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The Mathematics of Optimization
Christopher Makler
Stanford University Department of Economics
Econ 50: Lecture 7
Today's Agenda
- Solution functions and comparative statics
- Unconstrained optimization
- Constrained optimization and the Lagrange method
Solution Functions and Comparative Statics
Math problems with numbers
Math problems with parameters
Plot the line \(y = 5 - \frac{1}{2}x\)
10
5
4
3
2
7
6
1
9
12
11
2
1
4
3
6
5
8
7
x = 0 \Rightarrow y = 5
y = 0 \Rightarrow x = 10
Plot the line \(y = a - bx\)
x = 0 \Rightarrow y = a
y = 0 \Rightarrow x = \frac{a}{b}
a
\tfrac{a}{b}
y = a - bx
y = 5 - \tfrac{1}{2}x
Math problems with numbers
Math problems with parameters
Find the intersection of the lines \(y = 5 - \frac{1}{2}x\) and \(y = 2x\)
10
5
4
3
2
7
6
1
9
12
11
2
1
4
3
6
5
8
7
\text{Solve for }x: 5 - \tfrac{1}{2}x = 2x \Rightarrow x^* = 2
\text{Plug into }y = 2x \Rightarrow y^* = 4
Find the intersection of the lines \(y = a - bx\) and \(y = cx\)
a - bx = cx \Rightarrow x^*(a,b,c) = {a \over b + c}
y = cx \Rightarrow y^*(a,b,c) = \frac{ac}{b+c}
\tfrac{ac}{b + c}
\tfrac{a}{b+c}
Find the intersection of the lines \(y = 5 - \frac{1}{2}x\) and \(y = 2x\)
a
\tfrac{a}{b}
y = a - bx
y = cx
y = 5 - \tfrac{1}{2}x
y = 2x
X^* = (2,4)
X^*(a,b,c) = \left({a \over b+c},{ac \over b + c}\right)
[SOLUTIONS]
[SOLUTION FUNCTIONS]
x^*(a,b,c) = {a \over b + c}
y^*(a,b,c) = \frac{ac}{b+c}
What happens to the intersection when \(a\), \(b\), or \(c\) increases?
Comparative Statics
- When you solve a parameterized problem,
the result is a solution function: the endogenous variables as functions of the exogenous parameters
- Comparative statics is the analysis of the behavior of these solution functions: how does a change in the parameters of the problem affect its solution?
Unconstrained Optimization
Constrained Optimization
Think about maximizing each of these functions subject to the constraint \(0 \le x \le 10\).
Plot the graph on that interval; then find and plot the derivative \(f'(x)\) on that same interval.
Which function(s) reach their maximum in the domain [0, 10] at a point where \(f'(x) = 0\)?
f(x) = 5 + 4x - x^2
f(x) = 10 - |2-x|
f(x) = 9 - (x-11)^2
f(x) = 1 + \tfrac{1}{5}(x-5)^2
f(x) = 10 - x
f(x) = 3
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Sufficient conditions for an interior optimum characterized by \(f'(x)=0\) with constraint \(x \in [0,10]\)
- \(f'(0) > 0\)
- \(f'(10) < 0\)
- \(f'(x)\) continuous and strictly decreasing on \([0,10]\)
f(x)
f'(x)
x
x
10
0
10
Canonical Constrained Optimization Problem
f(x_1,x_2)
\text{s.t. }
g(x_1,x_2) = k
k - g(x_1,x_2) = 0
\mathcal{L}(x_1,x_2,\lambda)=
\displaystyle{\max_{x_1,x_2}}
f(x_1,x_2)
k - g(x_1,x_2)
+ \lambda\ (
)
Suppose \(g(x_1,x_2)\) is monotonic (increasing in both \(x_1\) and \(x_2\)).
Then \(k - g(x_1,x_2)\) is negative if you're outside of the constraint,
positive if you're inside the constraint,
and zero if you're along the constraint.
OBJECTIVE
FUNCTION
CONSTRAINT
\mathcal{L}(x_1,x_2,\lambda)=
f(x_1,x_2)
k - g(x_1,x_2)
+ \lambda\ (
)
\displaystyle{\partial \mathcal{L} \over \partial x_1} =
FIRST ORDER CONDITIONS
\displaystyle{\partial \mathcal{L} \over \partial x_2} =
\displaystyle{\partial \mathcal{L} \over \partial \lambda} =
\displaystyle{\partial f \over \partial x_1}
\displaystyle{\partial f \over \partial x_2}
k - g(x_1,x_2)
=0
- \lambda\ \times
\displaystyle{\partial g \over \partial x_1}
\displaystyle{\partial g \over \partial x_2}
- \lambda\ \times
=0
=0
3 equations,
3 unknowns
Solve for \(x_1\), \(x_2\), and \(\lambda\)
How does the Lagrange method work?
