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Summary and examples for section 14.7
Maximum and Minimum of Functions of Several Variables
One of the most important uses of calculus is for optimization. Indeed, you have seen optimization problems
reappear in a number of places. Undoubtedly, you recognize that in this age, manufacturing plants need to
maximize the quality of their products by minimizing defects. In order to design more and more efficient
spacecraft, engineers need to minimize the weight of a structure while maximizing its strength. The list goes
on and on. In this section, we introduce the mathematical basis for optimizing functions of several variables.
Carefully examine the surface
, shown in Figure 1a for -2 x 4 and -1 y 4 . From
the graph, notice that you can identify both a peak and a trough. We can zoom in to get a better view of the
peak (see Figure 1b for 0.9 x 1.1 and 0.9 y 1.1 ), we can zoom in to get a better view of the trough
(see Figure 1c for -1.1 x -0.9 and 0.9 y 1.1 ). Such points are referred to as local extrema, which we
define in the following.
Figure 1a: z = xe-x2/2-y3/3+y
Figure 1b: Local maximum
Figure 1c: Local minimum.
Look carefully at Figures 1b and 1c; it appears that at both local extrema, the tangent plane is horizontal. So,
both first partial derivatives are zero, in other words, the gradient is zero at the local extrema. However, as
we'll see, having zero first partial derivatives at a point does not by itself guarantee a local extremum.
DEFINITIONS (MAX, MIN, CRITICAL POINT, GRADIENT, SADDLE)
If f (a, b) f (x, y) for all (x, y) “near'' (a, b) , we call f (a, b) a local maximum.
If f (a, b) f (x, y) for all (x, y) “near'' (a, b) , we call f (a, b) a local minimum.
If f (a, b) f (x, y) for all (x, y) in the domain, we call f (a, b) a global maximum.
If f (a, b) f (x, y) for all (x, y in the domain, we call f (a, b) a global minimum.
Gradient of f at (a,b) noted grad f or ▼f(a,b)= <
,
>
The point (a, b) is a critical point of f (x, y) if either gradient of f is zero, or, one or both of
do not exist at (a, b).
and
The point (a,b) is a saddle if it is a critical point, which is no a local maximum or minimum (see below).
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EXAMPLE 1: FINDING LOCAL EXTREMA GRAPHICALLY
Find all critical points of
and analyze each critical point graphically.
1- First, we compute the first partial derivatives:
and
2- Notice that both derivatives exist for all (x, y). So, the only critical points are solutions of
3- Since exponentials are always positive, we have
We have
.
if and only if 1-x2 = 0, that is, when x = 1
if and only if x(-y2+1) = 0 , that is, when x = 0 or y =
1.
4-Notice that we need x = 1 and either x = 0 or y = 1.
However, if x = 0, then
so there are no critical points with x = 0
This leaves all combinations of x = 1 and y = 1 as critical points: (1, 1) , (-1, 1) , (1, -1) and (-1, -1) .
5- Keep in mind that the critical points are only candidates for local extrema; we must look further to
determine whether they correspond to extrema by using the second derivatives test (next) or the graphs.
6- Using graphs and contour map:We zoom in on each critical point in turn, to graphically identify any
local extrema. We have already seen (see Figures 1b and 1c) that f (x, y) has a local maximum at (1, 1) and
a local minimum at (-1, 1).
Figures 1d and 1e below show z = f (x, y) zoomed in on (1, -1) and (-1, -1), respectively. In Figure 1d,
notice that in the plane x = 1 (extending left to right), the point at (1, -1) is a local minimum. However, in
the plane y = -1 (extending back to front), the point at (1, -1) is a local maximum. This point is therefore not
a local extremum. We refer to such a point as a saddle point. (It looks like a saddle, doesn't it?) Similarly,
in Figure 1e, at (-1, -1) we have a saddle point.
Figure 1d: Saddle point at (1, -1).
Figure 1e: Saddle point at (-1, -1).
Figure 1f: contour map
To further explore this graphically, we show a contour plot of f (x, y) = xe-x2/2-y3/3+y in Figure 1f. Notice that
near the local maximum at (1, 1) and the local minimum at (-1, 1) the level curves look like concentric
circles. This corresponds to the paraboloid-like shape of the surface near these points (see Figures1band 1c).
Concentric ovals are characteristic of local extrema. Notice that, without the level curves labeled, there is
no way to tell from the contour plot which is the maximum and which is the minimum. Saddle points are
typically characterized by the hyperbolic-looking curves seen near (-1, -1) and (1, -1), (or in general
level curves that intersect as you can see from examples in the textbook).
