M38 Lec 021814

MATH 38: Mathematical Analysis IIIUnit 3: Differentiation of Functions of More than One Variable

I. F. Evidente

IMSP (UPLB)

Outline

1 Relative Extreme Function Values

2 Absolute Extrema of Functions of Two Variables

Figures taken from: J. Stewart, The Calculus: Early Transcendentals,Brooks/Cole, 6th Edition, 2008.

Outline



DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:

1 f is said to have a relative maximum at (x0, y0) if

there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .

2 f is said to have a relative minumum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B

3 f is said to have a relative extremum at (x0, y0) if either f has arelative maximum or a relative minimum at (x0, y0).


1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such that

f (x0, y0) f (x, y) for every (x, y) B .2 f is said to have a relative minumum at (x0, y0) if there exists an

open ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B



1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .





2 f is said to have a relative minumum at (x0, y0) if

there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B




2 f is said to have a relative minumum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such that

f (x0, y0) f (x, y) for every (x, y) B3 f is said to have a relative extremum at (x0, y0) if either f has a

relative maximum or a relative minimum at (x0, y0).




3 f is said to have a relative extremum at (x0, y0) if

either f has arelative maximum or a relative minimum at (x0, y0).

RemarkSuppose f has a relative minimum [maximum, extremum] at P .

The value f (x0, y0) is said to be a relative minimum [maximum,extremum] value of f .The point (x0, y0, f (x0, y0)) is called a relative minimum [maximum,extremum] point of f .


The value f (x0, y0) is said to be a relative minimum [maximum,extremum] value of f .

The point (x0, y0, f (x0, y0)) is called a relative minimum [maximum,extremum] point of f .


The value f (x0, y0) is said to be a relative minimum [maximum,extremum] value of f .The point (x0, y0, f (x0, y0)) is called a relative minimum [maximum,extremum] point of f .

Graphically, the graph of z = f (x, y) has peaks and trenches at relativemaximum and relative minimum points, respectively.

It is "relative" or "local" because the point is highest or lowest relative toits very close neighbors.Note that a function may have several relative extrema.


It is "relative" or "local" because the point is highest or lowest relative toits very close neighbors.

Note that a function may have several relative extrema.


It is "relative" or "local" because the point is highest or lowest relative toits very close neighbors.Note that a function may have several relative extrema.

Problem:How do we find the relative minima and relative maxima of f ?

DefinitionThe point (x0, y0) is a critical point of the function f (x, y) if

(x0, y0) dom f and either1 fx(x0, y0)= 0 and fy (x0, y0)= 0 (Type 1), or2 fx(x0, y0) or fy (x0, y0) does not exist (Type 2)

DefinitionThe point (x0, y0) is a critical point of the function f (x, y) if(x0, y0) dom f and either

1 fx(x0, y0)= 0 and fy (x0, y0)= 0 (Type 1), or2 fx(x0, y0) or fy (x0, y0) does not exist (Type 2)


1 fx(x0, y0)= 0 and fy (x0, y0)= 0 (Type 1), or

2 fx(x0, y0) or fy (x0, y0) does not exist (Type 2)


1 fx(x0, y0)= 0 and fy (x0, y0)= 0 (Type 1), or2 fx(x0, y0) or fy (x0, y0) does not exist (Type 2)

ExampleFind the critical points of f (x, y)= 3x22xy + y28y .

fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:

6x2y = 02x+2y 8 = 0

} 4x8= 0 x = 2 & y = 6

(2,6) is a critical point of f .


fx(x, y)= 6x

2y and fy (x, y)= 2x+2y 8Critical Points:

6x2y = 02x+2y 8 = 0

} 4x8= 0 x = 2 & y = 6



fx(x, y)= 6x2y

and fy (x, y)= 2x+2y 8Critical Points:

6x2y = 02x+2y 8 = 0

} 4x8= 0 x = 2 & y = 6



fx(x, y)= 6x2y and fy (x, y)=

2x+2y 8Critical Points:

6x2y = 02x+2y 8 = 0

} 4x8= 0 x = 2 & y = 6



fx(x, y)= 6x2y and fy (x, y)= 2x

+2y 8Critical Points:

6x2y = 02x+2y 8 = 0

} 4x8= 0 x = 2 & y = 6



fx(x, y)= 6x2y and fy (x, y)= 2x+2y

8Critical Points:

6x2y = 02x+2y 8 = 0

} 4x8= 0 x = 2 & y = 6



fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8

Critical Points:

