Maxima and Minima
of Functions of two variables
Objectives
1.To learn the definition of extreme values of a function of two variables
2.To study the method to locate the stationary points.
3. To optimize a function
4.To find the maximum and minimum values of a constrained
function.
Introduction
In this session we explore the maxima and minima of functions of two variables.
In Calculus of one variable, the derivative of a function is used as an effective tool for locating maximum and minimum values. We find out a point there in the domain of the function at which the tangent line to the curve becomes parallel to the independent variable axis. In this section we apply a similar concept to functions of two variables. Here we use the partial derivatives with respect to the variables of the function.
Let f(x,y) be a function of two independent variables x and y. We look for points where the surface z=f(x,y) has a horizontal tangent plane. Let (a,b) be a point of the domain of f having this property and E be a small neighbourhood of point (a,b). If (x,y) is another point in E, then f(x,y)<f(a,b) or f(x,y)>f(a,b). This point may give a maximum or a minimum value of the function. To understand and visualize this conceptbetter, it is enough to think of the graph of the function as a land area having peaks and valleys.
In the following, we give more precise definitions of these concepts.
Extreme Values:
The value f(a,b) of a function f of two variables is said to be an extreme value if for every point (x,y) other than (a,b) of small neighbourhood of (a,b) in the domain of definition of f, the difference f(x,y)f(a,b) keeps the same sign.
The extreme value f(a,b) is called a maximum or minimum if f(x,y)f(a,b) is negative or positive.
A Necessary Condition:
A necessary condition for f(x,y) to have an extreme value at (a,b) is that =0 and =0 at the point (a,b), provided these partial derivatives exist.
If f(a,b) is an extreme value of the function f(x,y) of two variables, then it must be an extreme value of both the functions f(x,b) and f(a,y) of one variable. But a necessary condition for the function f(x,b) of one variable to have extreme value at x=a is that the derivative fx(a,b)=0. That is, =0 at (a,b). Similarly, we obtain =0 at (a,b).
This point (a,b) at which =0 and =0 , is called a stationary point.
Let us consider a function defined by f(x,y) = x3+y33xy.
= 3x23y;= 3y23x
If the function has extreme values, then =0 and =0. This leads to the equations 3x23y=0 and 3y23x=0. Thus, the points (0,0) and (1,1) are the stationary points.
A Sufficient Condition:
Let =0 and = 0 at the point (a,b). Suppose that f(x,y) possesses continuous second order partial derivatives in a certain neighbourhood of (a,b) and that these derivatives at (a,b) are not all zero.
Let (a+h,b+k) be a point of this neighbourhood. By Taylor’s Theorem we have
f(a+h,b+k) = f(a,b) + Df + D2f + R3
where Df = at (a,b)
D2f = at (a,b)
and R3 consists of third and higher order terms.
But =0 and =0 imply that Df =0.
Let r = , s = and t = at the point (a,b).
Then D2f = h2 r + 2hks + k2 t =
=
Therefore,f(a+h,b+k)f(a,b) = + R3.
This shows that, if we choose h and k sufficiently small, the sign of f(a+h,b+k)f(a,b) depends on the expression and it has the same sign as r if rts2>0. Now, we consider the different possibilities.
Case I: Let rts2 >0.
If r >0, then f(a+h,b+k)f(a,b) >0, for small values of h and k. Hence f(a,b) is a minimum value of function f.
If r< 0, the function f has maximum at (a,b).
Case II :Let rts2 =0.
In this case the expression [(hr+ks)2+k2(rts2)] becomes a perfect square and so we cannot make a conclusion about the sign. Therefore, this is a doubtful case and requires further investigation.
Case III :Let rts2 <0.
Here, the expression [(hr+ks)2+k2(rts2)] is positive or negative according as (hr+ks)2 > k2(s2rt) or (hr+ks)2 < k2(s2rt). Hence f(a+h,b+k)f(a,b) does not keep the same sign for all values of h and k. Thus, in every neighbourhood of (a,b) there are points (x,y) where f(x,y)>f(a,b) and points where f(x,y)<f(a,b). This point (a,b) is called a saddle point.
Rule to calculate extreme value
Let f(x,y) possesses continuous second order partial derivatives in a certain neighbourhood of a point (a,b) and =0, =0 at (a,b). Then,
(i) f(a,b) is an extreme value of f(x,y) if rts2 >0. f(a,b) is minimum or maximum according as r>0 or r<0.
(ii)If rts2 <0, then f has neither maximum nor minimum at (a,b). In this case f has a saddle point.
(iii)If rts2 =0, we must find some other way to determine the behaviour of function f at (a,b).
