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Maximum and Minimum(Maximum and Minimum)

Dr. Amal Kumar Adak

Assistant Professor, Department of MathematicsG.D.College, Begusarai

e-mail id: [email protected]

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Dr. Amal Kumar Adak Maximum and Minimum

DefinitionExtreme Value: Let (a,b) be a point of the domain of definitionof a function f . Then f (a,b) is an extreme value of f , if for everypoint (x , y),[ other than (a,b)] of some neighbourhood of(a,b),the difference f (x , y)− f (a,b) keeps the same sign.The extreme value f (a,b) is called a maximum or a minimumvalue according as the sign of f (x , y)− f (a,b) is negative orpositive.

DefinitionExtreme Point: If f (a,b) is an extreme value of a function fthen (a,b) is called an extreme point.

DefinitionExtreme Value: Let (a,b) be a point of the domain of definitionof a function f . Then f (a,b) is an extreme value of f , if for everypoint (x , y),[ other than (a,b)] of some neighbourhood of(a,b),the difference f (x , y)− f (a,b) keeps the same sign.The extreme value f (a,b) is called a maximum or a minimumvalue according as the sign of f (x , y)− f (a,b) is negative orpositive.

DefinitionExtreme Point: If f (a,b) is an extreme value of a function fthen (a,b) is called an extreme point.

TheoremIf a function fof two independent variables x and yhas anextreme value at (a,b), then necessarily fx (a,b) = 0 andfy (a,b) = 0 provided, of course, fx and fy exist at (a,b).

Proof: If (a,b) is an extreme value of the function f (x , y) of twoindependent variables x and y , then, clearly, it is also anextreme value of the function f (x ,b) of one variable x for x = aand therefore its derivative fx (a,b) for x = a in case it exists,must necessarily be zero. Similarly we have fy (a,b) = 0.

TheoremIf a function fof two independent variables x and yhas anextreme value at (a,b), then necessarily fx (a,b) = 0 andfy (a,b) = 0 provided, of course, fx and fy exist at (a,b).

Proof: If (a,b) is an extreme value of the function f (x , y) of twoindependent variables x and y , then, clearly, it is also anextreme value of the function f (x ,b) of one variable x for x = aand therefore its derivative fx (a,b) for x = a in case it exists,must necessarily be zero. Similarly we have fy (a,b) = 0.

DefinitionStationary Point: A point (a,b) is said to be a stationary pointof a function fof two independent variables x and y iffx (a,b) = 0 and fy (a,b) = 0.

DefinitionSaddle Point or Critical Point: A stationary point which is notan extreme point is called a saddle point.

DefinitionStationary Point: A point (a,b) is said to be a stationary pointof a function fof two independent variables x and y iffx (a,b) = 0 and fy (a,b) = 0.

DefinitionSaddle Point or Critical Point: A stationary point which is notan extreme point is called a saddle point.

ExampleQ1. Show by an example that a stationary point may or may notbe an extreme point.

Ans: Let us consider a function f , where

f (x , y) ={

0, if either x = 0 or if y = 01, elsewhere.

Then fx (0,0) = limh→0

f (h,0)−f (0,0)h = lim

h→00−0

h = 0.

Similarly, fy (0,0) = 0.Therefore, (0,0) is a stationary point of f .But we see that there does not exist a deleted neighbourhoodof (0,0) where f (x , y)− f (0,0) is always positive and therefore(0,0) is not an extreme point of f .This example shows that a stationary point may or may not bean extreme point.

ExampleQ2. Show by an example that both of fx and fy do not exists butf has a minimum value at some point of f .

Ans: Let us consider a function f , where f (x , y) = |x |+ |y | forall (x , y) ∈ R2. Now f (h,0)−f (0,0)

h = |h|h

Since limh→0+

|h|h = 1 and lim

h→0−

|h|h = −1 , lim

f (h,0)−f (0,0)h does not

exist i.e., fx (0,0) does not exist.Similarly we can show that fy (0,0) does not exist.Since f (x , y)− f (0,0) = |x |+ |y | > 0 for all points in everydeleted neighbourhood of (0,0), f has a minimum value of f .

ExampleQ2. Show by an example that both of fx and fy do not exists butf has a minimum value at some point of f .

Ans: Let us consider a function f , where f (x , y) = |x |+ |y | forall (x , y) ∈ R2. Now f (h,0)−f (0,0)

h = |h|h

Since limh→0+

|h|h = 1 and lim

h→0−

|h|h = −1 , lim

f (h,0)−f (0,0)h does not

exist i.e., fx (0,0) does not exist.Similarly we can show that fy (0,0) does not exist.Since f (x , y)− f (0,0) = |x |+ |y | > 0 for all points in everydeleted neighbourhood of (0,0), f has a minimum value of f .

