JUST THE MATHS SLIDES NUMBER 14.10 PARTIAL DIFFERENTIATION 10 - - PDF document

“JUST THE MATHS” SLIDES NUMBER 14.10 PARTIAL DIFFERENTIATION 10 (Stationary values) for (Functions of two variables) by A.J.Hobson

14.10.1 Introduction 14.10.2 Sufficient conditions for maxima and minima

UNIT 14.10 PARTIAL DIFFERENTIATION 10 STATIONARY VALUES FOR FUNCTIONS OF TWO VARIABLES 14.10.1 INTRODUCTION The equation, z = f(x, y), will normally represent some surface in space, referred to cartesian axes, Ox, Oy and Oz

✲ ✻ ✟ ✟ ✟ ✟ ✟ ✙

z x y O

DEFINITION 1 The “stationary points”, on a surface whose equation is z = f(x, y), are defined to be the points for which ∂z ∂x = 0 and ∂z ∂y = 0.

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DEFINITION 2 The function, z = f(x, y), is said to have a “local max- imum” at a point, P(x0, y0, z0), if z0 is larger than the z co-ordinates of all other points on the surface, with equation z = f(x, y), in the neighbourhood of P. DEFINITION 3 The function, z = f(x, y), is said to have a “local min- imum” at a point, P(x0, y0, z0), if z0 is smaller than the z co-ordinates of all other points on the surface, with equation z = f(x, y), in the neighbourhood of P. Note: At a stationary point, P(x0, y0, z0), on the surface with equation z = f(x, y), each of the planes, x = x0 and y = y0, intersect the surface in a curve which has a stationary point at P.

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14.10.2 SUFFICIENT CONDITIONS FOR MAXIMA AND MINIMA The following conditions are stated without proof: (a) Sufficient conditions for a local maximum A point, P(x0, y0, z0), on the surface with equation z = f(x, y), is a local maximum if ∂z ∂x = 0, ∂z ∂y = 0, ∂2z ∂x2 < 0 or ∂2z ∂y2 < 0 and ∂2z ∂x2.∂2z ∂y2 −

    ∂2z

∂x∂y

   

2

> 0, when x = x0 and y = y0.

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(b) Sufficient conditions for a local minimum A point, P(x0, y0, z0), on the surface with equation z = f(x, y), is a local minimum if ∂z ∂x = 0, ∂z ∂y = 0, ∂2z ∂x2 > 0 or ∂2z ∂y2 > 0 and ∂2z ∂x2.∂2z ∂y2 −

    ∂2z

∂x∂y

   

2

> 0, when x = x0 and y = y0. Notes: (i) If ∂2z

∂x2 is positive (or negative) and also ∂2z ∂x2.∂2z ∂y2−

• ∂2z

∂x∂y

2 >

0, then ∂2z

∂y2 is automatically positive (or negative).

(ii) If ∂2z

∂x2.∂2z ∂y2−

• ∂2z

∂x∂y

2is negative at P, we have a “saddle-

point”, irrespective of what ∂2z

∂x2 and ∂2z ∂y2 are.

(iii) The values of z at the local maxima and local min- ima of the function z = f(x, y) may also be called the “extreme values” of the function, f(x, y).

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EXAMPLES

• 1. Determine the extreme values and the co-ordinates of

any saddle-points of the function, z = x3 + x2 − xy + y2 + 4. Solution (i) First, we determine ∂z

∂x and ∂z ∂y.

∂z ∂x = 3x2 + 2x − y and ∂z ∂y = −x + 2y. (ii) Secondly, we solve the equations, ∂z

∂x = 0 and ∂z ∂y = 0, for x and y.

3x2 + 2x − y = 0, − − − − − − − − −(1) −x + 2y = 0. − − − − − − − − − (2) Substituting equation (2) into equation (1) gives 3x2 + 2x − 1 2x = 0.

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That is, 6x2 + 3x = 0 or 3x(2x + 1) = 0. Hence, x = 0 or x = −1

2, with corresponding values,

y = 0, z = 4 and y = −1

4, z = −65 16, respectively.

