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Your business sells cupcakes in boxes of 10. The demand equation is

x = −5p + 100
In the above formula x is the number of boxes you sell in one month for a unit price (per box) of p dollars. The cost of producing x boxes is
C = $50 + 6x
Set up the profit function P in terms of an arbitrary number of boxes x alone.

User Jnoreiga
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2 Answers

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In Business, the following functions are important.
Revenue function = (price per unit) . (quantity of units) Symbols:
Cost function = (average cost per unit) . (quantity of units) Symbols:
Profit function = revenue − cost
Symbols:
Sometimes in a problem some of these functions are given. Note: Do not confuse p and P .
The price per unit p is also called the demand function p .
Marginal Functions:
The derivative of a function is called marginal function.
The derivative of the revenue function R(x) is called marginal revenue with notation: The derivative of the cost function C(x) is called marginal cost with notation:
The derivative of the profit function P(x) is called marginal profit with notation:
Example 1: Given the price in dollar per unit p = −3x2 + 600x , find: (a) the marginal revenue at x = 300 units. Interpret the result.
R=p.x
C=C.x
P=R−C
R′(x) = dR dx
C′(x) = dC dx
P′(x) = dP dx
revenue function: R(x) = p . x = (−3x2 + 600x) . x = −3x3 + 600x2 marginal revenue: R′(x) = dR = −9x2 + 1200x
dx
marginalrevenueat x=300 =⇒ R′(300)= dR =−9(300)2+1200(300)=−450000
dx x=300
Interpretation: If production increases from 300 to 301 units, the revenue decreases by 450 000 dollars.
(b) the marginal revenue at x = 100 units. Interpret the result. revenue function: R(x) = p . x = (−3x2 + 600x) . x = −3x3 + 600x2
marginal revenue: R′(x) = dR = −9x2 + 1200x dx
marginalrevenueat x=100 =⇒ R′(100)= dR =−9(100)2+1200(100)=30000 dx x=100
Interpretation: If production increases from 100 to 101 units, the revenue increases by 30 000 dollars.
User Milushov
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1 vote
Step by step equation
User MathieuMaree
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