grade 11 applied math

Question

A company is experimenting with the pricing on a calculator. They currently average 200 daily sales at a price of $10. Research suggests that if they raise the price of the calculator by 50¢ that they will make 5 fewer sales. It costs the company $4 to manufacture a calculator.

a) Find an equation for the revenue the company will make.

b) Given that Profit = Revenue – Cost, find an equation for the profit the company can make.

c) What price should the company charge for a calculator in order to maximize the profit?

Marks may be awarded as outlined below. If your teacher plans to use a different strategy to evaluate your work they will inform you before you start the assignment.
Unless you are instructed differently this assignment is worth 7 marks. Use the following information to guide your work:

2 marks for a revenue equation
2 marks for a profit equation
2 marks for showing work appropriately to find price to maximize profit
1 mark for finding the price that will maximize profit consistent with work

Answer 1

The demand decreases by 5 for every 50 cent price hike...

The slope of the demand line is -5/0.5 = -10

POint (10,200) is on the demand line , since 200 are sold at $10 each.

y = mx + b
200 = (-10)(10) + b
200 = -100 + b
b = 300

the demand line has equation y =D(x) = -10x + 300


The revenue function is R(x) = x*D(x) = x(-10x+300) where x is the price


Part B: the cost function is C(x) = 4x

So the profit function is P(x) = R(x) - C(x) = x(-10x+300) - 4x
= -10x^2 + 300x - 4x
= -10x^2 + 296x

Part C:

Sets the Derivative equal to zero so as to maximize:
dP/dx = -20x + 296 = 0
-20x = -296
x = -296/-20 = $14.80

If you don't like the calculus, then the max occurs at the average of the zeros
of the quadratic function. the zeros are -x( 10x - 296)=0
x=0 and x=296/10 = 29.6
The average of which is $14.80