100, 000 + $1.500q + $0.1q^2 What level of production will maximize profit At this level of production find marginal cost, average cost, price and profit. What level of production minimizes average cost? At this level of production find marginal cost, average cost, price and profit. A manufacturer can produce radios at a cost of $5 each and estimates that if they are sold for x dollars each, consumers will buy approximately f(x) = 1000e^-0.1x radios per week. If Che radios are priced at $12.75. about how many are sold per week? If consumers buy about 427 radios per week, at what price are they selling? Write a revenue function for this product What price should be charged in order to maximize revenue" How many radios will be sold at this price? What is the revenue?Explanation / Answer
Solution 5:
R = 3000q - 0.5q2 and C = 100000 + 1500q + 0.1q2
(a) Profit, P = R - C = -0.6q2 + 1500q - 100000 (which is a downward parabola )
For maximum profit, dp/dq = 0 => -1.2q +1500 = 0
=> 1500 = 1.2q => q = 1250
Also as it is a downward parabola, it will have a maxima at a fixed point and minima at negatve infinity. So profit will be maximum when 1250 units are produced.
As now level of production has been found, all other things can be found by simply putting the value of q in different relations.
So marginal cost = C(1251) - C(1250) = 1500 * 1 + 0.1(12512 - 12502) = 1500 + 0.1*2501 = $1750.1
Average cost = C/q = (100000 + 1500*1250 + 0.1*1250*1250) / 1250 = $1705
Price = 3000*1250 - 0.5*1250*1250 = $2968750
Profit = -0.6*1250*1250 + 1500*1250 - 100000 = $1037500
(b) For minimum cost, dc/dq = 0 => 1500+0.2q = 0 => q < 0 => cost has a minima which is negative (which is not possible as q is production level.) So cost will be minimum at q=0.
So for q = 0, marginal cost = 1500+0.1 = $1500.1
Average cost will not be defined, no price or profit will be there for 0 production.
Solution 6:
Cost, C = 5, Revenue,R = x. 1000e-0.1x
(a) x = 12.75 => No of units sold = 1000* e0.1*12.75 = 279.4
So 279 units will be sold.
(b) No of units sold = 427 = 1000e-0.1x => 0.427 = e-0.1x => ln(0.427) = -0.1x => -0.8509 = -0.1x => x= $8.509
(c) Revenue function has already been written, R = x.1000e-0.1x = 1000.x.e-0.1x
For maximum revenue, dr/dx = 0 => 1000(xe-0.1x.(-0.1) + e-0.1x) = 0 => 1000.e-0.1x(-0.1x + 1) = 0
As 1000.e-0.1x is not equal to zero, so -0.1x + 1 = 0 => x = 10
So $10 should be charged in order to maximize profit.
No of radios sold = 1000.e-0.1x = 1000*e-1 = 367.87
So 367 units will be sold.
Revenue = x.1000.e-0.1x = 10*367 =$3670.