Maxima and Minima. (a, b) of R if

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1 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 R of 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)

2 Example Prove that the function f (x, y) = x + y attains a maximum value 2 and minimum value 0 on the square R = { (x, y) 0 x 1, and 0 y 1 } Proof For any point P(x, y) in R, one has 0 x 1 and 0 y 1, so it follows that 0 = x + y = 2 So 0 f (x, y) 2, which satisfies the condition (i) Moreover f (0, 0) = = 0, and f (1, 1) = = 2 such that (0, 0) and (1, 1) are in R, in which the condition (ii) holds

3 Local Extreme Values Definition Let f (x) be a function defined in a domain D of R n (n = 2, 3), and a point x 0 D Function f is said to attain its local maximum at x 0, if there exists δ > 0 such that f (x 0 ) f (x) for all x B(x 0, δ) D Remarks (i) In general, a function can have many local maxima, and can have no local maxima in its domain (ii) The concept of local minimum is similarly defined xz-cross section z = f m (x, y) = (x 2 + y 2 ) 2 + mx z = f m (x, y) = (x 2 + y 2 ) 2 + mx xz-cross section

4 Existence Theorem of Global Extremum Theorem A continuous function always attains its global maximum and global minimum values, if its domain is closed and bounded Remarks (a) One can replace the condition on the domain by a simple but more restrictive one, such as, the domain consists of the points 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 continuous curve 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 are interior points; and (ii) those that are on the curve called boundary points

5 Example Let f (x, y) = x 2 + y 2 on the region R = { (x, y) x 2 + y 2 1 } Determine the absolute minimum of f on R Solution As R is closed and bounded region in R 2, so f attains its absolute minimum As the square root is non-negative, so f (x, y) 0 = f (0, 0) for all (x, y) in R The minimum of the function f on R is given by f (0, 0) = 0 However, f (x, 0) = x = x, which is not differentiable at x = 0, and hence f x 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

6 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 f x (a, b) and f y (a, b) exist, Then f x (a, b) = f y (a, b) = 0 The theorem gives only a necessary condition for existence of a local extreme point provided that all the first order partial derivatives exist It can happen that at a local extreme point, some of the first partial derivatives 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

7 Types of Absolute Extremum Theorem Suppose that f is continuous on the plane region R consisting of the points on and within a simple closed curve C If f (a, b) is either the absolute maximum or the absolute minimum value of f (x, y) on R, then (a, b) is either 1 An interior point of R at which f x (a, b) = f y (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) is called a critical point of the function f Proposition Any extreme value of the continuous function f on the plane region R, bounded by a simple closed curve, must occur at an interior critical point or at a boundary point

8 Notation For any given f, one write f (x, y) = f x (x, y)i + f y (x, y)j = (f x, f y ) 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 determine the possible interior critical points Example Define f (x, y) = x 2 + y 2 on R 2 It is easy to see that the partial derivatives f x and f y exist only for all (x, y) = (0, 0) Moreover, f (h,0) f (0,0) h = h 2 h = h h h < 0 So the limit lim which is equal to 1 if h > 0, and equal to 1 if f (h,0) f (0,0) h 0 h does not exist, ie f x (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 condition is not satisfied

9 Example Find the maximum and minimum values by the function f (x, y) = xy x y + 3 at the points of the triangular region R in the xy-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 R 2, so one knows that the continuous function f must attains extremum values at some points in R together with the boundary of AOB (Interior Critical Points) As f is a polynomial, so both f x and f y exist at any interior points of R According to the theorem above, to search for the interior extremum point of f, one needs to determine the interior critical points of f in AOB Then f (x, y) = (f x (x, y), f y (x, y)) = (y 1, x 1), so (x, y) = (1, 1) is the only critical point of f So one needs to consider the boundary points of R as well

10 Example Find the maximum and minimum values by the function f (x, y) = xy x y + 3 at the points of the triangular region R in the xy-plane with vertices at O(0, 0), A(2, 0) and B(0, 4) Solution (Boundary Points) Consider the boundary points of R, ie points on the boundary 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 + tab = (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 = 8t 2 + 6t + 1 for 0 t 1 k (t) = 16t + 6, so t = 6/16 = 3/8 is a critical point for k max k = max{k(0), k(1), k(3/8)} = max{3, 1, 17/8} = 3, and min k = min{k(0), k(1), k(3/8)} = 1 Finally, on the region R we have max f = max{17/8, f (1, 1) = 2} = 2, and min f = 1

