CHAPTER : Partial Derivatives.1 Definition of a Partial Derivative. Increments and Differential.3 Chain Rules.4 Local Etrema.5 Absolute Etrema 1
Chapter : Partial Derivatives.1 Definition of a Partial Derivative The process of differentiating a function of several variables with respect to one of its variables while keeping the other variables fied is called partial differentiation. The resulting derivative is a partial derivative of the function. See illustration
As an illustration, consider the surface area of a right-circular clinder with radius r and height h: r h We know that the surface area is given b S pr p rh. This is a function of two variables r and h. Suppose r is held fied while h is allowed to var. Then, ds p r dh r const. This is the partial derivative of S with respect to h. It describes the rate with which a clinder s surface changes if its height is increased and its radius is kept constant. Likewise, suppose h is held fied while r is allowed to var. Then, 3
ds dr h const. 4pr p h This is the partial derivative of S with respect to r. It represents the rate with which the surface area changes if its radius is increased and its height is kept constant. In standard notation, these epressions are indicated b Sh p r, S 4pr p h r Thus in general, the partial derivative of z f(, ) with respect to, is the rate at which z changes in response to changes in, holding constant. Similarl, we can view the partial derivative of z with respect to in the same wa. Note Just as the ordinar derivative has different interpretations in different contets, so does a partial derivative. We can interpret derivative as a rate of change and the slope of a tangent line. 4
5 Recall: Derivative of a single variable f is defined formall as, f f f ) ( ) ( lim ) ( 0 The definition of the partial derivatives with respect to and are defined similarl. Definition.1 If ), ( f z, then the (first) partial derivatives of f with respect to and are the functions f and f respectivel defined b f f f ), ( ), ( lim 0 f f f ), ( ), ( lim 0 provided the limits eist.
.1.1 Notation For z f (, ), the partial derivatives f and f are also denoted b the smbols: f, z, f (, ) z, f (, ) or f, z, f (, ) z, f (, ) or The values of the partial derivatives at the point (a, b) are denoted b f ( a, b) f ( a, b) f and f ( a, b) ( a, b) Note The stlized d smbol in the notation is called roundback d, curl d or del d. It is not the usual derivative d (dee) or (delta d). 6
Illustration Finding and evaluating partial derivative of a function of two variables Finding partial derivative of a function of three variables Finding partial derivative of an implicitl defined function Eample.7 If f (, ) 3 4, find i. f ii. f iii. f (1, ) Prompts/Questions What do the notations stand for? o Which variable is changing? o Which variable is held constant? Which variables give the value of a derivative? 7
Solution (a) For f, hold constant and find the derivative with respect to : f 3 3 4 (b) For f, hold constant and find the derivative with respect to : f 3 (c) f 3 3 (1, ) (1) (1) ( ) 3 8
For a function f (,, z) of three variables, there are three partial derivatives: f, f and The partial derivative f is calculated b holding and z constant. Likewise, for f and f z. Eample. Let f (a) f Solution (,, z) z (b) f f z 3 (c) f z, find: (a) f (,, z) (b) f (,, z) 4 z 3 (c) f z (,, z) 3z 9
The rules for differentiating functions of a single variable holds in calculating partial derivatives. Eample.3 Find f if f (, ) ln( ). Solution We treat as a constant and f as a composite function: f [ln( )] 1 1 1 (0 ( 1) ) 10
Eample.3a Determine the partial derivatives of the following functions with respect to each of the independent variables: (a) (b) z w ( 3 ze 37 ) 5 11
Eample.3b Determine the partial derivatives of the following functions with respect to each of the independent variables: a) z sin( 5) b) f (, ) cos 1
Eample.4 If z f ( ), show that z z 0 13
Eample.5 Find z if the equation z ln z defines z as a function of two independent variables and. Solution We differentiate both sides of the equation with respect to, holding constant and treating z as a differentiable function of : ( z) (ln z) ( ) ( ) z 1 z 1 0, constant z 1 z 1 z z z z 1 14
Eample.5a If cos( z) 3 z 0 defines z as a function of two independent variables and. Determine epressions for z and z in terms of, and z. 15
