. . . . . .

.. Maxima and Minima.

......

Definition. Let R be any region on the xy-plane, a function f (x, y)attains its absolute or global, maximum value M on R at the point(a, b) of R if(i) f (x, y) ≤ M for all points (x, y) in R, and(ii) M = f (a, b).

Similarly, we can define the global minimum as well..

......

Function f (x, y) attains the absolute or global, minimum value m on Rof at the point (a, b) of R if(i) f (x, y) ≥ m for all points (x, y) in R, and(ii) m = f (a, b).

Matb 210 in 2012

. . . . . .

.Example..

......

Prove that the function f (x, y) = x + y attains a maximum value 2 andminimum value 0 on the square R = { (x, y) | 0 ≤ x ≤ 1, and0 ≤ y ≤ 1 }.

Proof. For any point P(x, y) in R, one has 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, soit follows that

0 = 0 + 0 ≤ x + y ≤ 1 + 1 = 2.

So 0 ≤ f (x, y) ≤ 2, which satisfies the condition (i). Moreoverf (0, 0) = 0 + 0 = 0, and f (1, 1) = 1 + 1 = 2 such that (0, 0) and (1, 1)are in R, in which the condition (ii) holds.

Matb 210 in 2012

. . . . . .

.. Local Extreme Values.

......

Definition. Let f (x) be a function defined in a domain D ofRn (n = 2, 3), and a point x0 ∈ D. Function f is said to attain its localmaximum at x0, if there exists δ > 0 such that f (x0) ≥ f (x) for allx ∈ B(x0, δ) ∩ D.

Remarks. (i) In general, a function can have many local maxima, andcan have no local maxima in its domain.(ii) The concept of local minimum is similarly defined.xz-cross section z = fm(x, y) = (x2 + y2)2 + mx

xz-cross sectionz = fm(x, y) = (x2 + y2)2 + mx

Matb 210 in 2012

. . . . . .

.. Existence Theorem of Global Extremum

.

......

Theorem. A continuous function always attains its global maximumand global minimum values, if its domain is closed and bounded.

Remarks. (a) One can replace the condition on the domain by asimple but more restrictive one, such as, the domain consists of thepoints on and within a simple closed curve.(b) The proof of this theorem requires Heine-Borel cover theorem..

......

Definition. A simple closed curve is a parameterized continuouscurve r(t) for t ∈ [a, b], such that (i) r(a) = r(b), and the vectors r(t)are different from each other except when t = a and t = b.

.

......

Let D be consist of points on and within a simple and closed curve.One can prove that (i) the points in D that are not on the curve areinterior points; and (ii) those that are on the curve called boundarypoints.

Matb 210 in 2012

. . . . . .

.

......

Example Let f (x, y) =√

x2 + y2 on the regionR = { (x, y) |

√x2 + y2 ≤ 1 }. Determine the absolute minimum of f

on R.

Solution. As R is closed and bounded region in R2, so f attains itsabsolute minimum. As the square root is non-negative, sof (x, y) ≥ 0 = f (0, 0) for all (x, y) in R. The minimum of the function fon R is given by f (0, 0) = 0. However, f (x, 0) =

√x2 + 02 = |x|, which

is not differentiable at x = 0, and hence fx does not exist at (0, 0).Remark. It is an example that the absolute minimum occurs at which

f (x, y) does not have partial derivatives.

Matb 210 in 2012

. . . . . .

.Necessary Conditions for Local Extremum..

......

Theorem. Suppose that f (x, y) is a continuous on region D in plane. If(i) f (x, y) attains a local maximum value at an interior point (a, b) in D,and(ii) both the partial derivatives fx(a, b) and fy(a, b) exist,Then fx(a, b) = fy(a, b) = 0.

The theorem gives only a necessary condition for existence of a localextreme point provided that all the first order partial derivatives exist.It can happen that at a local extreme point, some of the first partialderivatives do not exist..

......

Definition. A point P in the domain of f is called a critical point of f , if(i) all first order partial derivatives of f at P exist and are equal to 0, or(ii) at least a first order partial derivatives of f does not exist.