6x2y = 02x+2y 8 = 0

} 4x8= 0 x = 2 & y = 6




6x2y = 02x+2y 8 = 0

} 4x8= 0 x = 2 & y = 6




6x2y = 02x+2y 8 = 0

}

4x8= 0 x = 2 & y = 6




6x2y = 02x+2y 8 = 0

} 4x8= 0

x = 2 & y = 6




6x2y = 02x+2y 8 = 0

} 4x8= 0 x = 2 &

y = 6




6x2y = 02x+2y 8 = 0

} 4x8= 0 x = 2 & y = 6


TheoremIf the function f (x, y) has a relative extremum at (x0,y0) and fx(x0, y0) andfy (x0, y0) exist, then

fx(x0, y0)= 0 and fy (x0, y0)= 0.

TheoremIf the function f (x, y) has a relative extremum at (x0,y0) and fx(x0, y0) andfy (x0, y0) exist, then fx(x0, y0)= 0 and fy (x0, y0)= 0.

The theorem tells us the following:

RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).


RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be

a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).


RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.

If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).


RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from

the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).


RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f .

(Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).


RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points)

Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).


RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?

The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).


RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).

Geometrically

Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.

fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:

z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)

What kind of plane is this? HORIZONTAL!

RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!

Geometrically


fx(x0, y0)= 0 and fy (x0, y0)= 0

Equation of tangent plane:




Geometrically



z = L(x, y)

z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)



Geometrically



z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)

z = f (x0, y0)



Geometrically






Geometrically




What kind of plane is this?

HORIZONTAL!


Geometrically






Geometrically





RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then

it must be horizontal!

Geometrically






DefinitionA point (x0, y0,z0) is said to be a saddle point of a surface if

there existtwo vertical planes such that the trace of in one plane has a relativemaximum at (x0, y0) and the trace of in the other plane has a relativeminimum at (x0, y0).

DefinitionA point (x0, y0,z0) is said to be a saddle point of a surface if there existtwo vertical planes such that

the trace of in one plane has a relativemaximum at (x0, y0) and the trace of in the other plane has a relativeminimum at (x0, y0).

DefinitionA point (x0, y0,z0) is said to be a saddle point of a surface if there existtwo vertical planes such that the trace of in one plane has a relativemaximum at (x0, y0) and

the trace of in the other plane has a relativeminimum at (x0, y0).

DefinitionA point (x0, y0,z0) is said to be a saddle point of a surface if there existtwo vertical planes such that the trace of in one plane has a relativemaximum at (x0, y0) and the trace of in the other plane has a relativeminimum at (x0, y0).

f (x, y)= x2 y2 has a saddle point at (0,0)

Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B

((x0, y0),r

)and fx(x0, y0)= fy (x0, y0)= 0.

Let

D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)

]2 .1 If D > 0 and

1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).

2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.


((x0, y0),r

)and fx(x0, y0)= fy (x0, y0)= 0. Let


]2 .

1 If D > 0 and1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).



((x0, y0),r

)and fx(x0, y0)= fy (x0, y0)= 0. Let


]2 .1 If D > 0 and




((x0, y0),r

)and fx(x0, y0)= fy (x0, y0)= 0. Let


]2 .1 If D > 0 and

1 fxx (x0, y0)> 0, then

f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).



((x0, y0),r

)and fx(x0, y0)= fy (x0, y0)= 0. Let


]2 .1 If D > 0 and

1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).

2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.


((x0, y0),r

)and fx(x0, y0)= fy (x0, y0)= 0. Let


]2 .1 If D > 0 and

1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then

f (x, y) has a relative maximum at (x0, y0).2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.


((x0, y0),r

)and fx(x0, y0)= fy (x0, y0)= 0. Let


]2 .1 If D > 0 and




((x0, y0),r

)and fx(x0, y0)= fy (x0, y0)= 0. Let


]2 .1 If D > 0 and


2 If D < 0, then

f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.


((x0, y0),r

)and fx(x0, y0)= fy (x0, y0)= 0. Let


]2 .1 If D > 0 and


2 If D < 0, then f (x, y) has a saddle point at (x0, y0).

3 If D = 0, then no conclusion can be made.


((x0, y0),r

)and fx(x0, y0)= fy (x0, y0)= 0. Let


]2 .1 If D > 0 and


2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then

no conclusion can be made.


((x0, y0),r

)and fx(x0, y0)= fy (x0, y0)= 0. Let


]2 .1 If D > 0 and



RemarkThis test is only for Type 1 critical points!If f is differentiable, then this test allows you to get ALL criticalpoints.