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THEOREM: SECOND DERIVATIVES TEST
Suppose that f (x, y) has continuous second-order partial derivatives in some open disk containing the point
(a, b) and that fx (a, b) = fy (a, b) = 0. Define the discriminant D for the point (a, b) by
D(a, b) = f xx(a, b)f yy(a, b)-[f xy(a, b)]2.
If D(a, b) > 0 and f xx(a, b) > 0 , then f has a local minimum at (a, b).
If D(a, b) > 0 and f xx(a, b) < 0 , then f has a local maximum at (a, b).
If D(a, b) < 0, then f has a saddle point at (a, b).
If D(a, b) = 0 , then no conclusion can be drawn.
It's important to make some sense of this result (in other words, to understand it and not just memorize it).
Note that to have D(a, b) > 0, we must have both f xx(a, b) > 0 and f yy(a, b) > 0 or both f xx(a, b) < 0 and
f yy(a, b) < 0 . In the first case, notice that the surface z = f (x, y) will be concave up in the plane y = b and
concave up in the plane x = a . In this case, the surface will look like an upward-shaped paraboloid near the
point (a, b). Consequently, the critical point (a, b) must be the location of a local minimum. In the second
case, both f xx(a, b) < 0 and f yy(a, b) < 0 . This says that the surface z = f (x, y) will be concave down in the
plane y = b and concave down in the plane x = a . So, in this case, the surface looks like a downward-shaped
paraboloid near the point (a, b) , which must then be the location of a local maximum. Observe that one way
to get D(a, b) < 0 is for f xx(a, b) and f yy(a, b) to have opposite signs (one positive and one negative). To
have opposite concavities in the planes x = a and y = b means that there is a saddle point at (a, b) , as in
Figures 1d and 1e.
EXAMPLE 2: USING THE DISCRIMINANT TO FIND LOCAL EXTREMA
Find and classify all critical points for f (x, y) = 2x2-y3-2xy.
1- We first compute the first partial derivatives: f x = 4x-2y and f y = -3y2-2x.
2- Since both derivatives are defined for all (x, y), the critical points are solutions of the two equations:
f x = 4x-2y = 0
and
f y = -3y2-2x = 0.
3- Solving the first equation for y, we get y = 2x . Substituting this into the second equation, we have
0= -3(2x)2-2x = -6x2-2x = -2x(6x+1), so that x = 0 or x = - . Thus with y = 2x = 0 or 4- The only two critical points are then (0, 0) and
5- To classify these points, we first compute the second partial derivatives: f xx = 4, f yy = -6y and f xy = -2
and then test the discriminant. We have D(0, 0) = (4)(0)-(-2)2 = -4 and
6- We conclude that there is a saddle point of f at (0, 0) , since D(0, 0) < 0.
7- Further, there is a local minimum at
since
and
.
The following example shows that the second derivatives test does not always help us to classify a critical point.
EXAMPLE 3: CLASSIFYING CRITICAL POINTS.
Find and classify all critical points for f (x, y) = x3-2y2-2y4+3x2y.
Here, we have f x = 3x2+6xy and f y = -4y-8y3+3x2. Since both f x and f y exist for all (x, y), the critical points are
solutions of the two equations:
f x = 3x2+6xy = 0 and f y = -4y-8y3+3x2 = 0 .
From the first equation, we have 3x2+6xy = 3x(x+2y) =0, so
x = 0 or x = -2y
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Substituting x = 0 into the second equation, we have -4y-8y3 = -4y(1+2y2)=0
The only (real) solution of this equation is y = 0. This says that for x = 0, we have only one critical point: (0, 0).
Substituting x = -2y into the second equation, we have -4y-8y3+3(4y2) = -4y(1+2y2-3y) = -4y(2y-1)(y-1)=0
The solutions of this equation are y = 0 , y = and y = 1 , with corresponding critical points (0, 0) , (-1,1/2)and (-2, 1)
.
To classify the critical points, we compute the second partial derivatives: f xx = 6x+6y, f yy = -4-24y2 and f xy = 6x
and evaluate the discriminant at each critical point. We have
D(0, 0) = (0)( -4) - (0)2 = 0,
We conclude that f has a saddle point at
,
, since
D(-2, 1) = (-6)(-28)-(-12)2 = 24.
.
Further, f has a local maximum at (-2, 1) since D(-2, 1) > 0 and fxx (-2, 1) < 0.
Unfortunately, the Theorem gives us no information about the critical point (0, 0), since D(0, 0) = 0 . However, notice
that in the plane y = 0 we have f (x, y) = x3. In two dimensions, the curve z = x3 has an inflection point at x = 0. This
shows that there is no local extremum at (0, 0). The surface near (0, 0) is shown in first Figure. Since the graphs are
not especially clear for, it's good that we have done the analysis!
z = f (x, y) near (0, 0).
z = f (x, y) near (-2, 1)
z = f (x, y) near