.1. Partial Derivative as a Slope To understand the concept let s take a look at the one-variable case: Curve C Secant f( + ) Tangent line f() P + At P, the tangent line to the curve C has slope f (). 16
The intersection of the plane 0 with the surface z f(, ). 17
The intersection of the plane 0 with the surface z f(, ). 18
Eample.6 Find the slope of the line that is parallel to the z-plane and tangent to the surface z at the point P (1, 3, ). Solution Given f (, ) WANT: f (1, 3) f (, ) ( ) 1 1 ( ) 1 (1 0) Thus the required slope, f 1 ( 1, 3) 1 3 1 3 9 4 19
.1.3 Partial Derivative as a Rate of Change The derivative of a function of one variable can be interpreted as a rate of change. Likewise, we can obtain the analogous interpretation for partial derivative. A partial derivative is the rate of change of a multi-variable function when we allow onl one of the variables to change. f Specificall, the partial derivative at ( 0, 0) gives the rate of change of f with respect to when is held fied at the value 0. 0
Eample.7 The volume of a gas is related to its temperature T and its pressure P b the gas law PV 10T, where V is measured in cubic inches, P in pounds per square inch, and T in degrees Celsius. If T is kept constant at 00, what is the rate of change of pressure with respect to volume at V 50? Solution WANT: P V T 00, V 50 Given PV 10T. P V 10T V P ( 10)(00) 4 (50) 5 V T 00, V 50 1
.1.4 Higher Order Partial Derivatives The partial derivative of a function is a function, so it is possible to take the partial derivative of a partial derivative. If z is a function of two independent variables, and, the possible partial derivatives of the second order are: second partial derivative taking two consecutive partial derivatives with respect to the same variable mied partial derivative - taking partial derivatives with respect to one variable, and then take another partial derivative with respect to a different variable
Standard Notations Given z f (, ) Second partial derivatives f f = f = f ( f ) = f = ( f ) = f = Mied partial derivatives f = f f = f ( f ) = f = ( f ) = f = 3
Remark The mied partial derivaties can give the same result whenever f, f, f, f and f are all continuous. Partial derivaties of the third and higher orders are defined analogousl, and the notation for them is similar. 3 f = f = f 4 f = f = f The order of differentiation is immaterial as long as the derivatives through the order in question are continuous. 4
Eample.8 3 3 Let z 7 5 6. Find the indicated partial derivatives. i. z ii. z z iii. iv. f (,1) Prompts/Questions What do the notations represent? What is the order of differentiation? o With respect to which variable do ou differentiate first? Solution Keeping fied and differentiating w.r.t., we z obtain 1 10. Keeping fied and differentiating w.r.t., we z obtain 5 18. (i) (ii) (iii) z z ( 5 18 ) 10 z z (1 10 ) 10 z z (1 10 ) 4 10 z 5
(iv) z f (,1) 10() 0 (,1) Eample.9 Determine all first and second order partial derivatives of the following functions: i. z sin cos ii. z e ( ) iii. f (, ) cos e Prompts/Questions What are the first partial derivatives of f? o Which derivative rules or techniques do ou need? How man secondorder derivatives are there? 6
. Increments and Differential..1 Functions of One Variable A Recap Tangent Line approimation f( 1 ) f T f( 0 ) P 0 1 If f is differentiable at 0, the tangent line at P ( 0, f ( 0)) has slope m f ( 0 ) and equation = f( 0 ) + f )( 0 ) ( 0
If 1 is near 0, then f( 1 ) must be close to the point on the tangent line, that is f ( 1 ) f ( 0) f ( 0)( 1 0) This epression is called the linear approimation formula. Incremental Approimation We use the notation for the difference 1 0 and the corresponding notation for f( 1 ) f( 0 ). Then the linear approimation formula can be written as f ( ) f ( 0) f ( 0 1 ) or equivalentl f ( ) 0 3
Definition. If f is differentiable and the increment is sufficientl small, then the increment, in, due to an increment of, in is given b d d or f f ( ) Note This version of approimation is sometimes called the incremental approimation formula and is used to stud propagation of error. 4
The Differential d is called the differential of and we define d to be, an arbitrar increment of. Then, if f is differentiable at, we define the corresponding differential of, d as d d d d or equivalentl df f ( ) d Thus, we can estimate the change f, in f b the value of the differential df provided d is the change in. f df 5