RemarkOne way to remember the formula for D:

The Jacobian of f is the matrix of partial derivatives of f[fxx fxyfyx fy y

]D is the determinant of the Jacobian of f .

TheoremLet f be a function in 2 variables and R a region on the xy-plane. If fxyand fyx are continuous on R, then fxy = fyx on that open ball.

RemarkOne way to remember the formula for D:The Jacobian of f is the matrix of partial derivatives of f[

fxx fxyfyx fy y

]

D is the determinant of the Jacobian of f .


RemarkOne way to remember the formula for D:The Jacobian of f is the matrix of partial derivatives of f[

fxx fxyfyx fy y

]D is the determinant of the Jacobian of f .


ExampleLocate all relative extrema and saddle points of

f (x, y)= 3x22xy + y28y

.

fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:

fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,

D = 6 22 2

= 124= 8> 0 (rel ext)Now

fxx(2,6)= 6> 0 f has a relative minimum at (2,6)


f (x, y)= 3x22xy + y28y

.

fx(x, y)= 6x2y and fy (x, y)=2x+2y 8

Critical Point: (2,6)Second Partial Derivatives:


D = 6 22 2

= 124= 8> 0 (rel ext)Now



f (x, y)= 3x22xy + y28y

.

fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point:

(2,6)Second Partial Derivatives:


D = 6 22 2

= 124= 8> 0 (rel ext)Now



f (x, y)= 3x22xy + y28y

.

fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)

Second Partial Derivatives:


D = 6 22 2

= 124= 8> 0 (rel ext)Now



f (x, y)= 3x22xy + y28y

.


fxx(x, y)=

6 fy y (x, y)= 2 fxy (x, y)= 2Thus,

D = 6 22 2

= 124= 8> 0 (rel ext)Now



f (x, y)= 3x22xy + y28y

.


fxx(x, y)= 6

fy y (x, y)= 2 fxy (x, y)= 2Thus,

D = 6 22 2

= 124= 8> 0 (rel ext)Now



f (x, y)= 3x22xy + y28y

.


fxx(x, y)= 6 fy y (x, y)=

2 fxy (x, y)= 2Thus,

D = 6 22 2

= 124= 8> 0 (rel ext)Now



f (x, y)= 3x22xy + y28y

.


fxx(x, y)= 6 fy y (x, y)= 2

fxy (x, y)= 2Thus,

D = 6 22 2

= 124= 8> 0 (rel ext)Now



f (x, y)= 3x22xy + y28y

.


fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)=

2Thus,

D = 6 22 2

= 124= 8> 0 (rel ext)Now



f (x, y)= 3x22xy + y28y

.


fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2

Thus,

D = 6 22 2

= 124= 8> 0 (rel ext)Now



f (x, y)= 3x22xy + y28y

.



D = 6 22 2

= 124= 8> 0 (rel ext)Now



f (x, y)= 3x22xy + y28y

.



D = 6

22 2

= 124= 8> 0 (rel ext)Now



f (x, y)= 3x22xy + y28y

.



D = 6 2

2 2= 124= 8> 0 (rel ext)

Nowfxx(2,6)= 6> 0

f has a relative minimum at (2,6)


f (x, y)= 3x22xy + y28y

.



D = 6 22

2

= 124= 8> 0 (rel ext)Now



f (x, y)= 3x22xy + y28y

.



D = 6 22 2

=

124= 8> 0 (rel ext)

Nowfxx(2,6)= 6> 0



f (x, y)= 3x22xy + y28y

.



D = 6 22 2

= 12

4= 8> 0 (rel ext)

Nowfxx(2,6)= 6> 0



f (x, y)= 3x22xy + y28y

.



D = 6 22 2

= 124=

8> 0 (rel ext)

Nowfxx(2,6)= 6> 0



f (x, y)= 3x22xy + y28y

.



D = 6 22 2

= 124= 8

> 0 (rel ext)

Nowfxx(2,6)= 6> 0



f (x, y)= 3x22xy + y28y

.



D = 6 22 2

= 124= 8> 0

(rel ext)

Nowfxx(2,6)= 6> 0



f (x, y)= 3x22xy + y28y

.



D = 6 22 2

= 124= 8> 0 (rel ext)

Nowfxx(2,6)= 6> 0



f (x, y)= 3x22xy + y28y

.



D = 6 22 2

= 124= 8> 0 (rel ext)Now

fxx(2,6)=

6> 0 f has a relative minimum at (2,6)


f (x, y)= 3x22xy + y28y

.