Matb 210 in 2012

. . . . . .

.Types of Absolute Extremum..

......

Theorem. Suppose that f is continuous on the plane region Rconsisting of the points on and within a simple closed curve C. Iff (a, b) is either the absolute maximum or the absolute minimum valueof f (x, y) on R, then (a, b) is either

...1 An interior point of R at which fx(a, b) = fy(a, b) = 0,

...2 An interior point of R where not both partial derivatives exist, or

...3 A point of the boundary curve C of R.

.

......

Definition. A point which satisfies the above conditions (1) or (2) iscalled a critical point of the function f .

.

......

Proposition. Any extreme value of the continuous function f on theplane region R, bounded by a simple closed curve, must occur at aninterior critical point or at a boundary point.

Matb 210 in 2012

. . . . . .

.

......

Notation. For any given f , one write∇f (x, y) = fx(x, y)i + fy(x, y)j = (fx, fy) at point (x, y).

.

......

If f has 1st order partial derivatives on its domain, then(a, b) is a critical point of f ⇐⇒ ∇f (a, b) = (0, 0).So one can think of this condition as a set of equations to determinethe possible interior critical points.

.

......

Example. Define f (x, y) =√

x2 + y2 on R2. It is easy to see that thepartial derivatives fx and fy exist only for all (x, y) ̸= (0, 0). Moreover,f (h,0)−f (0,0)

h =√

h2

h = |h|h which is equal to 1 if h > 0, and equal to −1 if

h < 0. So the limit limh→0

f (h,0)−f (0,0)h does not exist, i.e. fx(0, 0) does not

exist.However, the function f attains the global minimum value at(x, y) = (0, 0), so the theorem fails to apply to apply as the conditionis not satisfied.

Matb 210 in 2012

. . . . . .

.

......

Example. Find the maximum and minimum values by the functionf (x, y) = xy − x − y + 3 at the points of the triangular region R in thexy-plane with vertices at O(0, 0), A(2, 0) and B(0, 4).

Solution. (Existence) As R is a closed and bounded set in R2, so oneknows that the continuous function f must attains extremum values atsome points in R together with the boundary of △AOB.(Interior Critical Points) As f is a polynomial, so both fx and fy existat any interior points of R. According to the theorem above, to searchfor the interior extremum point of f , one needs to determine theinterior critical points of f in △AOB. Then∇f (x, y) = (fx(x, y), fy(x, y)) = (y − 1, x − 1), so (x, y) = (1, 1) is theonly critical point of f .So one needs to consider the boundary points of R as well.

Matb 210 in 2012

. . . . . .

.

......

Example. Find the maximum and minimum values by the functionf (x, y) = xy − x − y + 3 at the points of the triangular region R in thexy-plane with vertices at O(0, 0), A(2, 0) and B(0, 4).

Solution. (Boundary Points) Consider theboundary points of R, i.e. points on theboundary of ∆AOB.AO : g(x) = f (x, 0) = −x + 3 for 0 ≤ x ≤ 2.max g = 3, min g = 1.BO : h(y) = f (0, y) = −y + 3 for 0 ≤ y ≤ 4.max h = 3, min h = −1.AB : Parameterize segment AB by

r(t) =−→OA + t

−→AB = (2, 0) + t(−2, 4) = (2 − 2t, 4t) for 0 ≤ t ≤ 1.

AB : k(t) = f (2 − 2t, 4t) = (2 − 2t)(4t)− (2 − 2t)− 4t + 3 =

−8t2 + 6t + 1 for 0 ≤ t ≤ 1. k′(t) = −16t + 6, so t = 6/16 = 3/8 is acritical point for k.max k = max{k(0), k(1), k(3/8)} = max{3,−1, 17/8} = 3, andmin k = min{k(0), k(1), k(3/8)} = −1. Finally, on the region R wehave max f = max{17/8, f (1, 1) = 2} = 2, and min f = −1.

Matb 210 in 2012

. . . . . .

.

......