ExampleQ3. Show by an example that fx does not exist and fy exist but fhas a minimum at some pint.

Ans: Let us consider a function f , where f (x , y) = |x |+ y2 forall (x , y) ∈ R2.Here fx (0,0) = lim

f (h,0)−f (0,0)h = lim

|h|h .

This limit does not exist because limh→0+

|h|h = 1 and

limh→0−

|h|h = −1.

fy (0,0) = limh→0

f (0,h)− f (0,0)h

= limh→0

h2 − 0h

Since f (x , y)− f (0,0) = |x |+ y2 > 0 for all points in everydeleted neighbourhood of (0,0), f has a minimum value of f .

ExampleQ3. Show by an example that fx does not exist and fy exist but fhas a minimum at some pint.

Ans: Let us consider a function f , where f (x , y) = |x |+ y2 forall (x , y) ∈ R2.Here fx (0,0) = lim

f (h,0)−f (0,0)h = lim

|h|h .

This limit does not exist because limh→0+

|h|h = 1 and

limh→0−

|h|h = −1.

fy (0,0) = limh→0

f (0,h)− f (0,0)h

= limh→0

h2 − 0h

Since f (x , y)− f (0,0) = |x |+ y2 > 0 for all points in everydeleted neighbourhood of (0,0), f has a minimum value of f .

ExampleQ4. Let (a,b) be a point in the domain of definition f . When isf (a,b) said to be a maximum value of f (x , y) ?Let fx (a,b) = 0, fy (a,b) = 0 and fxx (a,b) = A, fxy (a,b) = B,fyy (a,b) = C. If AC − B2 > 0 and A < 0, and fxx , fxy , fyy all arecontinuous in some neighbourhood of (a,b), then show thatf (a,b) will be a maximum value of f (x , y).

Ans: 2nd Part: According to the given conditions, the functionf satisfies all the conditions of placeCityTaylor’s theorem insome neighbourhood of (a,b). Let (x , y) be a point of thisneighbourhood. Then we have by placeCityTaylor’s theorem,

f (x , y) = f (a,b) + (x − a) fx (a,b) + (y − b) fy (a,b)

+12![(x − a)2 fxx (a + θh,b + θk) + 2 (x − a) (y − b) fxy (a + θh,b + θk)

+ (y − b)2 fyy (a + θh,b + θk)], where 0 < θ < 1.

ExampleQ4. Let (a,b) be a point in the domain of definition f . When isf (a,b) said to be a maximum value of f (x , y) ?Let fx (a,b) = 0, fy (a,b) = 0 and fxx (a,b) = A, fxy (a,b) = B,fyy (a,b) = C. If AC − B2 > 0 and A < 0, and fxx , fxy , fyy all arecontinuous in some neighbourhood of (a,b), then show thatf (a,b) will be a maximum value of f (x , y).

Ans: 2nd Part: According to the given conditions, the functionf satisfies all the conditions of placeCityTaylor’s theorem insome neighbourhood of (a,b). Let (x , y) be a point of thisneighbourhood. Then we have by placeCityTaylor’s theorem,

f (x , y) = f (a,b) + (x − a) fx (a,b) + (y − b) fy (a,b)

+12![(x − a)2 fxx (a + θh,b + θk) + 2 (x − a) (y − b) fxy (a + θh,b + θk)

+ (y − b)2 fyy (a + θh,b + θk)], where 0 < θ < 1.

Therefore,

f (x , y)− f (a,b) =12![(x − a)2 fxx (a + θh,b + θk)

+ 2 (x − a) (y − b) fxy (a + θh,b + θk) + (y − b)2 fyy (a + θh,b + θk)], . . . (1)[∵ fx (a,b) = 0 = fy (a,b)] .

From (1) it is clear that the sign of f (x , y)− f (a,b) depends onthe sign of(x − a)2 fxx (a + θh,b + θk)+2 (x − a) (y − b) fxy (a + θh,b + θk)+ (y − b)2 fyy (a + θh,b + θk).Since the second order partial derivatives are continuous insome neighbourhood of (a,b),the sign of right hand side of (1) depends only on(x − a)2 fxx (a,b)+2 (x − a) (y − b) fxy (a,b)+ (y − b)2 fyy (a,b)Let Q =(x − a)2 fxx (a,b)+2 (x − a) (y − b) fxy (a,b)+ (y − b)2 fyy (a,b)=(x − a)2 A+2 (x − a) (y − b)B+(y − b)2 C.

Since A < 0 and AC − B2 > 0 i.e.,∣∣∣∣ A B

∣∣∣∣ > 0.