D = 6 22 2

= 124= 8> 0 (rel ext)Now

fxx(2,6)= 6

> 0 f has a relative minimum at (2,6)


f (x, y)= 3x22xy + y28y

.



D = 6 22 2

= 124= 8> 0 (rel ext)Now

fxx(2,6)= 6> 0



f (x, y)= 3x22xy + y28y

.



D = 6 22 2

= 124= 8> 0 (rel ext)Now


ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.

fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3

Critical Points:

4y 4x3 = 0 (1)4x4y3 = 0 (2)

} y = x

3 (1)x = y3 (2)

}


fx(x, y)=

4y 4x3 and fy (x, y)= 4x4y3

Critical Points:

4y 4x3 = 0 (1)4x4y3 = 0 (2)

} y = x

3 (1)x = y3 (2)

}


fx(x, y)= 4y 4x3

and fy (x, y)= 4x4y3

Critical Points:

4y 4x3 = 0 (1)4x4y3 = 0 (2)

} y = x

3 (1)x = y3 (2)

}


fx(x, y)= 4y 4x3 and

fy (x, y)= 4x4y3

Critical Points:

4y 4x3 = 0 (1)4x4y3 = 0 (2)

} y = x

3 (1)x = y3 (2)

}


fx(x, y)= 4y 4x3 and fy (x, y)=

4x4y3

Critical Points:

4y 4x3 = 0 (1)4x4y3 = 0 (2)

} y = x

3 (1)x = y3 (2)

}



Critical Points:

4y 4x3 = 0 (1)4x4y3 = 0 (2)

} y = x

3 (1)x = y3 (2)

}



Critical Points:

4y 4x3 = 0 (1)4x4y3 = 0 (2)

}

y = x3 (1)

x = y3 (2)}



Critical Points:

4y 4x3 = 0 (1)4x4y3 = 0 (2)

} y = x

3 (1)x = y3 (2)

}

y = x3 (1)x = y3 (2)

}

Substituting (1) into (2):

x = (x3)3x = x9

x9x = 0x(x81) = 0

x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0

x(x1)(x+1)(x2+1)(x4+1) = 0

x = 0 x =1 x = 1y = 0 y =1 y = 1

y = x3 (1)x = y3 (2)

}Substituting (1) into (2):

x = (x3)3x = x9

x9x = 0x(x81) = 0

x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0

x(x1)(x+1)(x2+1)(x4+1) = 0

x = 0 x =1 x = 1y = 0 y =1 y = 1

y = x3 (1)x = y3 (2)


x = (x3)3

x = x9x9x = 0

x(x81) = 0x(x41)(x4+1) = 0

x(x21)(x2+1)(x4+1) = 0x(x1)(x+1)(x2+1)(x4+1) = 0

x = 0 x =1 x = 1y = 0 y =1 y = 1

y = x3 (1)x = y3 (2)


x = (x3)3x = x9

x9x = 0x(x81) = 0

x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0

x(x1)(x+1)(x2+1)(x4+1) = 0

x = 0 x =1 x = 1y = 0 y =1 y = 1

y = x3 (1)x = y3 (2)


x = (x3)3x = x9

x9x = 0

x(x81) = 0x(x41)(x4+1) = 0

x(x21)(x2+1)(x4+1) = 0x(x1)(x+1)(x2+1)(x4+1) = 0

x = 0 x =1 x = 1y = 0 y =1 y = 1

y = x3 (1)x = y3 (2)


x = (x3)3x = x9

x9x = 0x(x81) = 0

x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0

x(x1)(x+1)(x2+1)(x4+1) = 0

x = 0 x =1 x = 1y = 0 y =1 y = 1

y = x3 (1)x = y3 (2)


x = (x3)3x = x9

x9x = 0x(x81) = 0

x(x41)(x4+1) = 0

x(x21)(x2+1)(x4+1) = 0x(x1)(x+1)(x2+1)(x4+1) = 0

x = 0 x =1 x = 1y = 0 y =1 y = 1

y = x3 (1)x = y3 (2)


x = (x3)3x = x9

x9x = 0x(x81) = 0

x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0

x(x1)(x+1)(x2+1)(x4+1) = 0

x = 0 x =1 x = 1y = 0 y =1 y = 1

y = x3 (1)x = y3 (2)


x = (x3)3x = x9

x9x = 0x(x81) = 0

x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0

x(x1)(x+1)(x2+1)(x4+1) = 0

x = 0

x =1 x = 1y = 0 y =1 y = 1

y = x3 (1)x = y3 (2)


