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Part I
Applications of Partial Derivatives
1 Extreme Values
Some important applications of the partial derivatives are Önding the maximum and the minimum values of a function of several variables, and determining the points where the extreme (maximum or minimum) values occur. In this section we learn how to Önd and classify the extreme values of functions of several variables.
DeÖnition 1 A function of two variables f (x; y) has a local maximum value at the point (a; b) in its domain, if f (a; b) f (x; y) for all points in its domain which are in an open disk centered at (a; b). If the inequality holds for all points in the domain of f , then f has a global or absolute maximum at the point (a; b). A function of two variables f (x; y) has a local minimum value at the point (a; b) in its domain, if f (a; b) f (x; y) for all points in its domain which are in an open disk centered at (a; b). If the inequality holds for all points in the domain of f , then f has a global or absolute minimum at the point (a; b).
In order to Önd the local extreme values of a function y = f (x) of a single variable, we consider the points where f 0 (x) = 0, called critical points, and the points where f 0 (x) does not exist, called the singular points. Similar ideas are used to Önd the local extreme values of a function f of several variables. We start by giving the deÖnitions for the functions of two variables.
DeÖnition 2 Let (a; b) be an interior point of the domain of a function f (x; y). If rf (a; b) = 0, then (a; b) is called a critical point of f ; if rf (a; b) does not exist, then (a; b) is called a singular point of f.
DeÖnition 3 A di§erentiable function f (x; y) has a saddle point at a critical point (a; b) if in every open ball centered at (a; b), there are points in the domain of f that satisfy f (x; y) > f (a; b), and there are points in the domain of f that satisfy f (x; y) < f (a; b).
In other words, a saddle point is a critical point which is neither a local maximum nor a local minimum point. The following theorem shows that the extreme values of a function can occur only at certain points.
Theorem 4 (Necessary conditions for the extreme values) A function f can have local or absolute extreme values only at a point P 0 only if P 0 is:
- a critical point, or
- a singular point, or
- a boundary point, of f.
Proof. We prove the theorem only for the functions of two variables. Assume that (a; b) be a point on the domain of f. Then there are two possibilities:
(a; b) is in the boundary of the domain of f (Boundary Point).
(a; b) is in the interior of the domain of f. For these two cases there are two possibilities:
a) rf (a; b) does not exist (Singular point). b) rf (a; b) exists. Finally, again there are two possibilities: i) rf (a; b) = 0 (Critical Point). ii) rf (a; b) 6 = 0. In this case f has a positive directional derivative in the direction of rf (a; b), and a negative directional derivative in the direction of rf (a; b), which shows that f cannot have an extreme value at (a; b).
Note that this theorem gives only the necessary conditions for the extreme values, it does not guarantee that a function has any extreme value. It only describes where to look if there are some extreme values. However, the next theorem gives the su¢ cient conditions for the extreme values, that is the theorem provides conditions which guarantee that the function will have absolute maximum and absolute minimum values.
Theorem 5 (Su¢ cient conditions for the extreme values) If f is continuous function of n variables which is deÖned on a closed and bounded domain, then there exist points in the domain where f takes maximum and minimum values.
In order to classify the critical points of a function of two variables, we have the following theorem:
Theorem 6 (Second Derivative Test) Suppose the function f (x; y) and its Örst and the second derivatives are continuous in a disk centered at (a; b). Suppose also that (a; b) is a critical point of f. Let Hessian of f be
f (x; y) =
f 11 f 12
f 21 f 22
= f 11 f 22 (f 12 )^2 , then
a) If f (a; b) > 0 and f 11 (a; b) > 0 (or f 22 (a; b) > 0 ), then f has a local minimum value at (a; b).
b) If f (a; b) > 0 and f 11 (a; b) < 0 (or f 22 (a; b) < 0 ), then f has a local maximum value at (a; b).
c) If f (a; b) < 0 , then f has a saddle point at (a; b).
d) If f (a; b) = 0, then the test is inconclusive, that is f may have a local minimum, or a local maximum, or a saddle point at (a; b).