Example. Find the maximum and minimum values of the functionf (x, y) = xy − x − y + 3 for the point on the unit discD = { (x, y) | x2 + y2 ≤ 1 }.

Solution. As f is a polynomial, it follows from the law of limits that itsfirst order partial derivatives exist on R2. For interior critical point, wehave ∇f (x, y) = (0, 0) if and only if(0, 0) = (fx(x, y), fy(x, y)) = (y − 1, x − 1), i.e. (x, y) = (1, 1). In thiscase, the point (1, 1) does not lie in the disc D. So both maximum andminimum values occur at the boundary points of D, which is the unitcircle C. Then we parameterize C : r(t) = (cos t, sin t) for 0 ≤ t ≤ 2π.Define the functiong(t) = f ◦ r(t) = f (cos t, sin t) = sin t cos t − sin t − cos t + 3, which isa differentiable function of one variable on the closed interval [0, 2π].One can use derivative test of one variable to find the maximum andminimum value of g, and hence that of f . Here we skip the details forthe sake of time.

Matb 210 in 2012

. . . . . .

.. Differentiable Functions

Let f be a function of single variable y = f (x), an increment ∆x ofinput x means x 7→ x + ∆x, then there is an increment in output ydefined as

∆y = f (x + ∆x)− f (x).Since derivative of f at x, exists, we adjust further by linearapproximation term, then we have

∆y = f (x + ∆x)− f (x) = f ′(x)∆x + ε(∆x) · ∆x,

where ε(∆x) =f (x + ∆x)− f (x)− f ′(x)∆x

∆xdenotes a small quantity

for the limit procedure will kill this term as ∆x → 0, that islim

∆x→0ε(∆x) = 0.

Matb 210 in 2012

. . . . . .

In view of the definition of∆y = f (x + ∆x)− f (x) = f ′(x)∆x + ε(∆x) · ∆x, and the condition

lim∆x→0

ε(∆x) = lim∆x→0

f (x + ∆x)− f (x)− f ′(x)∆x∆x

= 0, we introduce the

concept of differential dx, dy as mathematical object to represent therelationship of this increment relation,

dy = f ′(x)dx,

in which dx and dy can be interpreted as linear incrementsgeometrically.

Matb 210 in 2012

. . . . . .

Let z = f (x, y) be a function defined on domain D, and P(a, b) is apoint in D. Define the increment ∆f = f (a + ∆x, b + ∆y)− f (a, b) atP(a, b), which depends only on (∆x, ∆y). Like the one variable case,we subtract the linear part from increment, and then divide by√

∆x2 + ∆y2 to obtain

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

∆x2 + ∆y2,

which is a new function depending on (∆x, ∆y) only..

......

Definition. Let z = f (x, y) be a function defined on domain D, andP(a, b) is a point in D. f is called differentiable at a point P(a, b), if

lim(∆x,∆y)→(0,0)

ε(∆x, ∆y) = 0.

.

......

Definition. f is differentiable on its domain D, if f is differentiable atevery point in D.

.

......

Proposition. If f is a differentiable function, then all 1st order partialderivatives of f exist. However, the converse does not hold.

Matb 210 in 2012

. . . . . .

.. Differentiable Functions of n varibles.

......

Definition. The definition of differentiability of a function at a point canbe generalized to for function of more than 2 variables. Moreover, weassume that f has 1st order partial derivatives in D.Suppose that f = f (x) is a function of n variables x = (x1, · · · , xn),defined on the domain D ⊂ Rn. Let a = (a1, · · · , an) be a fixed pointin D instead of (a, b) ∈ R2, and h = (h1, · · · , hn) instead (∆x, ∆y).Define the following scalar function which only depends only h as

follows ε(h) =f (a + h)− f (a)−∇f (a) · h

∥h∥ . Then we f called

differentiable at a, if limh→0

ε(h) = 0.

.

......Remark. In general, it is very difficult to verify that a given function ofmulti-variables is differentiable.

Matb 210 in 2012

. . . . . .

.. Differential of a scalar function of n variables.

......