x = (x3)3x = x9

x9x = 0x(x81) = 0

x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0

x(x1)(x+1)(x2+1)(x4+1) = 0

x = 0 x =1

x = 1y = 0 y =1 y = 1

y = x3 (1)x = y3 (2)


x = (x3)3x = x9

x9x = 0x(x81) = 0

x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0

x(x1)(x+1)(x2+1)(x4+1) = 0

x = 0 x =1 x = 1

y = 0 y =1 y = 1

y = x3 (1)x = y3 (2)


x = (x3)3x = x9

x9x = 0x(x81) = 0

x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0

x(x1)(x+1)(x2+1)(x4+1) = 0

x = 0 x =1 x = 1y = 0

y =1 y = 1

y = x3 (1)x = y3 (2)


x = (x3)3x = x9

x9x = 0x(x81) = 0

x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0

x(x1)(x+1)(x2+1)(x4+1) = 0

x = 0 x =1 x = 1y = 0 y =1

y = 1

y = x3 (1)x = y3 (2)


x = (x3)3x = x9

x9x = 0x(x81) = 0

x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0

x(x1)(x+1)(x2+1)(x4+1) = 0

x = 0 x =1 x = 1y = 0 y =1 y = 1

First Partial Derivatives:


Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)

fxx(x, y)=12x2

0 12 12

fy y (x, y)=12y2

0 12 12

fxy (x, y)= 4

4 4 4

D

16 128 128

conclusion

saddle pt rel max rel max




fxx(x, y)=12x2 0

12 12

fy y (x, y)=12y2

0 12 12

fxy (x, y)= 4

4 4 4

D

16 128 128

conclusion





fxx(x, y)=12x2 0 12

12

fy y (x, y)=12y2

0 12 12

fxy (x, y)= 4

4 4 4

D

16 128 128

conclusion





fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2

0 12 12

fxy (x, y)= 4

4 4 4

D

16 128 128

conclusion





fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0

12 12

fxy (x, y)= 4

4 4 4

D

16 128 128

conclusion





fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12

12

fxy (x, y)= 4

4 4 4

D

16 128 128

conclusion





fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12

fxy (x, y)= 4

4 4 4

D

16 128 128

conclusion





fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12

fxy (x, y)= 4 4

4 4

D

16 128 128

conclusion





fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12

fxy (x, y)= 4 4 4

4

D

16 128 128

conclusion





fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12

fxy (x, y)= 4 4 4 4D

16 128 128

conclusion





fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12

fxy (x, y)= 4 4 4 4D 16

128 128

conclusion





fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12

fxy (x, y)= 4 4 4 4D 16 128

128

conclusion





fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12

fxy (x, y)= 4 4 4 4D 16 128 128

conclusion





fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12

fxy (x, y)= 4 4 4 4D 16 128 128

conclusion saddle pt

rel max rel max




fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12

fxy (x, y)= 4 4 4 4D 16 128 128

conclusion saddle pt rel max

rel max




fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12

fxy (x, y)= 4 4 4 4D 16 128 128

conclusion saddle pt rel max rel max

Outline



DefinitionLet f be a function of two variables x and y . Let R be a region on thexy-plane and (x0, y0) a point in R.

1 f is said to have an absolute maximum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.

2 f is said to have an absolute minimum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.

3 f is said to have an absolute extremum on R at (x0, y0) if either fhas an absolute maximum value or absolute minimum value on R at(x0, y0).

RemarkSuppose that f has an absolute maximum [minimum, extremum] at(x0, y0).

1 We say f (x0, y0) is the absolute maximum [minimum, extremum] valueof f on R.

2 What is the difference between a relative extremum point and anabsolute extremum point?

DefinitionLet R be a region in the xy-plane

1 R is bounded if it can be contained in some rectangle.2 R closed if R contains its boundary.


1 R is bounded if it can be contained in some rectangle.

2 R closed if R contains its boundary.


1 R is bounded if it can be contained in some rectangle.2 R closed if R contains its boundary.

Theorem (Extreme Value Theorem)If f (x, y) is continuous on a closed and bounded set R, then

f has both anabsolute maximum and an absolute minimum on R.

Theorem (Extreme Value Theorem)If f (x, y) is continuous on a closed and bounded set R, then f has both anabsolute maximum and an absolute minimum on R.

