The price p and the quantity x sold of a certain product obey the demand equation p equals negative one fifth x plus 30 comma 0 less than or equals x less than or equals 150 (a) Express the revenue R as a function of x. (b) What is the revenue if 70 units are sold? (c) What quantity x maximizes revenue? What is the maximum revenue? (d) What price should the company charge to maximize revenue?

The correct answer is a) R = = - + 30x; b) $ 1120; c) Maximum quantity is 75 units with maximum revenue being $1125; d) p = $15.

Step-by-step explanation:

Demand equation: p = - x + 30 , 0 x 150 ; where p is the price of the product and x is the quantity sold.

a) Revenue function by the problem is given to be R= p × x = - + 30x.

b) Revenue at x = 70 is given by 2100 - 980 = $ 1120.

c) For maximizing the R we differentiate it with respect to x and equate it to zero.

⇒ - x + 30 =0

⇒ x = 75.

As the second order derivative is negative at this point, this is the value of x that maximizes the revenue.

Maximum Revenue is at x = 75 and is equal to $1125.

d) Price charged by the company for maximum revenue is $15.

first, we define the variables:

x: number of years after 1950

f (x): amount of vinyl sold.

then, with the variables defined, we have:

68594 vinyl records were sold in 1958 > f (8) = 68594

91299 vinyl records were sold in 1961 > f (11) = 91299

38720 vinyl records were sold in 1952 > f (2) = 38720

161743 vinyl records were sold in 1967 > f (17) = 161743

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step-by-step explanation:

$30.32

step-by-step explanation:

65.00*.12=2.80

65.00-2.80=57.20

57.20*1.06=60.632

60.632/2=30.316