Definition. Let f = f (x1, · · · , xn) be a differentiable function (or least1st oder the partial derivatives of f exist) on D ⊂ Rn, then thedifferential of f , denoted by df , which can be reduced to thedifferentials of the coordinate functions defined on D as well:

df (a) = ∇f (a) · dx =n

∑i=1

fxi(a)dxi ♡.

.

......

Remark. Though we can not give a very concise definition ofdifferential of a scalar function f in this course, but one shouldcompare it with the linear approximation of f in the later section. Inthis case, the local variation of f can be described linearly in terms ofvariation dxi of position changes.

Matb 210 in 2012

. . . . . .

.

......

Proposition. (i) If f is a polynomial function, then f is differentiable;(ii) If f and g are differentiable functions defined on D, such that thenthe following functions s(x) = f (x) + g(x), d(x) = f (x)− g(x) ,p(x) = f (x) · g(x) and q(x) = f (x)

g(x) are differentiable on D, providedg(x) ̸= 0 for all x in D.(iii) In particular, a rational function is a differentiable at every point inthe domain D.

.

......

Proposition. Let f be a scalar differentiable function defined in adomain D of Rn with range R, and g is a scalar differentiable function

defined on the set R, i.e. Df−→ R

g−→ R. Then the compositefunction g ◦ f is a scalar differentiable function defined on D.

.

......

Remark. In fact, all the elementary functions are differentiable ontheir domain, which it follows from the two propositions above that wehave a large class of differentiable functions.

Matb 210 in 2012

. . . . . .

.

......

Example. Let f (x, y) = x2y2

x2+y2 for all (x, y) ̸= (0, 0), and f (0, 0) = 0.

Prove that f is differentiable on R2.

Solution. From the proposition above, one knows that f isdifferentiable on R2 \ {(0, 0)}. It remains to show that f isdifferentiable at P(0, 0). For this, one has to check that the first partialderivative exist. Indeed fx(0, 0) = lim

h→0

f (0+h,0)−f (0,0)h = lim

h→00−0

h = 0,

and similarly fy(0, 0) = 0. Now we need to check that

|ε((∆x, ∆y)| =∣∣∣∣ f (0+∆x,0+∆y)−f (0,0)−fx(0,0)·∆x−fy(0,0)·∆y√

(∆x)2+(∆y)2

∣∣∣∣ =∣∣∣∣ (∆x)2(∆y)2

(∆x)2+(∆y)2 · 1√(∆x)2+(∆y)2

∣∣∣∣ ≤ ( (∆x)2+(∆y)2)2

4( (∆x)2+(∆y)2 )3/2 =√(∆x)2 + (∆y)2.

In fact, the inequality is due to AM-GM inequality ab ≤(

a+b2

)2. Then

it follows from the sandwich theorem of limit thatlim

(∆x,∆y)→(0,0)ε(∆x, ∆y) = 0.

Matb 210 in 2012

. . . . . .

.

......

Definition. For any differentiable f and a point P(a, b) in its domain, letL(x, y) = f (a, b) + fx(a, b)(x − a) + fy(a, b)(y − b)

= f (a, b) +∇f (a, b) · (x − a, y − b)be the linear approximation of f at the point P(a, b).

Remark. For any fixed P(a, b), the function L(x, y) associated to afunction f at P(a, b) has a graph z = L(x, y) which is the tangent plane..

......

Example. Approximate the number√(3.2)2 + (3.9)2 using the linear

approximation to the function f (x, y) =√

x2 + y2 at (3, 4).

Solution. ∇f (x, y) = ( x√x2+y2

, y√x2+y2

), and ∇f (3, 4) = (3/5, 4/5), so

the linear approximation of f (x, y) at (3, 4) is given byL(x, y) = f (3, 4) +∇f (3, 4) · (x − 3, y − 4)= 5 + 3

5 (x − 3) + 45 (y − 4). Then the number

√(3.2)2 + (3.9)2 can be

approximated byL(3.2, 3.9) = 5 + 3

5 × 0.2 − 45 × 0.1 = 5 + 3

25 − 225 = 5.04.

