38 2. MULTIVARIATE AND CONSTRAINED OPTIMIZATION
Examine the formula for D.a; b/ and see if you can tell why this is true.
Example 2.3 Find and classify the critical point of
h.x; y/ D 4 C 3x C 5y 3x
2
C xy 2y
2
Solution:
e partials are f
x
D 3 6x C y and f
y
D 5 C x 4y. Setting these equal to zero we
obtain the simultaneous system of equations:
6x y D 3
x C 4y D 5
Solve:
6x C 24y D 30
23y D 33
y D 33=23
x C 132=23 D 115=23
x D 17=23
x D 17=23
Now compute D.x; y/. We see f
xx
D 6, f
yy
D 4 and f
xy
D 1. So,
D D .6/.4/ 1
2
D 23:
Since D > 0, the critical point is an optimum of some sort. e fact that f
xx
< 0 or the fact
that f
yy
< 0 suffices to tell us this point is a maximum.
˙
Notice that in Example 2.3 the equation is a quadratic with large negative squared terms and
a small mixed term (xy). e fact that its sole critical point is a maximum means that it is a
paraboloid opening downward. In the next example we look at a quadratic with a large mixed
term.
2.1. OPTIMIZATION WITH PARTIAL DERIVATIVES 39
Example 2.4 Find and classify the critical points of
q.x; y/ D x
2
C 6xy Cy
2
14x 10y C 3:
Solution:
Start by computing the needed partials and the discriminant.
f
x
.x; y/ D 2x C6y 14
f
y
.x; y/ D 6x C2y 10
f
xx
D 2
f
yy
D 2
f
xy
D 6
D.x; y/ D 2 2 6
2
D 32
ese calculations show that any critical point we find will be a saddle point since D < 0. To
find the critical point we solve the first partials equal to zero obtaining the system
2x C 6y D 14
6x C 2y D 10
6x C 18y D 42 3x line one
16y D 32 Line 3 minus line two
y D 2
2x C 12 D 14 Substitute
2x D 2
x D 1
So we see that the function has, as its sole critical point, a saddle point at (1,2).
˙
40 2. MULTIVARIATE AND CONSTRAINED OPTIMIZATION
e last two examples have been bivariate quadratic equations with a constant discriminant. is
is a feature of all bivariate quadratics. All functions of this kind have a single critical point and
a constant discriminant. For the next example we will tackle a more challenging example.
Example 2.5 Find and classify the critical points of f .x; y/ D x
4
C y
4
16xy C6:
Solution:
Start by computing the needed partials and the discriminant.
f
x
.x; y/ D 4x
3
16y
f
y
.x; y/ D 4y
3
16x
f
xx
.x; y/ D 12x
2
f
yy
.x; y/ D 12y
2
f
xy
.x; y/ D 16
D.x; y/ D f
xx
f
yy
f
2
xy
D.x; y/ D 144x
2
y
2
256
Next, we need to find the critical points by solving the system
4x
3
D 16y
4y
3
D 16x
or
x
3
D 4y
y
3
D 4x
Solve the second equation to get y D
3
p
4x and plug into the first, obtaining:
x
3
D 4
3
p
4x
x
9
D 256x Cube both sides
2.1. OPTIMIZATION WITH PARTIAL DERIVATIVES 41
x
9
256x D 0
x.x
8
256/ D 0
x D 0; ˙
8
p
256
x D 0; ˙2
Since the equation system is symmetric in x and y we may deduce that y D 0; ˙2 as well.
Referring back to the original equations, it is not hard to see that, when x D 0, y D 0; when
x D 2 so must y; when x D 2 so must y. is gives us three critical points: .2; 2/; .0; 0/;
and .2; 2/. Next, we check the discriminant at the critical points.
D.2; 2/ D 2304 256 D 2048 > 0
Since f
xx
.2; 2/ > 0, this point is a minimum.
D.2; 2/ D 2048 > 0
Another minimum, and
D.0; 0/ D 256;
so this point is a saddle point.
˙
One application of multivariate optimization is to minimize distances. For that we should state,
or re-state, the definition of distance.
Knowledge Box 2.4
e definition of distance
If p D .x
0
; y
0
/ and q D .x
1
; y
1
/ are points in the plane, then the distance be-
tween p and q is
d.p; q/ D
p
.x
0
x
1
/
2
C .y
0
y
1
/
2
:
If r D .x
0
; y
0
; z
0
/ and s D .x
1
; y
1
; z
1
/ are points in space, then the distance
between r and s is
d.r; s/ D
p
.x
0
x
1
/
2
C .y
0
y
1
/
2
C .z
0
z
1
/
2
:
42 2. MULTIVARIATE AND CONSTRAINED OPTIMIZATION
e definition in Knowledge Box 2.4 can be extended to any number of dimensions – the dis-
tance between two points is the square root of the sum of the squared differences of the individual
coordinates of the point. For this text only two- and three-dimensional distances are required.
With the definition of distance in place, we can now pose a standard type of problem.
Example 2.6 What point on the plane z D 3x C 2y C4 is closest to the origin?
Solution:
e origin is the point .0; 0; 0/, while a general point .x; y; z/ on the plane has the form
.x; y; 3x C 2y C4/. at means that the distance we are trying to minimize is given by
d D
p
.x 0/
2
C .y 0/
2
C .3x C 2y C4 0/
2
D
p
10x
2
C 12xy C5y
2
C 24x C16y C16
e partial derivatives of this function are:
d
x
D
20x C 12y C 24
2
p
10x
2
C 12xy C5y
2
C 24x C16y C16
d
y
D
12x C 10y C16
2
p
10x
2
C 12xy C5y
2
C 24x C16y C16
We need to solve the simultaneous equation in which each of these partials is zero. Remember a
fraction is zero only where its numerator is zero. is maxim gives us the simultaneous equations:
20x C 12y C 24 D 0
12x C 10y C16 D 0
Simplify
5x C 3y D 6
6x C 5y D 8
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