Let us find out the extreme values of the function f(x,y) = x3+y33xy. It has partial derivatives = 3x23y ; = 3y23x and stationary points (0,0) and (1,1). Now, =6x ; = 3 ; =6y.
Therefore, at the point (1,1), we have r=6>0, s=3 and t=6.
Here rts2 = 27>0. Hence the function f has minimum at the point (1,1) and minimum value = 1.
Example
Find the extreme values of the function f(x,y) = x3+y3+3x23y28.
Solution: The first partial derivatives are = 3x2+6x and = 3y26y.
For extreme values, put =0 and =0. Therefore, 3x2+6x=0 and 3y26y=0 together imply that x=0, x=2 and y=0, y=2. Thus we have four points (0,0), (0,2), (2,0) and (2,2) as stationary points.
Now, second order partial derivatives
= 6x+6 ; = 0 ; = 6y6
At the point (0,0), r=6, s=0, t=6 and rts2 = 36 <0. So (0,0) is a saddle point.
At the point (0,2), r=6, s=0, t=6 and rts2 =36 >0. Thus, the function has a minimum value 12at (0,2).
At (2,0), r=6, s=0, t=6 and rts2 =36 >0. Since r<0, f(2,0) = 4 is a maximum value of the function.
At (2,2), r=6, s=0, t=6 and rts2 = 36 <0. Therefore, the point (2,2) is a saddle point.
Let us see the generalized version of the necessary condition for a function of three or more variables to have stationary points.
The necessary condition for a function f(x1,x2,…,xn) of variables x1, x2,…xn to have extreme value at (a1,a2,…,an) is that the partial derivatives =0 for all i=1,2,..,n, at (a1,a2,…,an), in case they exist.
Suppose we are asked to find the extreme values of the function f(x,y,z) = x2+y2+z2 subject to the condition x2z21=0.
Applying the necessary condition =0, =0 and =0 , we obtain x=0, y=0, z=0. This gives (0,0,0) as a stationary point. But this does not lie on the given surface x2z21=0.
In order to locate the stationary point subject to the given condition, let us take z2= x21 and obtain f(x,y,z) as a function of independent variables x and y.
That is, h(x,y)=x2+y2+(x21).
The stationary points are obtained from =0 and =0. Here (0,0) is a stationary point which is not on the given surface x2z21=0. So we consider f(x,y,z) as a function of y and z using the given condition x2=z2+1. Let g(y,z) = (z2+1)+y2+z2. Then =0 and = 0 give the stationary points (1,0,0).
This method of finding stationary points is a fairly long messy process. We take a look at another way of locating stationary points of a constrained function. In this powerful method, a new function F is defined using the function f(x,y,z) and the constraint (x,y,z)=0.
Let F(x,y,z) = f(x,y,z) + (x,y,z), where the variables x,y,z are considered as independent. In the above example,
F(x,y,z) = x2+y2+z2 + (x2z21).
The stationary points of F(x,y,z) are located using the condition
=0, =0 and =0
Therefore,2x+2x=0 ; 2y=0 ; and 2z2z=0.
That is,(1+)x=0 ; y=0 ; and (1)z=0.
But, on the given surface (x,y,z)=0, the co-ordinate x0.
Therefore,1+ = 0 and hence = 1.
This shows that z = 0. Thus, from the condition x2z21=0, we obtain x = 1. Hence the stationary points are (1,0,0).
Lagrange developed this method in 1755. The constant is called the Lagrange multiplier.
Lagrange’s Method of Multipliers to find the stationary points of the function f(x1,…,xn;u1,…,um) of n+m variables which are connected by the equations r (x1,…,xn; u1,…,um ) = 0, where r = 1,2,..,m.
Define a function F = f + 11+……+mm and consider all the variables x1,…,xn; u1,…,um as independent. The stationary points of f may be found by determining the stationary points of the function F using the conditions
=0, ….., =0 ; =0, …., =0.
We discuss this method for a function of four variables x,y,u,v connected by two relations.
Let f(x,y,u,v) = 0 be a function of 2+2 variables which are connected by the conditions 1 (x,y,u,v) = 0 and 2 (x,y,u,v) = 0 . If f has stationary points, then =0, = 0, = 0 and = 0.
Therefore, dx + dy + du + dv = 0
But dx + dy + du + dv = 0
anddx + dy + du + dv = 0
Multiplying these equations by 1 and 2 respectively and adding to previous equation, we have
+++=0
Therefore, since 1 and 2 are arbitrary, we obtain the system of equations
=0,
= 0,
= 0,
= 0
Now define a new function F(x,y,u,v) = f + 11+ 22 , where x,y,u,v are considered as independent variables. For stationary points of F we must have =0, = 0, = 0 and = 0. This leads to the equations
=0,
= 0,
= 0,
= 0
This system of equations is the same as the system of equations obtained for the function f(x,y,u,v). Hence the stationary points of f are the same as that of the newly defined function F.