Matb 210 in 2012

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.

......

Definition. In general, if f is a function defined in a domain D inRn (n = 2, 3), and P(x0) is a point in D, then the linear approximationof f at P is given by

L(x) = f (x0) +∇f (x0) · (x − x0),

= f (x0) +∂f

∂x1(x0) · (x1 − a1) +

∂f∂x2

(x0) · (x2 − a2) + · · ·+ ∂f∂xn

(x0) ·(xn − an),where ∇f (x0) = ( ∂f

∂x1(x0),

∂f∂x2

(x0), · · · , ∂f∂xn

(x0) ), x0 = (a1, · · · , an),and x = (x1, · · · , xn).

.

......

Example. The function f (x, y) = 6x2 − 2x3 + y3 + 3y2 has n criticalpoints, then n = .

Solution. ∇f (x, y) = (12x − 6x2, 3y2 + 6y), ∇f (x, y) = (0, 0) if andonly 0 = 6x(x + 2) and 0 = 2y(y + 2), it follows that(x, y) = (0, 0), (−2, 0), (0,−2), (−2,−2) are the critical points of f .Then n = 4.

Matb 210 in 2012

. . . . . .

.Chain Rule I..

......

If w = f (x, y) has continuous first-order partial derivatives and thatr(t) = (g(t), h(t)) is a differentiable curve in the domain of f , then (i)the composite function F(t) = w ◦ r(t) = f (g(t), h(t)) is adifferentiable function of t, and (ii) its derivative is given bydFdt

=∂w∂x

· dxdt

+∂w∂y

· dydt

= ∇f (r(t)) · r′(t).

3

P(x¸, y¸, z¸)T

y

x

z

Remark. We omit the proof.

Matb 210 in 2012

. . . . . .

.

......

Example. Use the chain rule to finddzdt

at t = 1 if

z = arctan(yx2), x = x(t), y(t) = t2, anddxdt

(1) = x(1) = 1.

Solution. First∂z∂x

=1

1 + (yx2)2 · ∂

∂x(yx2) =

2xy1 + (yx2)2 , and

∂z∂y

=x2

1 + (yx2)2

dzdt

=∂z∂x

· dxdt

+∂y∂y

· dydt

=2xy

1 + (yx2)2 · dxdt

+x2

1 + (yx2)2 · 2t. When

t = 1, we have x(1) = x′(1) = 1, y(1) = 1, and y′(1) = 2. Thendzdt

∣∣∣t=1

=22· 1 +

12· 2 = 1 + 1 = 2.

Matb 210 in 2012

. . . . . .

.Chain Rule II..

......

Theorem. Suppose that w = f (x, y, z) is a differentiable function,where x = x(u, v), y = y(u, v), z = z(u, v), where the coordinatefunctions are parameterized by differentiable functions. Then thecomposite function w(u, v) = f ( x(u, v), (u, v), (u, v) ) is adifferentiable function in u and v, such that the partial functions aregiven by

∂w∂u

=∂w∂x

· ∂x∂u

+∂w∂y

· ∂y∂u

+∂w∂z

· ∂z∂u

;

∂w∂v

=∂w∂x

· ∂x∂v

+∂w∂y

· ∂y∂v

+∂w∂z

· ∂z∂v

.

Matb 210 in 2012

. . . . . .

.

......

Example. Suppose that w = f (x, y) has continuous first-order partialderivatives and that the coordinates (x, y) are functions of the othervariables r, θ such as in polar coordinates, i.e x(r, θ) = r cos θ, andy(r, θ) = r sin θ. By means of the new coordinates, w becomes afunction in the variables r, θ, Namely, by means of composition, wehave w(r, θ) = w(x(r, θ), y(r, θ)), for example f (r cos θ, r sin θ) in termsof polar coordinates. The partial derivative ∂w

∂r can be evaluated bytreating the variable θ as a fixed constant, so it follows from the chain

rule stated above that∂w∂r

=∂w∂x

· ∂x∂r

+∂w∂y

· ∂y∂r

.

Matb 210 in 2012