The foc and the complementary slackness conditions

The FOC and the complementary slackness conditions (KT1 and KT2) altogether are known
as the Kuhn - Tucker conditions (KT), from the names of two mathematicians that studied
their properties.
Example
We describe the consumption problem using the Kuhn – Tucker conditions: since we already
have some economic intuition on what the solution must look like, we can better see what the
conditions do, and how to solve them. Consider (H) with one physical commodity, 1 S X + =  ,
f(x) = U(x
0
,…, x
S
), g = (g
0
, g
1
,…, g
S+1
) where g
0
(x
0
,…, x
S
) = p w – p
0
x
0
– … – p
S
x
S
is the
budget constraint, g
1
(x
0
,…, x
S
) = x
0
, g
2
(x
0
,…, x
S
) = x
1
… through g
S+1
(x
0
,…, x
S
) = x
S
are the
non- negativity constraints, x ≥ 0.
We construct the Lagrangean:
L(x
0
,…, x
S
, λ
0
,…, λ
S+1
) = U(x
0
,…, x
S
) + λ
0
(p w – p
0
x
0
– … – p
S
x
S
) + Σ
j>0
λ
j
x
j-1
.
We compute the FOCs:
∂L/∂ x
0
: ∂ U(x
0
,…, x
S
) /∂ x
0
– λ
0
p
0
+ λ
1
= 0
…
∂L/∂ x
S
: ∂ U(x
0
,…, x
S
) /∂ x
S
– λ
0
p
S
+ λ
S+1
= 0
We then write the complementary slackness conditions:
min(λ
0
, p w – p
0
x
0
– … – p
S
x
S
) = 0
min(λ
1
, x
0
) = 0
…
min(λ
S+1
, x
S
) = 0
Once we set up the system of equations, we use it to study properties of optimal solution to
the programming problem, see how they change when we modify parameters of the problem
(such as endowments, income and prices), and at times we use them to solve explicitly for a
solution (with many variables, you need a computer to do so).
For a simple numerical example, let S = 2 and U(x
0
, x
1
, x
2
) = x
0
+ (1/2)ln x
1
+ (1/2)ln x
2
, while p
1
= p
2
= 1, and p
0
= ½. Also assume that w
0
= 0, and w
1
= 6, w
2
= 4. Then the KT conditions become (here
we use λ
0
= λ and λ
j
= μ
j-1
for j > 0):

1 – ½ λ + μ
0
= 0
1/2x
1
– λ + μ
1
= 0
1/2x
2
– λ + μ
2
= 0
min(λ, 10 – ½ x
0
– x
1
– x
2
) = 0
min(μ
0
, x
0
) = 0
min(μ
1
, x
1
) = 0
min(μ
2
, x
2
) = 0
Since 1/2x→ +∞ as x→ 0, the solution to this system must have x
1
> 0 and x
2
> 0. Then, from the min
equations, we have μ
1
= μ
2
= 0. Since 1/2x > 0 for x > 0, λ > 0 and ½ x
0
+ x
1
+ x
2
= 10. From the
second and third equation, x
1
= x
2
= x. We have a subsystem of four equations and four unknowns, x
0
,
x, λ and μ
0
:
1 – ½ λ + μ
0
= 0
1/2x – λ = 0
½ x
0
+ 2x = 10
min(μ
0
, x
0
) = 0
Suppose that μ
0
> 0. Then x
0
= 0, and x = 5, λ = 1/10, so that μ
0
+ 1 – 1/20 = 0 has no solution. Then,
μ
0
= 0, λ = 2, x = ¼ and x
0
= 19. Not surprisingly, since prices tomorrow in today’s terms are higher,
lots of consumption of the physical good occurs today, and because of risk aversion, the individual
(fully) insures himself.
A (bit of) geometry
The intuition for the recipe is most easily illustrated for the case when n = 1 (a one-
dimensional maximization problem) and there are two constraints, or m = 2.
Also assume that:
- f, g
1
and g
2
are continuously differentiable functions;
- the set satisfying the two constraints is the finite union of disjoint intervals;
- there is no point x at which both constraints bind simultaneously;
- if g
j
(x) = 0 some j, then the derivative of g
1
is non-zero.