RemarkIf f and R satisfy the conditions of EVT, the absolute extrema occur at

1 critical points inside R

2 points on the boundary of R

RemarkIf f and R satisfy the conditions of EVT, the absolute extrema occur at

1 critical points inside R2 points on the boundary of R

ProcedureFinding Absolute Extrema on a Region R satisfying EVT:

1 Find the critical numbers of f (x, y) in R.

Evaluate f (x, y) at thecritical numbers.

2 Evaluate f (x, y) at the vertices of R (if any).3 Convert f (x, y) to a function of one variable g (x) or g

(y)on the

boundaries of R.4 Find the critical numbers of g and evaluate f (x, y) at these numbers.

Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.

5 Compare the values obtained.


1 Find the critical numbers of f (x, y) in R. Evaluate f (x, y) at thecritical numbers.


(y)on the






2 Evaluate f (x, y) at the vertices of R (if any).

3 Convert f (x, y) to a function of one variable g (x) or g(y)on the







(y)on the

boundaries of R.

4 Find the critical numbers of g and evaluate f (x, y) at these numbers.Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.





(y)on the




ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}

fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:

(12 ,0

)At critical point inside R: f

(12 ,0

)= 14At vertices: noneAt boundary: f

(12 ,

p152

)= f

(12 ,

p152

)= 334

CONCLUSION: f has an absolute minimum at(12 ,0

)and absolute

maxima at(12 ,

p152

)and

(12 ,

p152

)with values 14 and 334 , respectively.


fx(x, y)=

2x1 and fy (x, y)= 4yCritical Point:

(12 ,0


(12 ,0


(12 ,

p152

)= f

(12 ,

p152

)= 334


)and absolute

maxima at(12 ,

p152

)and

(12 ,

p152



fx(x, y)= 2x1

and fy (x, y)= 4yCritical Point:

(12 ,0


(12 ,0


(12 ,

p152

)= f

(12 ,

p152

)= 334


)and absolute

maxima at(12 ,

p152

)and

(12 ,

p152



fx(x, y)= 2x1 and

fy (x, y)= 4yCritical Point:

(12 ,0


(12 ,0


(12 ,

p152

)= f

(12 ,

p152

)= 334


)and absolute

maxima at(12 ,

p152

)and

(12 ,

p152



fx(x, y)= 2x1 and fy (x, y)=

4y

Critical Point:(12 ,0


(12 ,0


(12 ,

p152

)= f

(12 ,

p152

)= 334


)and absolute

maxima at(12 ,

p152

)and

(12 ,

p152



fx(x, y)= 2x1 and fy (x, y)= 4y

Critical Point:(12 ,0


(12 ,0


(12 ,

p152

)= f

(12 ,

p152

)= 334


)and absolute

maxima at(12 ,

p152

)and

(12 ,

p152




(12 ,0

)

At critical point inside R: f(12 ,0


(12 ,

p152

)= f

(12 ,

p152

)= 334


)and absolute

maxima at(12 ,

p152

)and

(12 ,

p152




(12 ,0


(12 ,0

)=

14At vertices: noneAt boundary: f

(12 ,

p152

)= f

(12 ,

p152

)= 334


)and absolute

maxima at(12 ,

p152

)and

(12 ,

p152




(12 ,0


(12 ,0

)= 14

At vertices: noneAt boundary: f

(12 ,

p152

)= f

(12 ,

p152

)= 334


)and absolute

maxima at(12 ,

p152

)and

(12 ,

p152




(12 ,0


(12 ,0

)= 14At vertices:

noneAt boundary: f

(12 ,

p152

)= f

(12 ,

p152

)= 334


)and absolute

maxima at(12 ,

p152

)and

(12 ,

p152




(12 ,0


(12 ,0

)= 14At vertices: none

At boundary: f(12 ,

p152

)= f

(12 ,

p152

)= 334


)and absolute

maxima at(12 ,

p152

)and

(12 ,

p152




(12 ,0


(12 ,0


(12 ,

p152

)= f

(12 ,

p152

)= 334


)and absolute

maxima at(12 ,

p152

)and

(12 ,

p152


ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).

fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11


fx(x, y)=

3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11


fx(x, y)= 3y 6

and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11


fx(x, y)= 3y 6 and

fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11


fx(x, y)= 3y 6 and fy (x, y)=

3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11


fx(x, y)= 3y 6 and fy (x, y)= 3x3

Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11


fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)