Example 7 Find and classify all critical points of the function
f (x; y) =
x^2 + y^2 3 xy + 7x + 2:
Solution. First we Önd the Örst and the second partial derivatives of the function:
f 1 (x; y) = x 3 y + 7; f 2 (x; y) = 2 y 3 x; f 11 (x; y) = 1 ; f 12 (x; y) = 3 ; f 22 (x; y) = 2 :
Then we Önd the critical points:
f 1 (x; y) = 0 ) x 3 y + 7 = 0; f 2 (x; y) = 0 ) 2 y 3 x = 0:
Since f ( 2 ; 0) = 4 < 0 , f has a saddle point at ( 2 ; 0).
Consider the critical point (
f 11 (
f 12 (
f 22 (
Since f (
0 , and f 11 (
< 0 , f has a local maximum point at
(
2 Extreme Values of Functions DeÖned on Restricted Do-
mains
In this section we look for maximum and minimum values of functions with side conditions (con- straints). We know that if a continuous function f is deÖned on a closed and bounded domain D, then there exist absolute maximum and minimum values at some points in the domain. Remember that extreme values can occur only at critical points, singular points and boundary points. In order to Önd these points one can apply the procedure described below:
- Determine the critical and singular points of f on the interior of D.
- Determine the points on the boundary of D where f may have extreme values. For this purpose it is possible to
i parametrize the whole boundary and express f as a function of the parameter. ii parametrize the parts of the boundary and express f as a function of the parameters. In this case the end points of the pieces should be considered. iii apply Lagrange multipliers method.
- Evaluate f at all the points found in the previous steps.
Example 9 Let f (x; y) = (x + 1)^2 + (y + 1)^2 + 1. Find the absolute maximum and the absolute minimum values of f on the triangular region bounded by the lines x = 0, y = 0, and x + y = 4.
Solution. First, we seek the critical and singular points in the interior of the domain D.
f 1 (x; y) = 2 (x + 1) = 0 ) 2 x + 2 = 0 =) x = 1 ; f 2 (x; y) = 2 (y + 1) = 0 ) 2 y + 2 = 0 =) y = 1 ;
Thus P 0 ( 1 ; 1) is the only critical point in D since x = 1 < 0 , y = 1 < 0 and x + y = 1 1 = 2 > 4. As P 0 is inside the triangular region it is a candidate point for the absolute extreme values. Now, consider the boundary of the region. Let us consider the boundary as the union of three curves, C 1 , C 2 and C 3. The points on C 1 : C 1 : x = 0; 4 y 0 ;
f (0; y) = 1 + (y + 1)^2 + 1 = y^2 + 2y + 3 = g(y):
In order to Önd the extreme values of g(y), we should di§erentiate g to Önd its critical points on [ 4 ; 0]. g^0 (y) = 2y + 2 = 0 ) 2 y + 2 = 0 ) y = 1 2 [ 4 ; 0] :
We should also consider the end points (0; 4) and (0; 0) on C 1. Therefore the candidate points are P 1 (0; 1), P 2 (0; 4) and P 3 (0; 0). The points on C 2 : C 2 : y = 0; 4 x 0 ; f (x; 0) = (x + 1)^2 + 1 + 1 = x^2 + 2x + 3 = h(x). In order to Önd the extreme values of h(x), we should di§erentiate h to Önd its critical points on [ 4 ; 0]. h^0 (x) = 2x + 2 = 0 ) 2 x + 2 = 0 ) x = 1 2 [ 4 ; 0].
