Transcript Additional Topic for Ch.16: Optimal Price for Maximizing

Additional Topic for Ch.16:
Optimal Price for
Maximizing Revenue
Warin Chotekorakul
How to get an optimal price?

Two main ways:
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Calculus: taking the 1st derivative
Algebra: substituting constants
1st way: Calculus

Taking only the 1st derivative of the
revenue function, then we can get the
optimal price.
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y = f(x) = 3x2 + 9x + 4
y = f(x) = -4x2 + 9x + 5
y = f(x) = 7x2 - 6x + 6
2nd way: Algebra

Substituting constants or parameters into
the formulae:

( x = -b , y = 4ac – b2)
2a
4a
EXAMPLE

a)
b)
The demand for the product of a firm varies with the
price that the firm charges for the product. The firm
estimates that annual total revenue R (stated in
$1,000s) is a function of the price p (stated in
dollars). Specifically,
R = f(p) = - 50p2 + 500p
Determine the price which should be charged in
order to maximize total revenue.
What is the maximum value of annual total revenue?
SOLUTION: Calculus

The 1st derivative of R = f(p) = - 50p2 + 500p is
R’ = f’(p) = -100p + 500
If we set f’(p) = 0, then
-100p + 500 = 0
-100p = -500
p=5
a) Therefore, a max revenue occurs at price = 5
SOLUTION (2) : Calculus
b) The max value of R is found by substituting p = 5 into
revenue function
R = f(p) = -50p2 + 500p
R = f(5) = -50(5)2 + 500(5)
= -1,250 + 2,500 = 1,250
Thus, annual total revenue is expected to be maximized at
$1,250 (1,000s) or 1.25 million when the firm charges $5
per unit.
SOLUTION (3) : Algebra

R = f(p) = - 50p2 + 500p Substitute constants or
parameters of the function into the formulae:
( x = -b , y = 4ac – b2)
2a
4a
 x = -(500) , y = 4(-50)(0) – (500)2
2(-50)
4(-50)
Then, p = 5; R = 1,250
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What if, the question asks for max profit?

We knew Revenue = Price * Quantity sold
Quantity sold = Revenue/Price

Then, we can calculate max profit by
Profit = (P-Cost)* Quantity sold