f f( 0 + ) T f( 0 ) P d 0 0 + d = d is the rise of f (the change in ) that occurs relative to d d is the rise of tangent line relative to d The true change: f f ) f ( ) ( 0 0 The differential estimate: df f ( ) d 6
.. Functions of Two Variables Let z f (, ), where and are independent variables. If is subject to a small increment (or a small error) of, while remains constant, then the corresponding increment of z in z will be z z Similarl, if is subject to a small increment of, while remains constant, then the corresponding increment of z in z will be z z It can be shown that, for increments (or errors) in both and, z z z 7
The formula for a function of two variables ma be etended to functions of a greater number of independent variables. For eample, if w f (,, z) of three variables, then w w w w z z Definition.3 Let z f (, ) where f is a differentiable function and let d and d be independent variables. The differential of the dependent variable, dz is called the total differential of z is defined as dz df (, ) f (, ) d f (, ) d Thus, z dz provided d is the change in and d is the change in. 8
9
Eample.9 3 Let f (, ). Compute z and dz as (, ) changes from (, 1) to (.03, 0.98). Solution z = f(.03, 0.98) f(, 1) (.03) 3 3 (.03)(0.98) (0.98) 3 [() (1) 1 3 ] = 0.77906 dz f (, ) d f (, ) d (6 ) ( 3 ) At (, 1) with = 0.03 and = 0.0, dz ( 5)(0.03) ( 1)( 0.0) 0.77 30
Eample.10 A clindrical tank is 4 ft high and has a diameter of ft. The walls of the tank are 0. in. thick. Approimate the volume of the interior of the tank assuming that the tank has a top and a bottom that are both also 0. in. thick. Solution WANT: interior volume of tank, V KNOW: radius, r = 1 in., height, h = 48 in. V dv Vrdr Vhdh, dr 0. dh Volume of tank, V r h V r rh and V h r V V r dr V h dh (rh) dr ( r ) dh 31
Since r = 1 in., h = 48 in., and dr 0. dh we have, V (1)(48)( 0.) (1) ( 0.) 3 814.3in Thus the interior volume of the tank is V ( 1) (48) 814.3 0,900.4 in 3 Eample.11 Suppose that a clindrical can is designed to have a radius of 1 in. and a height of 5 in. but that the radius and height are off b the amounts dr = 0.03 and dh = 0.1. Estimate the resulting absolute, relative and percentage changes in the volume of the can. 3
Solution WANT: Absolute change, V dv Absolute change, dv V r Relative change, V dv V V dv Percentage change, 100 V dr V h dh rhdr r dh (1)(5)(0.03) (1) ( 0.1) 0. Relative change, dv V 0. r h 0. (1) (5) 0.04 Percentage change, dv V 100 0.04 100 4% 33
Eample.1 1. The dimensions of a rectangular block of wood were found to be 100 mm, 10 mm and 00 mm, with a possible error of 5 mm in each measurement. Find approimatel the greatest error in the surface area of the block and the percentage error in the area caused b the errors in the individual measurements.. The pressure P of a confined gas of volume V and temperature T is given b the formula T P k where k is a V constant. Find approimatel, the maimum percentage error in P introduced b an error of 0.4% in measuring the temperature and an error of 0.9% in measuring the volume. 34
Eample.13 The radius and height of a right circular cone are measured with errors of at most 3% and % respectivel. Use differentials to estimate the maimum percentage error in computing the volume. 35
..3 Eact Differential In general, an epression of the form, M (, ) d N(, ) d is known as an eact differential if it is a total differential of a function f(, ). Definition.4 The epression M (, ) d N(, ) d is an eact differential if Md Nd f d f d df Note The function f is found b partial integration. 36
37 Test for Eactness The differential form Nd Md is eact if and onl if N M B similar reasoning, it ma be shown that dz z P d z N d z M ),, ( ),, ( ),, ( is an eact differential when z N P, P z M, M N
Eample see illustration 38
.3 Chain Rule.3.1 Partial Derivatives of Composite Functions Recall: The chain rule for composite functions of one variable If is a differentiable function of and is a differentiable function of a parameter t, then the chain rule states that d dt d d d dt The corresponding rule for two variables is essentiall the same ecept that it involves both variables. Note The rule is used to calculate the rate of increase (positive or negative) of composite functions with respect to t. 1
Assume that z f (, ) is a function of and and suppose that and are in turn functions of a single variable t, (t), (t) Then z f ( ( t), ( t)) is a composition function of a parameter t. dz Thus we can calculate the derivative dt and its relationship to the derivatives z, z d d, and is given b the dt dt following theorem. Theorem.1 If z f (, ) is differentiable and and are differentiable functions of t, then z is a differentiable function of t and dz dt z d dt z d dt