It finds the point along the constraint where the
level set of the objective function passing through that point
is tangent to the constraint
\mathcal{L}(x_1,x_2,\lambda)=
f(x_1,x_2)
k - g(x_1,x_2)
+ \lambda\ (
)
\displaystyle{\partial \mathcal{L} \over \partial x_1} =
FIRST ORDER CONDITIONS
\displaystyle{\partial \mathcal{L} \over \partial x_2} =
\displaystyle{\partial \mathcal{L} \over \partial \lambda} =
\displaystyle{\partial f \over \partial x_1}
\displaystyle{\partial f \over \partial x_2}
k - g(x_1,x_2)
=0
- \lambda\ \times
\displaystyle{\partial g \over \partial x_1}
\displaystyle{\partial g \over \partial x_2}
- \lambda\ \times
=0
=0
\displaystyle{\Rightarrow \lambda\ = {{\partial f /\partial x_1} \over {\partial g/\partial x_1}}}
\displaystyle{\Rightarrow \lambda\ = {{\partial f /\partial x_2} \over {\partial g/\partial x_2}}}
\displaystyle{\Rightarrow {{\partial f /\partial x_1} \over {\partial f/\partial x_2}} = {{\partial g /\partial x_1} \over {\partial g/\partial x_2}}}
TANGENCY
CONDITION
CONSTRAINT
Example: Fence Problem
You have 40 feet of fence and want to enclose the maximum possible area.
Example: Fence Problem
You have 40 feet of fence and want to enclose the maximum possible area.
f(x_1,x_2)
\text{s.t. }
g(x_1,x_2) = k
k - g(x_1,x_2) = 0
\mathcal{L}(L,W,\lambda)=
\displaystyle{\max_{x_1,x_2}}
L \times W
40 - 2L - 2W
+ \lambda\ (
)
OBJECTIVE
FUNCTION
CONSTRAINT
L \times W
40 - 2L - 2W = 0
2L + 2W = 40
\mathcal{L}(L,W,\lambda)=
L \times W
40 - 2L - 2W
+ \lambda\ (
)
\displaystyle{\partial \mathcal{L} \over \partial L} =
FIRST ORDER CONDITIONS
\displaystyle{\partial \mathcal{L} \over \partial W} =
\displaystyle{\partial \mathcal{L} \over \partial \lambda} =
W
L
40 - 2L - 2W
=0
- \lambda\ \times
2
2
- \lambda\ \times
=0
=0
\displaystyle{\Rightarrow \lambda\ = {W \over 2}}
\displaystyle{\Rightarrow \lambda\ = {L \over 2}}
W = L
TANGENCY
CONDITION
CONSTRAINT
2L + 2W = 40
W = L
TANGENCY
CONDITION
CONSTRAINT
2L + 2W = 40
Meaning of the Lagrange multiplier
Suppose you have \(F\) feet of fence instead of 40.
\displaystyle{\partial \mathcal{L} \over \partial L} =
\displaystyle{\partial \mathcal{L} \over \partial W} =
\displaystyle{\partial \mathcal{L} \over \partial \lambda} =
W
L
F - 2L - 2W
=0
- \lambda\ \times
2
2
- \lambda\ \times
=0
=0
\displaystyle{\Rightarrow \lambda\ = {W \over 2}}
\displaystyle{\Rightarrow \lambda\ = {L \over 2}}
W = L
TANGENCY
CONDITION
CONSTRAINT
2L + 2W = F
SOLUTION
FUNCTIONS
L^*(F) = {F \over 4}
W^*(F) = {F \over 4}
\lambda^*(F) = {F \over 8}
Meaning of the Lagrange multiplier
SOLUTION
FUNCTIONS
L^*(F) = {F \over 4}
W^*(F) = {F \over 4}
\lambda^*(F) = {F \over 8}
Maximum enclosable area as a function of F:
A^*(F) =
L^*(F) \times W^*(F)
= {F \over 4} \times {F \over 4}
= {F^2 \over 16}
Meaning of the Lagrange multiplier
In a constrained optimization problem,
the constraint may be determined by a parameter:
how much labor you have, how much money you have,
how many units of a good you want to produce, etc.
The Lagrange multiplier tells you how much
the optimized value of the objective function will change
due to a change in that parameter.
"How much more utility could you get
if you had another hour of labor,
and used it optimally?"
Next Steps
- Tomorrow night: math homework due
- Rest of this week: use these techniques
to solve for the optimal division of labor
between two competing uses
Meaning of the Lagrange multiplier
SOLUTION FUNCTIONS
L^* = {F \over 4}
W^* = {F \over 4}
\lambda^* = {F \over 8}
Econ 50 | Lecture 07
By Chris Makler
Econ 50 | Lecture 07
Math Review: Solution Functions and Optimization