The stationary points are thus (0, 0, 4) and

• −1

2, −1 4, −65 16

• .

(iii) Thirdly, we evaluate ∂2z

∂x2, ∂2z ∂y2 and ∂2z ∂x∂y at each

stationary point. ∂2z ∂x2 = 6x + 2, ∂2z ∂y2 = 2, and ∂2z ∂x∂y = −1. (a) At the point (0, 0, 4), ∂2z ∂x2 = 2, ∂2z ∂y2 = 2, and ∂2z ∂x∂y = −1. Hence, ∂2z ∂x2 > 0, ∂2z ∂y2 > 0, and ∂2z ∂x2.∂2z ∂y2−

    ∂2z

∂x∂y

   

2

= 3 > 0 and, therefore, the point (0, 0, 4) is a local minimum, with z having a corresponding extreme value of 4.

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(b) At the point

• −1

2, −1 4, 65 16

• ,

∂2z ∂x2 = −1, ∂2z ∂y2 = 2, and ∂2z ∂x∂y = −1. Hence, ∂2z ∂x2 < 0, ∂2z ∂y2 > 0, and ∂2z ∂x2.∂2z ∂y2−

    ∂2z

∂x∂y

   

2

= −3 < 0 and, therefore, the point

• −1

2, −1 4, −65 16

• is a saddle-

point.

• 2. Determine the stationary points of the function,

z = 2x3 + 6xy2 − 3y3 − 150x, and determine their nature. Solution ∂z ∂x = 6x2 + 6y2 − 150 and ∂z ∂y = 12xy − 9y2. Hence, we solve the simultaneous equations, x2 + y2 = 25, − − − − − − − − (1) y(4x − 3y) = 0. − − − − − − − − − (2) From equation (2), y = 0 or 4x = 3y. Putting y = 0 in equation (1) gives x = ±5. Stationary points occur at (5, 0, −500) and (−5, 0, 500).

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Putting x = 3

4y into (1) gives y = ±4, x = ±3.

Stationary points occur at (3, 4, −300) and (−3, −4, 300). To classify the stationary points, ∂2z ∂x2 = 12x, ∂2z ∂y2 = 12x − 18y, and ∂2z ∂x∂y = 12y.

Point

∂2z ∂x2 ∂2z ∂y2 ∂2z ∂x∂y ∂2z ∂x2.∂2z ∂y2 −

• ∂2z

∂x∂y 2

Nature (5, 0, −500) 60 60 positive minimum (−5, 0, 500) −60 −60 positive maximum (3, 4, −300) 36 −36 48 negative saddle-pt. (−3, −4, 300) −36 36 −48 negative saddle-pt.

Note: The conditions used in the examples above are only suf- ficient conditions; that is, if the conditions are satisfied, we may make a conclusion.

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Outline Proof of the Sufficient Conditions From Taylor’s theorem for two variables, f(a + h, b + k) − f(a, b) = hfx(a, b) + kfy(a, b)+ 1 2

• h2fxx(a, b) + 2hkfxy(a, b) + k2fyy(a, b)
• + . . ,

where h and k are small compared with a and b. fx means ∂f

∂x, fy means ∂f ∂y, fxx means ∂2f ∂x2, fyy means ∂2f ∂y2

and fxy means

∂2f ∂x∂y.

If fx(a, b) = 0 and fy(a, b) = 0, the conditions for a local minimum at the point (a, b, f(a, b)) will be satisfied when the second term on the right-hand side is positive; and the conditions for a local maximum at this point are satisfied when the second term on the right is negative. We assume that later terms of the Taylor series expansion are negligible.

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Also, it may be shown that a quadratic expression of the form, Lh2 + 2Mhk + Nk2, is positive when L > 0 or N > 0 and LN − M 2 > 0; but negative when L < 0 or N < 0 and LN − M 2 > 0. If LN − M 2 < 0, it may be shown that the quadratic expression may take both positive and negative values. Finally, replacing L, M and N by fxx(a, b), fyy(a, b) and fxy(a, b) respectively, the sufficient conditions for local maxima, local minima and saddle-points follow.

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