This document describes a personal finance software program called PLF that helps users analyze and predict personal financial decisions. It uses market data to create demand, revenue, cost and profit functions. These functions are used to find the optimal price point that maximizes profit. The software represents demand as a quadratic function and assumes the product has a monopoly in the market. It graphs demand, revenue, cost and profit to determine the optimal price is $419.68, selling 666,000 units will result in maximum profit of $34.3 million. Sensitivity analysis shows a 1% lower demand decreases profit by $10k and a 2% higher cost decreases profit by $7.03 million.

1. Danny Gruen

2.  PLF (Personal Learning Finance)
 Programmed in SQL
 Specific for PC
◦ Mac coming soon...

3.  Product:
◦ PLF – Program for end-user to analyze and predict
personal financial decisions
 Objectives
◦ Maximize profit
◦ Use market data to create demand, revenue, cost
and profit functions
◦ Use the functions to find the optimal price at which
to produce and sell and at the maximum profit
◦ See how changes in cost, price, and demand affect
profit

4.  Demand is a quadratic function
 Our market data given is accurate
 Our product is in a monopoly market

5.  Demand
◦ D(q) : price per unit (dependent on q)
◦ q: number of units sold
 Revenue
◦ R(q): quantity multiplied by price per unit
◦ R(q) = q x D(q)
 Cost
◦ C(q): comprised of two components → fixed cost
(constant) and variable cost (dependent on q)
◦ C(q) = Fixed Cost + VC(q)
 Profit
◦ P(q): revenue minus total cost
◦ P(q) = R(q) – C(q)

6.  Marginal Profit
◦ MP(q): Profit generated from extra unit
sold; MP(q) = MR(q) – MC(q)
 Marginal Cost
◦ MC(q) = Cost incurred for producing the
extra unit; MC(q) = C’(q)
 Marginal Revenue
◦ MR(q): The revenue generated from one
extra unit sold
◦ MR(q) = R’(q)

8. - Our demand graph is a graphical representation of the quadratic
demand function through plotting the market data points
K’s

9. •R(q) = D(q) x q
•R(q) =-.0002605115x3-.0503324462x2+568.7833351581x
•Peak of graph is Max Revenue

10. Derivative of Revenue graph
•Where graph crosses x-axis this is our
max revenue

11. The cost graph depicts the sum of the fixed and variable costs at
different quantities or levels of production
Cost Function: C(q) = Fixed cost +variable cost
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
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13848001670124.
8005001420155.
5000850270.
)(
qq
qq
qq
qC

12. •Derivative of Cost function

13. •The points where the two graphs meet are the break even
points, where the profit = 0, and when revenue = cost
•The maximum profit is represented by the greatest distance
between the revenue and cost

14. P(q) = R(q) – C(q); MP(q) = MR(q) – MC(q)
The profit graphs shows the relationship between the number of units sold
and the profit earned at these various quantities
The maximum profit is $34.30 million at 666,000 procedures.

15. Where the graph crosses the x-
axis is where we achieve max
profit

16. Max Profit occurs where MC(q) = MR(q)
MR(q) = MC(q) at q=666

17.  What is the optimal price to set our product
at?
◦ We determined that our optimal price would be
$419.68
 How many pills can we sell at this price?
◦ We determined that we could sell 666,000 pills
 What is the maximum profit?
◦ Maximum profit = $34.30 million

18.  We must find the revenue at the profit
maximizing point:
◦ R(q) = optimal price x expected quantity sold at
maximum price
= $419.68 x 666= $279.54 million
Subtract total possible revenue from the revenue at
profit maximizing point:
◦ Consumer Surplus:
)666*419(78333.56805033.00026.
.666
0
2
 dxxx
=195548725.06

19. A 1% decrease in demand yields:
•Optimal price of $419.20
•Quantity of 961.06??
•Profit would decrease by $10
thousand
•Total profit of $34.29 million

20. A 2% increase in marginal cost
yields:
 Optimal price of $419.68
 Quantity of 666 thousand
 Decreased the profit by $7.03
million
 total profit of $25.99 million

22. Questions?