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.

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.

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