At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11


fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)=

1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11


fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1

At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11


fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:

f (0,0)= 7 f (0,5)= 8 f (3,0)= 11


fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)=

7 f (0,5)= 8 f (3,0)= 11


fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7

f (0,5)= 8 f (3,0)= 11


fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)=

8 f (3,0)= 11


fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8

f (3,0)= 11


fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)=

11


fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11

ExampleGiven: f (x, y)= 3xy 6x3y +7

At boundary x = 0:

g (y)= f (0, y) = 3y +7g (y) = 3

no critical pointsAt boundary y = 0:

g (x)= f (x,0) = 6x+7g (x) = 6

no critical points


At boundary x = 0:

g (y)=

f (0, y) = 3y +7g (y) = 3


g (x)= f (x,0) = 6x+7g (x) = 6

no critical points


At boundary x = 0:

g (y)= f (0, y) =

3y +7g (y) = 3


g (x)= f (x,0) = 6x+7g (x) = 6

no critical points


At boundary x = 0:

g (y)= f (0, y) = 3y +7

g (y) = 3


g (x)= f (x,0) = 6x+7g (x) = 6

no critical points


At boundary x = 0:

g (y)= f (0, y) = 3y +7g (y) = 3


g (x)= f (x,0) = 6x+7g (x) = 6

no critical points


At boundary x = 0:

g (y)= f (0, y) = 3y +7g (y) = 3

no critical points

At boundary y = 0:

g (x)= f (x,0) = 6x+7g (x) = 6

no critical points


At boundary x = 0:

g (y)= f (0, y) = 3y +7g (y) = 3


g (x)= f (x,0) = 6x+7g (x) = 6

no critical points


At boundary x = 0:

g (y)= f (0, y) = 3y +7g (y) = 3


g (x)=

f (x,0) = 6x+7g (x) = 6

no critical points


At boundary x = 0:

g (y)= f (0, y) = 3y +7g (y) = 3


g (x)= f (x,0) =

6x+7g (x) = 6

no critical points


At boundary x = 0:

g (y)= f (0, y) = 3y +7g (y) = 3


g (x)= f (x,0) = 6x+7

g (x) = 6

no critical points


At boundary x = 0:

g (y)= f (0, y) = 3y +7g (y) = 3


g (x)= f (x,0) = 6x+7g (x) = 6

no critical points


At boundary y =53x+5:

g (x) = f(x,5

3x+5

)= 3x

(53x+5

)6x3

(53x+5

)+7

= 5x2+24x8g (x) = 10x+24

Critical points:

10x+24= 0 x = 125

& y = 1

f(125 ,1

)= 20.8



g (x) = f(x,5

3x+5

)

= 3x(53x+5

)6x3

(53x+5

)+7

= 5x2+24x8g (x) = 10x+24

Critical points:

10x+24= 0 x = 125

& y = 1

f(125 ,1

)= 20.8



g (x) = f(x,5

3x+5

)= 3x

(53x+5

)6x3

(53x+5

)+7

= 5x2+24x8g (x) = 10x+24

Critical points:

10x+24= 0 x = 125

& y = 1

f(125 ,1

)= 20.8



g (x) = f(x,5

3x+5

)= 3x

(53x+5

)6x3

(53x+5

)+7

= 5x2+24x8

g (x) = 10x+24

Critical points:

10x+24= 0 x = 125

& y = 1

f(125 ,1

)= 20.8



g (x) = f(x,5

3x+5

)= 3x

(53x+5

)6x3

(53x+5

)+7

= 5x2+24x8g (x) = 10x+24

Critical points:

10x+24= 0 x = 125

& y = 1

f(125 ,1

)= 20.8



g (x) = f(x,5

3x+5

)= 3x

(53x+5

)6x3

(53x+5

)+7

= 5x2+24x8g (x) = 10x+24

Critical points:

10x+24= 0

x = 125

& y = 1

f(125 ,1

)= 20.8



g (x) = f(x,5

3x+5

)= 3x

(53x+5

)6x3

(53x+5

)+7

= 5x2+24x8g (x) = 10x+24

Critical points:

10x+24= 0 x = 125

&

y = 1

f(125 ,1

)= 20.8



g (x) = f(x,5

3x+5

)= 3x

(53x+5

)6x3

(53x+5

)+7

= 5x2+24x8g (x) = 10x+24

Critical points:

10x+24= 0 x = 125

& y = 1

f(125 ,1

)= 20.8


At critical point: f (1,2)= 1At vertices: f (0,0)= 7 f (0,5)=8 f (3,0)=11At boundary y =53x+5: f

(125 ,1

)= 20.8

CONCLUSION: The function has an absolute maximum at(125 ,1

)and an

absolute minimum at (3,0) with values 20.8 and 11, respectively.