Example 1
Find the stationary points of the function 3x+4y subject to the condition x2+y2=1.
Solution:Here f(x,y) = 3x+4y and the condition (x,y) = x2+y21.
Define F(x,y) = (3x+4y) + (x2+y21).
and imply that 3+2x=0 and 4+2y=0.
Therefore, x = and y = .
From the condition x2+y2=1, we have . Therefore, and hence , . This gives the co-ordinates of the stationary points.
Example 2
Obtain the stationary points of the function f(x,y,z) = x2+2yz2 subject to the constraints 2xy=0 and y+z=0.
Solution:Define F(x,y,z) = x2+2yz2 + 1 (2xy) + 2 (y+z).
The necessary conditions , and give us the equations
2x+21 = 0, 21+2 = 0 and 2z+2 = 0.
That is,x+1 = 0, 12 = 2 and 2 = 2z.
Substituting the values of 1 and 2 , we get the equation x2z = 2.
Then, from the given condition, we have 2x+z = 0. Solving these equations subject to the given conditions we get , y = and z = .
Therefore, is a stationary point of the given function
Assignments
1.Find the extreme values of the function f(x,y) = x3y2(1xy).
2.Find the maximum and minimum values of the function
sin x + sin y + sin(x+y).
3.Obtain the saddle points of the function f(x,y) = x2+xy+3x+2y+5.
4.Find a point with in a triangle such that the sum of the squares of its distances from the three vertices is a minimum.
5. Show that the function f(x,y)=2x43x2y+y2 has neither maximum nor a minimum at (0,0).
6.Find the stationary points of x2+y2+z2 subject to the conditions and z=x+y.
7.Find the greatest and smallest values that the function f(x,y) = xy takes on the ellipse
QUIZ
1.Stationary point of function f(x,y) = x2+y2
(a) (1,0)(b) (0,1)(c) (0,0)(d) none
2.At (0,0) the function f(x,y) = y2x2 has ------
(a) maximum value(b) minimum value
(c) saddle point(d) none.
3.f(x) = x2+kxy+y2. For what value of k, is the second derivative test not helpful to locate extreme values?.
(a) 1(b) 2(c) 3(d) none
4.The point on the ellipse x2+2y2=1 where f(x,y) = xy has its extreme value.
(a) (1,0)(b) (1,0) (c) (d)
5.The maximum value of xy, subject to the constraint x+y=16 is
(a) 8(b) 16(c) 32(d) 64
Quiz Answers
1. c2. c 3. b4. d5. d
Glossary
Neighbourhood
For a small positive number , the set of all points whose distance from a point P is less than is called a -neighbourhood of point P.
Surface
The graph of the function z=f(x,y) consisting of all points of the form (x,y,z) where (x,y) is in the domain of function f can be interpreted as a surface.
Necessary condition
A condition which is a logical consequence of a given statement is a necessary condition.
Sufficient condition
A condition from which a given statement logically follows is a sufficient condition.
FAQS
1.What is Taylor’s Theorem for functions of two variables?.
Ans. If a function f(x,y) of two independent variables x and y possesses continuous partial derivatives of order n in any domain E of a point (a,b) and if (a+h,b+k) is any point of E, then there exists a positive number 0<<1, such that
Where and , 0<<1.
2.How do we explain necessary and sufficient conditions?
Ans.Consider any two statements p and q. Suppose statement p implies q. Then statement q is a necessary condition for statement p. The statement p is a sufficient condition for the statement q.
3.What is the necessary condition for a function of one variable to have extreme values?
Ans.If f(c) is an extreme value of a function f(x) then at x=c, in case it exists.
References
Books
1.S. C. Malik & Savita Arora, Mathematical Analysis(4th Edn), New Age International Publishers.
2.Murray R Spiegal, Theory and Problems of Advanced Calculus, Schaum Publishing Co., New York.
Websites
Summary
A continuous function defined on a closed and bounded region in the xy-plane attains its maximum value and its minimum value. It is important to be able to locate these extreme values.
In this section, a method to find the extreme values of a function of two variables is explained. By examining the partial derivatives, the stationary points of a function are located and a sufficient condition for a function to have extreme values is also derived. This unit includes the Lagrange’s method of multipliers to find the stationary points of constraint functions.
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