Unconstrained maximization/minimization with several variables
Ok, so now the obvious question. What do we do when the problem involves more than one variable?
Minimise/maximise
f ( x1 , x 2 , .... , x n )
subject to ( x1 , x 2 , .... , x n ) ∈ Rn
Going back to 1st year calculus, we note that
Theorem 6
x* is a stationary point if
∂f ( x*)
= 0
∂xi
∀i
Then, x* = (x*1, x*2, … , x*n) is called a stationary point and may be a local minimum, local maximum, or
neither.
In words, a point x* = (x*1, x*2, … , x*n) is a potential extrema if all partial derivatives of the function are
equal to 0 at x*.
Now, of course, we must determine what type of extreme point x* is. Fortunately, if we go back to our
earlier work on the Hessian and its relationship to concave and convex functions, we can develop
appropriate tests. Define Hk to be the kth leading principal minor of the Hessian matrix. (The leading
principal minor of an n × n matrix is the determinant of the k × k matrix obtained by deleting the last
(n − k ) rows and columns of the matrix.)
Theorem 7
If Hk(x*) > 0 for k = 1, 2, …, n then x* is a local minimum.
Theorem 7’
If Hk(x*) is non-zero for k = 1, 2, …, n and has the same sign as (–1)k then x* is a local maximum.
Theorem 7”
If Hn(x*) ≠ 0 and the conditions of Theorems 7 and 7’ do not apply, then x* is neither a local minimum
nor maximum. Instead, we may label x* as a saddle point.
Note
If Hn(x*) = 0 for x*, then the above tests are inconclusive and we cannot determine of the point is a
minimum, maximum or saddle point.
Example 1
Find all local maxima, minima, and saddle points for
f ( x1 , x 2 , x3 ) = x1 x 2 + x 2 x3 + x1 x3 .
df
= x 2 + x3
dx1
df
= x1 + x 3
dx 2
df
= x1 + x 2
dx 3
Setting these three equations to 0, we find that the only candidate point is x* = (0, 0, 0).
The Hessian for this matrix is
0 1 1 
H ( x1 , x2 , x3 ) = 1 0 1
1 1 0
The 1st principal minors are: 0, 0, 0.
(Remember these are the diagonal entries of the Hessian.)
The first leading principal minor is 0.
The 2nd principal minors are: –1, –1, –1
(Confirm this).
The second leading principal minor is –1.
The 3rd principal minor is 0(-1) – 1(-1) + 1(1) = 2.
(Confirm this by co-factoring along the top row.)
Thus we note that since H2 < 0, the function cannot be a minimum. Since H3 = 2 is greater than 0, the
function cannot be a maximum. We may therefore label it as a saddle point, according to Theorem 7”.
Example 2 (Collusive Duopoly Model)
There are two firms producing widgets. It costs the 1st firm q1 dollars to produce q1 widgets. The 2nd firm
can produce q2 widgets at a cost of 0.5q22. A total of q widgets are to be produced and consumers will
pay $200 – q for each widget. If the two firms collude to maximize their profits, how many widgets
should each firm produce?
Total sales:
q1 + q2
Total revenue: [200 – (q1 + q2)](q1 + q2)
Total cost:
q1 + 0.5q22
Total profit:
Revenue – Costs
[200 – (q1 + q2)] (q1 + q2) – (q1 + 0.5q22)
200q1 + 200q2 – q12 – q1q2 – q1q2 – q22 – q1 – 0.5q22
199q1 + 200q2 – q12 – 2q1q2 – 1.5q22
df
= 199 − 2q1 − 2q 2
dq1
df
= 200 − 2q1 − 3q 2
dq 2
Setting these equations to zero and solving, using Gaussian elimination, we get q2 = 1, q1 = 98.5 as the
only candidate for extreme point.
 − 2 − 2
H (q1 , q2 ) = 

 − 2 − 3
The 1st principal minors are –2, -3. The 1st leading principal minor is –2.
The 2nd principal minors are 2 (confirm this). The 2nd leading principal minor is +2.
Therefore, this function satisfies the tests for a local maximum. Note also that this function satisfies the
test for a concave function. Therefore our solution is also a global maximum.
Example 3
Find all local maxima, minima, and saddle points for f ( x1 , x 2 ) = x14 + 3 x1 x 2 + x 23 .
df
= 4 x13 + 3 x 2
dx1
df
= 3x1 + 3x 22
dx 2
(1)
(2)
From (2) we have x1 = − x 22 . Substituting into (1) we find 4( − x 22 ) 3 + 3 x 2 = 0 or x2 = 0.944 and
x1 = −0.891 . Thus (–0.891, 0.944) is a candidate point along with (0, 0).
The Hessian for this matrix is:
12 x 2
H ( x1 , x 2 ) =  1
 3
3 

6 x2 
The 1st principal minors are: 12x12, 6x2. (Remember these are the diagonal entries of the Hessian.)
The first leading principal minor is 12x12.
Evaluated at (–0.891, 0.944) we have the 1st leading principal minor = 9.533.
Evaluated at (0, 0) 1st leading principal minor is 0.
The 2nd principal minor is: 72x12x2 – 9.
Evaluated at (–0.891, 0.944), the second leading principal minor is 45.
Evaluated at (0, 0), 2nd leading principal minor is –9.
At (0, 0) the tests indicate a saddle point.
At (–0.891, 0.944) both leading principal minors are positive.
This is a local minimum. (Why isn’t this a global minimum?)