11 Example Find the maximum and minimum values of the function f (x, y) = xy x y + 3 for the point on the unit disc D = { (x, y) x 2 + y 2 1 } Solution As f is a polynomial, it follows from the law of limits that its first order partial derivatives exist on R 2 For interior critical point, we have f (x, y) = (0, 0) if and only if (0, 0) = (f x (x, y), f y (x, y)) = (y 1, x 1), ie (x, y) = (1, 1) In this case, the point (1, 1) does not lie in the disc D So both maximum and minimum values occur at the boundary points of D, which is the unit circle C Then we parameterize C : r(t) = (cos t, sin t) for 0 t 2π Define the function g(t) = f r(t) = f (cos t, sin t) = sin t cos t sin t cos t + 3, which is a differentiable function of one variable on the closed interval [0, 2π] One can use derivative test of one variable to find the maximum and minimum value of g, and hence that of f Here we skip the details for the sake of time

12 Differentiable Functions Let f be a function of single variable y = f (x), an increment x of input x means x x + x, then there is an increment in output y defined as y = f (x + x) f (x) Since derivative of f at x, exists, we adjust further by linear approximation 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 denotes a small quantity x for the limit procedure will kill this term as x 0, that is ε( x) = 0 lim x 0

13 In view of the definition of y = f (x + x) f (x) = f (x) x + ε( x) x, and the condition f (x + x) f (x) f lim ε( x) = (x) x lim = 0, we introduce the x 0 x 0 x concept of differential dx, dy as mathematical object to represent the relationship of this increment relation, dy = f (x)dx, in which dx and dy can be interpreted as linear increments geometrically

14 Let z = f (x, y) be a function defined on domain D, and P(a, b) is a point in D Define the increment f = f (a + x, b + y) f (a, b) at P(a, b), which depends only on ( x, y) Like the one variable case, we subtract the linear part from increment, and then divide by x 2 + y 2 to obtain ε( x, y) = f (a + x, b + y) f (a, b) f x(a, b) x f y (a, b) y, x 2 + y 2 which is a new function depending on ( x, y) only Definition Let z = f (x, y) be a function defined on domain D, and P(a, b) is a point in D f is called differentiable at a point P(a, b), if lim ε( x, y) = 0 ( x, y) (0,0) Definition f is differentiable on its domain D, if f is differentiable at every point in D Proposition If f is a differentiable function, then all 1st order partial derivatives of f exist However, the converse does not hold

15 Differentiable Functions of n varibles Definition The definition of differentiability of a function at a point can be generalized to for function of more than 2 variables Moreover, we assume that f has 1st order partial derivatives in D Suppose that f = f (x) is a function of n variables x = (x 1,, x n ), defined on the domain D R n Let a = (a 1,, a n ) be a fixed point in D instead of (a, b) R 2, and h = (h 1,, h n ) instead ( x, y) Define the following scalar function which only depends only h as f (a + h) f (a) f (a) h follows ε(h) = Then we f called h differentiable at a, if lim ε(h) = 0 h 0 Remark In general, it is very difficult to verify that a given function of multi-variables is differentiable

16 Differential of a scalar function of n variables Definition Let f = f (x 1,, x n ) be a differentiable function (or least 1st oder the partial derivatives of f exist) on D R n, then the differential of f, denoted by df, which can be reduced to the differentials of the coordinate functions defined on D as well: df (a) = f (a) dx = n i=1 f xi (a)dx i Remark Though we can not give a very concise definition of differential of a scalar function f in this course, but one should compare it with the linear approximation of f in the later section In this case, the local variation of f can be described linearly in terms of variation dx i of position changes

17 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 then the 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) are differentiable on D, provided g(x) g(x) = 0 for all x in D (iii) In particular, a rational function is a differentiable at every point in the domain D Proposition Let f be a scalar differentiable function defined in a domain D of R n with range R, and g is a scalar differentiable function defined on the set R, ie D f R g R Then the composite function g f is a scalar differentiable function defined on D Remark In fact, all the elementary functions are differentiable on their domain, which it follows from the two propositions above that we have a large class of differentiable functions

