Exam cheat sheet week 7
5.2.1 & 5.2.2
Optimization function with one variable
Profit Profit = Revenue- costs
optimizati
Most profitable output quantity (y) is a
on is also:
stationary point of (π).
Marginal output rule. (answers where min max) (maximal profit)
MR’ (y) = how much R will approx. change if
y changes 1 unit
MC’ (y) = extra costs if y changes approx. 1
unit
Production rule. (answers profit -/+) (means positive profit
AR (Y) = avg. R at (P)roduction (y) = Tot R-P possible)
AC (Y) = avg. C at (P)roduction (y) = Tot C-P
Supply function. (answers for each price, the
output quantity that maximizes profit)
P ≥ min AC (y) results in profit
P < min AC (y) results in loss
So how to find the Optimal Profit: NOTE: Since a maximum can be
1. π(Y) = R(Y) – C(Y) OR π’(Y) = below 0 we need to find whether
R’(Y) – C’(Y) the profit is positive or negative
2. Stationary point π’(y) = 0 OR R’(Y) = C’(Y)
3. Find MAX using sign chart fill SP neighborhood into the
π’(y)
4. Find X coordinate fill SP into π(Y) = R(Y) – C(Y)
How to find whether the profit is positive or negative
1. AR (Y) ≥ AC (Y)
2. Calc AR and AC using AR = R(y) / y and AC= C(y) / y
3. Y (AR(y) – AC(y)) if > 0 then positive profit
How to find for each price the output quantity that
maximizes profit (Supply function)
1. Combining the previous two questions.
2. Read of the sign chart
for where profit is
positive and where
its negative. P ≥ min AC (y) results in profit P < min AC
(y) results in loss
, Exam cheat sheet week 8
5.5 & 5.6
Constrained (beperkt) Optimization (3) methods (3) applications
In general Optimize (objective function) z(x,y)
Subject to (Constrained) g(x,y) = k
Where (Domain) x in D1 and Y in D2
Feasible point All (x,y) that satisfy the constrained and the domain
Method 1: Substitution (Not always possible to rewrite constrained)
1. Rewrite constrained to x = … OR y=…
2. Substitute result in objective function. (now you’re left with extremum problem with one variable)
3. Find stationary points z’(x or y)= 0
4. Use sign chart to find the maximum of z’
5. Find corresponding coordinate fill in SP into z (step 2)
Method 2: First-order condition for a constrained extremum
(always possible)
1. Find the coordinate/interior point (x,y) using the system.
2. Calculate the function value at the boundary points and the
interior.
3. Step 2 gives answer to min or max.
Method 3: Langrange 2+ constraints (don’t use this 1 or 2 will always work)
1.
Application 1: Utility Maximization Good X with price Px
Example: Good Y with price Py
Maximize U (X,Y)
Subject to PxX + PyY = 1
Where X ≥ 0, Y
≥0
Mrs = U’x/U’y
Application 2: Cost Minimization
L = Labor, wL = price of 1
Example:
labor
K = Capital, rK = price of 1
capital
Minimize C (L,K) = wL+rK
Subject to P(L,K) = y
Where L≥0,K
Application 3: Optimal Portfolio
≥0
Selection
Mrts = P’l / P’k
5.2.1 & 5.2.2
Optimization function with one variable
Profit Profit = Revenue- costs
optimizati
Most profitable output quantity (y) is a
on is also:
stationary point of (π).
Marginal output rule. (answers where min max) (maximal profit)
MR’ (y) = how much R will approx. change if
y changes 1 unit
MC’ (y) = extra costs if y changes approx. 1
unit
Production rule. (answers profit -/+) (means positive profit
AR (Y) = avg. R at (P)roduction (y) = Tot R-P possible)
AC (Y) = avg. C at (P)roduction (y) = Tot C-P
Supply function. (answers for each price, the
output quantity that maximizes profit)
P ≥ min AC (y) results in profit
P < min AC (y) results in loss
So how to find the Optimal Profit: NOTE: Since a maximum can be
1. π(Y) = R(Y) – C(Y) OR π’(Y) = below 0 we need to find whether
R’(Y) – C’(Y) the profit is positive or negative
2. Stationary point π’(y) = 0 OR R’(Y) = C’(Y)
3. Find MAX using sign chart fill SP neighborhood into the
π’(y)
4. Find X coordinate fill SP into π(Y) = R(Y) – C(Y)
How to find whether the profit is positive or negative
1. AR (Y) ≥ AC (Y)
2. Calc AR and AC using AR = R(y) / y and AC= C(y) / y
3. Y (AR(y) – AC(y)) if > 0 then positive profit
How to find for each price the output quantity that
maximizes profit (Supply function)
1. Combining the previous two questions.
2. Read of the sign chart
for where profit is
positive and where
its negative. P ≥ min AC (y) results in profit P < min AC
(y) results in loss
, Exam cheat sheet week 8
5.5 & 5.6
Constrained (beperkt) Optimization (3) methods (3) applications
In general Optimize (objective function) z(x,y)
Subject to (Constrained) g(x,y) = k
Where (Domain) x in D1 and Y in D2
Feasible point All (x,y) that satisfy the constrained and the domain
Method 1: Substitution (Not always possible to rewrite constrained)
1. Rewrite constrained to x = … OR y=…
2. Substitute result in objective function. (now you’re left with extremum problem with one variable)
3. Find stationary points z’(x or y)= 0
4. Use sign chart to find the maximum of z’
5. Find corresponding coordinate fill in SP into z (step 2)
Method 2: First-order condition for a constrained extremum
(always possible)
1. Find the coordinate/interior point (x,y) using the system.
2. Calculate the function value at the boundary points and the
interior.
3. Step 2 gives answer to min or max.
Method 3: Langrange 2+ constraints (don’t use this 1 or 2 will always work)
1.
Application 1: Utility Maximization Good X with price Px
Example: Good Y with price Py
Maximize U (X,Y)
Subject to PxX + PyY = 1
Where X ≥ 0, Y
≥0
Mrs = U’x/U’y
Application 2: Cost Minimization
L = Labor, wL = price of 1
Example:
labor
K = Capital, rK = price of 1
capital
Minimize C (L,K) = wL+rK
Subject to P(L,K) = y
Where L≥0,K
Application 3: Optimal Portfolio
≥0
Selection
Mrts = P’l / P’k
