1. Steps in Solving Optimization Problems:
Calculus Notes 4.7: Optimization Problems
4.8: Applications to Business and Economics
4.7: Optimization Problems
Steps in Solving Optimization Problems:
Understand the problem: What is known? Unknown? Want do you want to find?
Draw a Diagram: to visualize and see if things make sense.
Introduce Notation: label and define variables.
Write equations in terms of one variable: re-write everything into one variable.
Find relationships between the variables to simplify into one.
Use methods of sections 4.1 and 4.3 to find absolute maximum or minimum value of the function.
First Derivative Test for Absolute Extreme Values: Suppose that c is a critical number of a continuous function f defined on an interval.
If f ‘ (x) > 0 for all x < c and f ‘ (x) < 0 for all x > c, then f (c) is the absolute maximum value of f.
If f ‘ (x) < 0 for all x < c and f ‘ (x) > 0 for all x > c, then f (c) is the absolute minimum value of f.

2. The Average Cost Function: With C(x)=cost function
Calculus Notes 4.7: Optimization Problems
4.8: Applications to Business and Economics
4.8: Applications to Business and Economics
The Average Cost Function:
With C(x)=cost function
x is the units of a certain product.
If average cost is a minimum, then marginal cost = average cost.
Demand Function:
With p(x)=price per unit that the company can charge if it sells x units.
R(x) is the revenue function (or sales function).
Marginal revenue function is R ‘ (x).
Profit Function:
With P(x) is the profit function.
The marginal profit function is P ‘ (x).
If the profit is a maximum, then marginal revenue = marginal cost.

3. The unknowns are the length and width of the large fenced area.
Calculus Notes 4.7: Optimization Problems
4.8: Applications to Business and Economics
Example 1: Ben Dover wants to use 480 yards of fencing to create a fenced, rectangular area in his pasture. The fenced area will be divided into 6 smaller areas as shown in the diagram. What should be the dimensions of the entire fenced rectangle so that the total area of the fenced portion is the greatest possible? What is that area? y x
The unknowns are the length and width of the large fenced area.
The quantity to be maximized is the area of the large rectangle.
A picture is already given.
Let A=the area of the large rectangle. We do NOT say simple “A=area”, because we also have six smaller areas in addition to the area of the large rectangle! Also, let x=length of the large rectangle, and y=width of the large rectangle, so that our diagram looks something like the above with the dimensions labeled.

4. Calculus Notes 4.7: Optimization Problems
4.8: Applications to Business and Economics
Example 1: Ben Dover wants to use 480 yards of fencing to create a fenced, rectangular area in his pasture. The fenced area will be divided into 6 smaller areas as shown in the diagram. What should be the dimensions of the entire fenced rectangle so that the total area of the fenced portion is the greatest possible? What is that area? y x
A=xy
The other information in the problem, which we have not yet used, is that the total amount of fencing will be 480 yards.
7. Substitute to get:

5. Calculus Notes 4.7: Optimization Problems
4.8: Applications to Business and Economics
Example 1: Ben Dover wants to use 480 yards of fencing to create a fenced, rectangular area in his pasture. The fenced area will be divided into 6 smaller areas as shown in the diagram. What should be the dimensions of the entire fenced rectangle so that the total area of the fenced portion is the greatest possible? What is that area? y x
Maximize the Area function now.
Domain will be [0,160]. Why?
The derivative is:
Set A’=0, we get x=80.
1st derivative test, so at x=80 we have a maximum.
10. Now find y and the area. + — A’ 80
160

6. Profit=Revenue-Cost: Revenue=(Units Sold)(Price Per Unit):
Calculus Notes 4.7: Optimization Problems
4.8: Applications to Business and Economics
Summary of Formulas:
Profit=Revenue-Cost:
Revenue=(Units Sold)(Price Per Unit):
Average Cost=
Revenue from selling one more unit Marginal Revenue=
Profit from selling one more unit Marginal Profit=
Cost of producing one more unite Marginal Cost=

7. Example 2: Given the cost function Find the marginal cost when x=50;
Calculus Notes 4.7: Optimization Problems
4.8: Applications to Business and Economics
Example 2: Given the cost function
Find the marginal cost when x=50;
Find the marginal profit at x=50, if the price per unit is $20.
Marginal cost is C’(x).
(b) Marginal Profit is P’(x)
Cost of producing 51st unit is $18, and the profit from producing the 51st unit is $2.

8. Calculus Notes 4.7: Optimization Problems
4.8: Applications to Business and Economics
Example 3: Given the cost function
and the demand function (the price function)
Find the number of units produced in order to have maximum profit.
Step 1: Write an equation.
Step 2: Differentiate.
Step 3: Find critical numbers.
Step 4: Apply 2nd Derivative Test.
Since x=350 is the only relative maximum, it is the absolute maximum.
Step 5: Write a solution.
PS 4.7 pg.283 #1, 2, 7, 14, 34, 46 (6)
PS 4.8 pg.293 #1, 6, 9, 12, 20 (5)
Producing 350 units will lead to a maximum profit.