18 Example Let f (x, y) = x2 y 2 for all (x, y) = (0, 0), and f (0, 0) = 0 x 2 +y 2 Prove that f is differentiable on R 2 Solution From the proposition above, one knows that f is differentiable on R 2 \ {(0, 0)} It remains to show that f is differentiable at P(0, 0) For this, one has to check that the first partial f (0+h,0) f (0,0) 0 0 derivative exist Indeed f x (0, 0) = lim h 0 h = lim h 0 h = 0, and similarly f y (0, 0) = 0 Now we need to check that ε(( x, y) = f (0+ x,0+ y) f (0,0) f x(0,0) x f y (0,0) y = ( x) 2 +( y) 2 ( ( x)2 +( y) 2 ) 2 ( x) 2 +( y) 2 ( x)2 ( y) 2 1 ( x) 2 +( y) 2 = ( x) 4( ( x) 2 +( y) 2 ) 2 + ( y) 2 3/2 ) 2 In fact, the inequality is due to AM-GM inequality ab Then it follows from the sandwich theorem of limit that ε( x, y) = 0 lim ( x, y) (0,0) ( a+b 2

19 Definition For any differentiable f and a point P(a, b) in its domain, let L(x, y) = f (a, b) + f x (a, b)(x a) + f y (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 a function f at P(a, b) has a graph z = L(x, y) which is the tangent plane Example Approximate the number (32) 2 + (39) 2 using the linear approximation to the function f (x, y) = x 2 + y 2 at (3, 4) x Solution f (x, y) = (, x 2 +y 2 y ), and f (3, 4) = (3/5, 4/5), so x 2 +y2 the linear approximation of f (x, y) at (3, 4) is given by L(x, y) = f (3, 4) + f (3, 4) (x 3, y 4) = (x 3) (y 4) Then the number (32) 2 + (39) 2 can be approximated by L(32, 39) = = = 504

20 Definition In general, if f is a function defined in a domain D in R n (n = 2, 3), and P(x 0 ) is a point in D, then the linear approximation of f at P is given by L(x) = f (x 0 ) + f (x 0 ) (x x 0 ), = f (x 0 ) + f x (x 1 0 ) (x 1 a 1 ) + f x (x 2 0 ) (x 2 a 2 ) + + f x n (x 0 ) (x n a n ), where f (x 0 ) = ( f f f x (x 1 0 ), x (x 2 0 ),, x n (x 0 ) ), x 0 = (a 1,, a n ), and x = (x 1,, x n ) Example The function f (x, y) = 6x 2 2x 3 + y 3 + 3y 2 has n critical points, then n = Solution f (x, y) = (12x 6x 2, 3y 2 + 6y), f (x, y) = (0, 0) if and only 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

21 Chain Rule I If w = f (x, y) has continuous first-order partial derivatives and that r(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 a differentiable function of t, and (ii) its derivative is given by df dt = w x dx dt + w y dy dt = f (r(t)) r (t) z T P(x,y,z ) y x Remark We omit the proof

22 Example Use the chain rule to find dz at t = 1 if dt z = arctan(yx 2 ), x = x(t), y(t) = t 2, and dx (1) = x(1) = 1 dt Solution First z x = (yx 2 ) 2 2xy x (yx2 ) = 1 + (yx 2 ) 2, and z y = x (yx 2 ) 2 dz dt = z x dx dt + y y dy dt = 2xy 1 + (yx 2 ) 2 dx dt + x (yx 2 2t When ) 2 t = 1, we have x(1) = x (1) = 1, y(1) = 1, and y (1) = 2 Then dz = 2 dt t= = = 2

23 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 coordinate functions are parameterized by differentiable functions Then the composite function w(u, v) = f ( x(u, v), (u, v), (u, v) ) is a differentiable function in u and v, such that the partial functions are given by w u w v = w x x u + w y y u + w z z u ; = w x x v + w y y v + w z z v

24 Example Suppose that w = f (x, y) has continuous first-order partial derivatives and that the coordinates (x, y) are functions of the other variables r, θ such as in polar coordinates, ie x(r, θ) = r cos θ, and y(r, θ) = r sin θ By means of the new coordinates, w becomes a function in the variables r, θ, Namely, by means of composition, we have w(r, θ) = w(x(r, θ), y(r, θ)), for example f (r cos θ, r sin θ) in terms of polar coordinates The partial derivative w r can be evaluated by treating the variable θ as a fixed constant, so it follows from the chain w rule stated above that r = w x x r + w y y r