Therefore the matrix[

A BB C

]is negative definite. It follows

that the quadratic form Q is negative definite. Thus, the sign ofright hand side of (1) is negative. Since (x , y) is arbitrary, thefunction f has a maximum at (a,b).

Note1: If fxx (a,b) fyy (a,b)− [fxy (a,b)]2 < 0 then f has neithera maximum nor a minimum at (a,b).

Note2: The matrix A =

[fxx fxyfxy fyy

]is called a Hessian matrix.

Let fx (a,b) = 0 = fy (a,b). Further, let us suppose that fpossesses continuous second order partial derivatives in acertain neighbourhood of (a,b) and that these derivatives at(a,b) viz. fxx (a,b) , fxy (a,b) , fyy (a,b) are not all zero. Thenf (a,b) is not an extreme value if the Hessian matrix evaluatedat (a,b) is indefinite and f (a,b) is a maximum or minimumvalue of f if the Hessian matrix is negative definite or positivedefinite. If the Hessian matrix is semi-definite then the furtherinvestigation is necessary.

Note1: If fxx (a,b) fyy (a,b)− [fxy (a,b)]2 < 0 then f has neithera maximum nor a minimum at (a,b).

Note2: The matrix A =

[fxx fxyfxy fyy

]is called a Hessian matrix.

Let fx (a,b) = 0 = fy (a,b). Further, let us suppose that fpossesses continuous second order partial derivatives in acertain neighbourhood of (a,b) and that these derivatives at(a,b) viz. fxx (a,b) , fxy (a,b) , fyy (a,b) are not all zero. Thenf (a,b) is not an extreme value if the Hessian matrix evaluatedat (a,b) is indefinite and f (a,b) is a maximum or minimumvalue of f if the Hessian matrix is negative definite or positivedefinite. If the Hessian matrix is semi-definite then the furtherinvestigation is necessary.

Extreme values of a function of three variables:

Rule: For a function f , sufficient conditions for (a,b, c) to be anextreme point are that (i) df (a,b, c) = fxdx + fydy + fzdz = 0,so that fx = fy = fz = 0,(ii) d2f (a,b, c) =fxx (dx)2 + fyy (dy)2 + fzz (dz)2 +2fxydxdy +2fyzdydz +2fzxdzdxkeeps the same sign for arbitrary values of dx ,dy ,dz; theextreme point being a maxima or a minima according as thesign of d2f is negative or positive. The point will be neither amaxima nor a minima if d2f does not keep the same sign; andrequires further investigation, if d2f keeps the same sign butvanishes at some points of a neighbourhood of (a,b, c).

The conditions that d2f keeps the same sign may be stated interms of matrices, as follows:

Consider the matrix

fxx fxy fxzfyx fyy fyzfzx fzy fzz

. d2f will always be

positive if and only if the three principal minors fxx ,∣∣∣∣ fxx fxy

fyx fyy

∣∣∣∣,∣∣∣∣∣∣fxx fxy fxzfyx fyy fyzfzx fzy fzz

∣∣∣∣∣∣ are all positive, and d2f will be always negative

if and only if their signs are alternately negative or positive.

ExampleQ5. Find the maxima and minima of the function f wheref (x , y) = x3 + y3 − 3x − 12y + 20 for (x , y) ∈ R2.

Ans: We have fx (x , y) = 3x2 − 3 and fy (x , y) = 3y2 − 12.Therefore fx (x , y) = 0 when x = ±1 and fy (x , y) = 0 wheny = ±2.Thus the function has four stationary points: (1,2),(−1,2),(1,−2), (−1,−2).Now fxx (x , y) = 6x , fxy (x , y) = 0, fyy (x , y) = 6y .Since fxx (1,2) = 6 andfxx (1,2) fyy (1,2)− [fxy (1,2)]2 = 72 > 0, (1,2) is a point ofminima of the function f and f has a minimum value 2 at (1,2).

Since fxx (−1,2) fyy (−1,2)− [fxy (−1,2)]2 = −72 < 0, f has aneither maximum nor minimum at (−1,2).Since fxx (1,−2) fyy (1,−2)− [fxy (1,−2)]2 = −72 < 0, f has aneither maximum nor minimum at (1,−2).Since fxx (−1,−2) = −6 < 0 andfxx (−1,−2) fyy (−1,−2)− [fxy (−1,−2)]2 = 72 > 0, (−1,−2) isa point of maxima of the function f and f has a maximum value38 at (−1,−2).

ExampleQ6. Find the maxima and minima of the functionf (x , y) = 8x3 + 27y3 − 6x − 36y + 25.

Ans: The stationary points are(12 ,

),(−1

2 ,−23

),(−1

2 ,−23

)f has a minimum value. At

2 ,−23

)f has a

maximum value. f has neither maximum nor minimum at(−1

2 ,−23

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