We should also consider the end points ( 4 ; 0) and (0; 0) on C 2. Therefore the candidate points are P 4 ( 1 ; 0), P 5 ( 4 ; 0) and P 3 (0; 0). The points on C 3 : C 3 : y = 4 x; 4 x 0 ; f (x; 4 x) = (x + 1)^2 + ( 4 x + 1)^2 + 1 = 2x^2 + 8x + 11 = k(x). In order to Önd the extreme values of k(x), we should di§erentiate k to Önd its critical points on [ 4 ; 0]. k^0 (x) = 4x + 8 = 0 ) 4 x + 8 = 0 ) x = 2 2 [ 4 ; 0] :
We should also consider the end points ( 4 ; 0) and (0; 4) on C 3. Therefore the candidate points are P 6 ( 2 ; 2), P 5 ( 4 ; 0) and P 2 (0; 4). Now we will evaluate f at all the candidate points:
f (P 0 ) = f ( 1 ; 1) = 1, f (P 1 ) = f (0; 1) = 2, f (P 2 ) = f (0; 4) = 11,
f (P 3 ) = f (0; 0) = 3, f (P 4 ) = f ( 1 ; 0) = 2, f (P 5 ) = f ( 4 ; 0) = 11,
f (P 6 ) = f ( 2 ; 2) = 3:
Example 12 Find the maximum and minimum values of the function f (x; y; z) = x 2 y z on the sphere x^2 + y^2 + z^2 = 6.
Solution. Our aim is to
minimize f (x; y; z) = x 2 y z subject to the constraint g(x; y; z) = x^2 + y^2 + z^2 6 = 0
We should Önd the critical points of the Lagrangian function
L(x; y; z; ) = f (x; y; z) + g(x; y; z) = x 2 y z + (x^2 + y^2 + z^2 6)
Lx = 1 + 2x = 0 ) x =
as 6 = 0, x 6 = 0
Ly = 2 + 2y = 0 ) y =
as 6 = 0, y 6 = 0
Lz = 1 + 2z = 0 ) z =
as 6 = 0, z 6 = 0
L = x^2 + y^2 + z^2 6 = 0:
Using the last equation we obtain
6 =
4 ^2
^2
4 ^2
^2
2 ^2
^2 =
or =
For =
, (x; y; z) = (1; 2 ; 1) and f (1; 2 ; 1) = 2 is the minimum value of f on the sphere.
For =
, (x; y; z) = ( 1 ; 2 ; 1) and f ( 1 ; 2 ; 1) = 2 is the maximum value of f on the sphere.
Example 13 Find the maximum and minimum values of f (x; y) = x^2 + 2y^2 on the closed disk x^2 + y^2 4.
Solution. f is continuous and the domain D is inside the circle and boundary of the circle which is closed and bounded. Therefore f has maximum and minimum values at some points of the domain. First, we seek the critical and singular points in the interior of D. f 1 (x; y) = 2x if f 1 (x; y) = 0 then x = 0,
f 2 (x; y) = 4y if f 2 (x; y) = 0 then y = 0
thus P 0 (0; 0) is the only critical point which is in D. Now, consider the boundary of D. To look for candidate points on the whole boundary we can use Lagrange multipliers method:
L(x; y; ) = f (x; y) + g(x; y) = x^2 + 2y^2 + (x^2 + y^2 4) @L @x
= 2 x + 2x = 2x(1 + ) = 0; @L @y
= 4 y + 2y = 2y(2 + ) = 0;
@L @ = x^2 + y^2 4 = 0:
Consider the Örst equation. There are two possibilities:
i. x = 0, then by third equation y^2 = 4 and hence y = 2 or y = 2 =) P 1 (0; 2) and P 2 (0; 2).
ii. = 1 , then by the second equation y = 0 and by third equation x^2 = 4 and hence x = 2 or x = 2 =) P 3 ( 2 ; 0), and P 4 (2; 0).
Thus P 1 (0; 2), P 2 (0; 2), P 3 ( 2 ; 0), and P 4 (2; 0) are the candidate points on the boundary. Now we should evaluate f at all the candidate points: f (P 0 ) = 0, f (P 1 ) = 8, f (P 2 ) = 8, f (P 3 ) = 4, and f (P 4 ) = 4. Therefore f has a maximum value 8 at points (0; 2), and (0; 2), and a minimum value 0 at the point (0; 0).
