2.1. OPTIMIZATION WITH PARTIAL DERIVATIVES 43
Solve:
30x C 18y D 36
30x C 25y D 40
7y D 4
y D 4=7
5x 12=7 D 42=7
5x D 30=7
x D 6=7
So we find a single critical point .6=7; 4=7/ Considering the geometry of a plane, we see
that it has a unique closest approach to the origin. So, computing z D 3x C 2y C 4 D 18=7
8=7 C28=7 D 2=7, gives us that the point on the plane closest to the origin is
.6=7; 4=7; 2=7/:
˙
Clever students will have noticed that when we were minimizing
p
10x
2
C 12xy C5y
2
C 24x C128 C 16
the numerators of the partial derivatives were exactly the partial derivatives of
10x
2
C 12xy C5y
2
C 24x C128 C 16. is observation is an instance of a more general
shortcut.
Knowledge Box 2.5
Minimization of distance—a shortcut
Suppose that we are optimizing a function g.x; y/ D
p
f .x; y/ or g.x; y; z/ D
p
f .x; y; z/. en the critical points for optimization of g and f are the same,
and the second derivative tests agree. e actual values of the function are not,
which means care is required.
44 2. MULTIVARIATE AND CONSTRAINED OPTIMIZATION
Why does this shortcut work? It is because if 0 < a < b, then 0 <
p
a <
p
b. So optimizing the
value finds where the optimum of the square root of the value is as well.
Example 2.7 Minimize:
g.x; y/ D
p
x
2
C y
2
C 3
Solution:
In this case g.x; y/ D
p
f .x; y/ where f .x; y/ D x
2
C y
2
C 3, so the shortcut applies.
Finding the relevant partials we see that
f
x
.x; y/ D 2x
f
y
.x; y/ D 2y
which is easy to see has a critical point at (0,0). e second derivative test shows that:
f
xx
f
yy
f
2
xy
D 2 2 0 D 4 > 0
So the critical point is an optimum. Since f
xx
D 2 > 0, it is a minimum. is means the
minimum value of g.x; y/ is at the point .0; 0;
p
3/.
˙
Definition 2.1 A function m.x/ is monotone increasing if, whenever a < b and m.x/ exists on
Πa; b , then m.a/ < m.b/.
Notice that m.x/ D
p
x is monotone increasing. In fact the shortcut in Knowledge Box 2.5
works for any monotone increasing function. ese functions include:
e
x
,
ln
.x/
,
tan
1
.x/
,
x
n
when n is odd, and
n
p
x.
Knowledge Box 2.6
A test for a function being monotone increasing
We know that a function is increasing if its first derivative is positive. If a func-
tion exists and has a positive derivative on the interval [a,b] it is a monotone
increasing function on [a,b].
2.1. OPTIMIZATION WITH PARTIAL DERIVATIVES 45
Example 2.8 Show that y D ln.x/ is monotone increasing where it exists.
Solution:
According to Knowledge Box 2.6, a function is increasing when its first derivative is positive.
e function y D ln.x/ only exists on the interval .0; 1/. Its derivative is y
0
D 1=x which is
positive for any positive x. us, ln.x/ is increasing everywhere that it exists and so is a monotone
increasing function.
˙
Using an extension of the shortcut from Knowledge Box 2.5 with these other monotone func-
tions will make the homework problems much easier. Let’s practice.
Example 2.9 Find the point of closest approach to the origin on the plane
3x 4y z D 4:
Solution:
First find the distance between the origin .0; 0; 0/ and a generic point .x; y; 3x 4y 4/
on the plane. We get that
d.x; y/ D
p
x
2
C y
2
C 9x
2
24xy C 16y
2
32y 24x C 16
d.x; y/ D
p
10x
2
24xy C 17y
2
32y 24x C 16
Using the distance minimization shortcut, find the critical points of
d
2
.x; y/ D 10x
2
24xy C 17y
2
32y 24x C 16
which is the relatively simple quadratic case.
f
x
.x; y/ D 20x 24y 24
f
y
.x; y/ D 24x C 34y 32
f
xx
.x; y/ D 20
f
yy
.x; y/ D 34
f
x;y
.x; y/ D 24
D.x; y/ D 20 34 .24/
2
D 104
46 2. MULTIVARIATE AND CONSTRAINED OPTIMIZATION
So we see that D > 0 and f
xx
> 0 meaning we will find a minimum distance. Now we must
solve for the critical point.
20x 24y D 24
24x C 34y D 32
5x 6y D 6
12x C 17y D 16
60x 72y D 72
60x C 85y D 80
13y D 152
y D 152=13
5x 912=13 D 78=13
5x D 990=13
x D 198=13
So the critical point is roughly at x D 15:23, y D 11:69, making the point of closest approach
.15:23; 11:69; 2:93/. Not a lot easier – but we avoided all sorts of square roots.
˙
PROBLEMS
Problem 2.10 Find the critical points for each of the following functions, and use the second
derivative test to classify them as maxima, minima, or saddle points. Also, find the value of the
function at the critical point.
1. f .x; y/ D x
2
C y
2
6x 4y C 4
2. g.x; y/ D 3x
2
y Cy
3
3x
2
3y
2
C 2
3. h.x; y/
D
12
x
2
C
3xy
y
2
C
2x
y
C
1
4. q.x; y/ D x
2
C xy Cy
2
C 4x 5y C 2
5. r.x; y/ D x
4
C y
4
4xy C 1
6. s.x; y/
D
x
3
C
2y
3
4xy
2.1. OPTIMIZATION WITH PARTIAL DERIVATIVES 47
Problem 2.11 Show that
f .x; y/ D x
4
C y
4
is an example of a function where the second derivative test yields no useful information.
Problem 2.12 Suppose that
f .x; y/ D ax
2
C bxy C cy
2
C dx C ey C f:
Show that this function has at most one critical point. When it does have a critical point, derive
rules based on the constants a-f for classifying that critical point.
Problem 2.13 Find the point on the plane
x C y C z D 4
closest to the origin.
Problem 2.14 Find the point on the plane
z D x=3 Cy=4 C 1
closest to the origin.
Problem 2.15 Find the point on the tangent plane of
f .x; y/ D x
2
C y
2
C 1
at .2; 1; 6/ that is closest to the origin.
Problem 2.16 Find the critical points for each of the following functions, and use the second
derivative test to classify them as maxima, minima, or saddle points. Also, find the value of the
function at the critical point.
1. f .x; y/ D ln.x
2
C y
2
C 4/
2. g.x; y/ D e
x
4
Cy
4
36xyC6
3. h.x; y/ D tan
1
.x
3
C 2y
3
4xy/
4. r.x; y/ D .x
2
C xy C3y
2
4x C3y C 1/
5
5. s.x; y/ D
p
x
2
C y
2
C 14
6. q.x; y/ D tan
1
e
x
2
CxyCy
2
2x3yC1
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