Optimization: Maximums & Minimums
The Graphical Representations
Optimization: Procedures
• Find “Extreme Value” of Function (Min or Max)
• To Do So:
– Set Up Objective Function
– Take Derivative, Set = 0 k
i i S
0 (so slope =0)
– Solve Derivative Equation
Profit Example: TP = TR – TC (Total Profit Equation)
Given: P = 10 ‐ .05Q (demand curve)
TC = 120 + 2Q +.05Q2 (cost curve)
Then You Can Determine Total Revenue & Profit
Then You Can Determine Total Revenue & Profit
TR = [10‐.05Q ]*Q= 10Q ‐.05Q2 (since TR = P*Q) TP = [10Q ‐ .05Q2] ‐ [120 + 2Q +.05Q2]
TP = 8Q ‐ .1Q2 ‐ 120
1
TR = [10Q ‐ .05Q2] (shape?)
Total Revenue
500
400
300
200
100
0
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
TC = 120 +2Q +.05Q2
(shape?)
Total Cost
600
500
400
300
200
100
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
TR & TC
Total Rev & Total Cost
600
500
400
300
200
100
0
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
2
Profitability: Losses
Total Rev & Total Cost
600 500 400 300 200 100 Losses
0 0 2 4 6 810121416182022242628303234363840424446485052545658606264666870
Profitability: Breakeven
Total Rev & Total Cost
600
500
400
300
200
Breakeven
100
0
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
Profitability: Positive Profits
Total Rev & Total Cost
600
500
400
Profit
300
200
100
0
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
3
The Profit Function
(shape?)
TP = 8Q ‐ .1Q2 ‐ 120
Total Profit
50
0
-50
-100
-150
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
Profit Function: Losses
TP = 8Q ‐ .1Q2 ‐ 120
Total Profit
50
0
-50
-100
Losses
-150
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
Profit Function: Breakeven
TP = 8Q ‐ .1Q2 ‐ 120
Total Profit
50
0
-50
Breakeven
-100
-150
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
4
Profit Function: Positive Profits
TP = 8Q ‐ .1Q2 ‐ 120
Total Profit
50
Profits
0
-50
-100
-150
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
Profit Function: Max Profits
TP = 8Q ‐ .1Q2 ‐ 120
Total Profit
50
50 0 ‐50 Profit
MAX !
‐100 ‐150 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70
NOTE: Alternative Approach
TR = 10Q ‐ .05Q2
TC = 120 +2Q+.05Q2
dTR = 10 ‐ .1Q = MR dTC = 2 + .1Q = MC
dQ dQ
Q
Q
MR
10 ‐ .1Q
= MC
= 2 + .1Q
Q = 40
5
AR = 10 – .05Q
TR = 10Q ‐ .05Q2 
MR = 10 ‐ .1Q
Ave & Marginal Revenue
10
9
8
7
6
5
4
3
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
Ave Rev
Marg Rev
TC = 120 +2Q+.05Q2  MC = 2 + .1Q
Ave & Marginal Revenue
10
8
6
AR = 10 – .05Q
4
MR = 10 ‐ .1Q
2
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
Ave Rev
Marg Rev
Marg Cost
MR = MC  Profit Max!
Ave & Marginal Revenue
10
MC = 2 + .1Q
8
6
AR = 10 – .05Q
4
2
Max !
MR = 10 ‐ .1Q
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
Ave Rev
Marg Rev
Marg Cost
6
Wrap Up:
• Many Optimization Applications in Business
• Basic Mechanics Straightforward
• 2 Math Approaches:
Set Up Objective Function
Take Derivative Set = 0
Take Derivative, Set = 0 Solve Derivative Equation
Check Sign of 2nd Derivative
Determine Marginal Revenue
Determine Marginal Cost Set MR =MC
• 2 Graphical Approaches:
Total Rev & Total Cost
600
Ave & Marginal Revenue
500
10
400
300
8
200
100
6
0
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
Total Profit
4
50
0
2
Max !
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
-50
Ave Rev
Marg Rev
Marg Cost
-100
-150
0 2 4 6 8 10121416182022242628303234363840424446485052545658606264666870
7