At critical point: f (1,2)= 1At vertices: f (0,0)= 7 f (0,5)=8 f (3,0)=11At boundary y =53x+5: f

(125 ,1

)= 20.8CONCLUSION: The function has an absolute maximum at

(125 ,1

)and an

absolute minimum at (3,0) with values 20.8 and 11, respectively.

RemarkLet f be a function of two variables and R = {(x, y) | x > 0 and y > 0}.Suppose f is continuous on R and (x0, y0) R is the only critical point off .

1 If f (x, y) is very large when either x or y are very close to zero or verylarge, then f has an absolute minimum at (x0, y0).

2 If f (x, y) is "very negative" when either x or y are very close to zeroor very large, then f has an absolute maximum at (x0, y0).

ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.

Minimize: Surface AreaConstraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000

l+ 1000

wRegion: {(l ,w) | l > 0, w > 0}


Minimize:

Surface AreaConstraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000

l+ 1000

wRegion: {(l ,w) | l > 0, w > 0}


Minimize: Surface Area

Constraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000

l+ 1000

wRegion: {(l ,w) | l > 0, w > 0}


Minimize: Surface AreaConstraint:

V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000

l+ 1000

wRegion: {(l ,w) | l > 0, w > 0}


Minimize: Surface AreaConstraint: V = 500 cm3

ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000

l+ 1000

wRegion: {(l ,w) | l > 0, w > 0}


Minimize: Surface AreaConstraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500

ORMinimize: A(l ,w)= lw + 1000

l+ 1000

wRegion: {(l ,w) | l > 0, w > 0}


Minimize: Surface AreaConstraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000

l+ 1000

wRegion: {(l ,w) | l > 0, w > 0}

Example

Minimize: A(l ,w)= lw + 1000l

+ 1000w

Region: l > 0, w > 0

Al (l ,w)=w 1000

l2and Aw (l ,w)= l 1000

w2

Critical Point: l = 10 and w = 10.

Example


+ 1000w


Al (l ,w)=

w 1000l2

and Aw (l ,w)= l 1000w2


Example


+ 1000w


Al (l ,w)=w

1000l2

and Aw (l ,w)= l 1000w2


Example


+ 1000w


Al (l ,w)=w 1000

l2

and Aw (l ,w)= l 1000w2


Example


+ 1000w


Al (l ,w)=w 1000

l2and

Aw (l ,w)= l 1000w2


Example


+ 1000w


Al (l ,w)=w 1000

l2and Aw (l ,w)=

l 1000w2


Example


+ 1000w


Al (l ,w)=w 1000

l2and Aw (l ,w)= l

1000w2


Example


+ 1000w


Al (l ,w)=w 1000

l2and Aw (l ,w)= l 1000

w2


Example


+ 1000w


Note that A(l ,w) is continuous on the R.Hold w constant:

liml

lw + 1000l

+ 1000w

=

liml0+

lw + 1000l

+ 1000w

=

Hold l constant:limw lw +

1000

l+ 1000

w=

limw0+

lw + 1000l

+ 1000w

=

Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.

Example


+ 1000w


Note that A(l ,w) is continuous on the R.

Hold w constant:liml

lw + 1000l

+ 1000w

=

liml0+

lw + 1000l

+ 1000w

=


1000

l+ 1000

w=

limw0+

lw + 1000l

+ 1000w

=


Example


+ 1000w



liml

lw + 1000l

+ 1000w

=

liml0+

lw + 1000l

+ 1000w

=


1000

l+ 1000

w=

limw0+

lw + 1000l

+ 1000w

=


Example


+ 1000w



liml

lw + 1000l

+ 1000w

=

liml0+

lw + 1000l

+ 1000w

=

Hold l constant:

limw lw +

1000

l+ 1000

w=

limw0+

lw + 1000l

+ 1000w

=


Example


+ 1000w



liml

lw + 1000l

+ 1000w

=

liml0+

lw + 1000l

+ 1000w

=


1000

l+ 1000

w=

limw0+

lw + 1000l

+ 1000w

=


Example


+ 1000w



liml

lw + 1000l

+ 1000w

=

liml0+

lw + 1000l

+ 1000w

=


1000

l+ 1000

w=

limw0+

lw + 1000l

+ 1000w

=

Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.

The minimum surface area is 300cm2.

Example


+ 1000w



liml

lw + 1000l

+ 1000w

=

liml0+

lw + 1000l

+ 1000w

=


1000

l+ 1000

w=

limw0+

lw + 1000l

+ 1000w

=


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M38 Lec 021814

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