If a total cost function is C (x) = .0001x^3 - .06x^2 + 12x + 100, is the margin
ID: 2881224 • Letter: I
Question
If a total cost function is C (x) = .0001x^3 - .06x^2 + 12x + 100, is the marginal cost increasing, decreasing, or not changing at x = 100? Find the minimum marginal cost. the revenue function for a one-product firm is R (x) = 200 - 1600/x + 8 - x. Find the value of x that results in maximum revenue. the revenue function for a particular product is R (x) = x (4 - .0001x). Find the largest possible revenue. A one-product firm estimates that its daily total cost function (in suitable units) is C(x) = x^3 - 6x^2 + 13x + 15 and its total revenue function is R (x) = 28x. Find the value of x that maximizes the daily profit. A small tie shop sells ties for $3.50 each. the daily cost function is estimated to be C(x) dollars, where x is the number of ties sold on a typical day and C (x) = .0006x^3 - .03x^2 + 2x + 20. Find the value of x that will maximize the store's daily profit. the demand equation for a certain commodity is p = 1/12 x^2 - 10x + 300, 0 lessthanorequalto x lessthanorequalto 60. Find the value of x and the corresponding price p that maximize the revenue.Explanation / Answer
Profit = Revenue - Cost
= 28 x - ( x^3-6x^2+13x+15 ) =
- x^3 + 6x^2 - 13x + 28 x - 15 =
- x^3 + 6x^2 + 15 x - 15 = Profit = P(x)
max profit, dP/dx = 0
- 3 x^2 + 12x + 15 = 0....................or
3x^2 - 12 x - 15 = 0
( 3x - 15 )( x + 1) = 0..............so
3x = 15....................or
x = -1.............so
Max profit occurs when x = 5
2) Asked to maximize Revenue not Profit..................
Price // Units // Revenue
4.00 // 10,000 // 40,000
4.40 // 8,000 // 35,200
P // x // xP
NOW // $ 4.00 // 10,000 // $ 40,000.00
1 price hike // $ 4.40 // 8,000 // $ 35,200.00
x price reduction // 4 - x // 10000 + 2000 x // R
R = ( 4 - x ) * ( 10000 + 2000 x ) =
40,000 - 10,000 x - 2000 x^2 + 8,000 x =
- 2000 x^2 - 2000 x + 40,000
-2000( x^2 + x - 20) = R
dR/ dx = 2x + 1
x = price change = - 1/2 = - $0.5
Max Revenue at Price = (4 - 0.5) = $ 3.5
2b)
Profit = Revenue - Cost
Let T be the price when unit sold is x:
= 3.5 x - (1000 + 0.6 x) = P(x)
= ( 3.5 - 0.6) x - 1000
Max profit when , dp/dx = 0
Price = 3.5 - 0.6 = $ 2.9
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