Chain Rule one parameter z f (, ) Dependent variable z z Intermediate variable d dt d dt t Independent variable dz dt z d dt z d dt 3
4 Chain Rule one parameter dt dz z w dt d w dt d w dt dw w z w dt dz dt d w Dependent variable Intermediate variable Independent variable t z dt d ),, ( z f w
Chain Rule two parameters f () d d r s r s r d d r, s d d s 5
6 Theorem. Let ), ( s r and ), ( s r have partial derivatives at r and s and let ), ( f z be differentiable at (, ). Then )), ( ),, ( ( s r s r f z has first derivatives given b r z r z r z s z s z s z
Eample.14 3 Suppose that z where t and dz t. Find. dt Solution WANT: z dz dt z d dt z d dt 3 z 3 and t d dt t d t dt z 3 Hence, dz dt z d dt z d dt (3 )() ( 3 )(t) 6(t) ( t ) (t) (t) 40t 3 4 7
Eample.15 Suppose that z where dz cos and sin. Find when d Solution WANT: dz d From the chain rule with in place of t, dz z d z d d d d we obtain dz d 1 ( ) 1 1 ( ( )( ) 1 sin ) ( 1)(cos ) 8
When, we have cos 0 and sin 1 Substituting = 0, = 1, dz formula for ields dt in the dz d 1 (1)(1)( 1) 1 (1)(1)(0) 1 9
Eample.16 Let z 4 where = uv and = u 3 z z v. Find and. u v 10
Eample.16a Suppose that w z where = sin and z e. Use an appropriate form of dw the chain rule to find. d 10
Eample.17 Find w s rs e, if w 4 z where r s ln and z = rst. t 3 11
.3. Partial Derivatives of Implicit Functions The chain rule can be applied to implicit relationships of the form F (, ) 0. Differentiating F (, ) 0 with respect to gives F d d F d d F F d In other words, 0 d Hence, d d F F In summar, we have the following results. Theorem.3 If F (, ) 0 defines implicitl as a differentiable function of, then d F d F 0 1
Theorem.3 has a natural etension to functions z f (, ), of two variables. Theorem.4 If F (,, z) 0 defines z implicitl as a differentiable function of and, then z F F z z and F F z Eample.18 If is a differentiable function of such that d find. d 3 4 3 0 13
Solution KNOW: Let d d F F 3 F(, ) 4 3 F 3 8 3. Then and F 4 3 d d F F (3 4 8 3) 3 Alternativel, differentiating the given function implicitl ields 3 8 4 d d d d (3 4 3 3 8 3 d d 3) d d which agrees with the result obtained b Theorem.3. 14 0
Eample.19a If sin( ) cos( ) determine d. d 15
Eample.19b If z z 5 determine z z epressions for and. d d 16
.5 Local Etrema Focus of Attention What is the relative etremum of a function of two variables? What does a saddle point mean? What is a critical point of a function of two variables? What derivative tests could be used to determine the nature of critical points? In this section we will see how to use partial derivatives to locate maima and minima of functions of two variables. First we will start out b formall defining local maimum and minimum: Definition.5 A function of two variables has a local maimum at (a, b) if f(, ) f( a, b ) when (, ) is near (a, b). The number f( a, b ) is called a local maimum value. If f(, ) f( a, b ) when (, ) is near ( ab, ), then f( a, b ) is a local minimum value. 1
Note The points (, ) is in some disk with center (a, b). Collectivel, local maimum and minimum are called local etremum. Local etremum is also known as relative etremum. The process for finding the maima and minima points is similar to the one variable process, just set the derivative equal to zero. However, using two variables, one needs to use a sstem of equations. This process is given below in the following theorem: Theorem.5 If f has a local maimum or minimum at (a, b) and the first-order partial derivatives of f eist at this point, then f ( a, b ) 0 and f ( a, b ) 0.
Definition.6 A point (a, b) is called a critical point of the function z f(, ) if f ( a, b ) 0 and f ( a, b ) 0 or if one or both partial derivatives do not eist at (a, b). 3
Relative Ma Point ( a, b, f ( a, b )) is a local maimum Relative Min. Point ( a, b, f ( a, b )) is a local minimum Saddle Point 4
Point ( a, b, f ( a, b )) is a saddle point Remark The values of z at the local maima and local minima of the function z f(, ) ma also be called the etreme values of the function, f(, ). Eample.33 Discuss the nature of the critical point for the following surfaces: i. z ii. z 1 iii. z Prompts/Questions Where can relative etreme values of f(, ) occur? o What are critical points? How do ou decide the nature of critical points? 5
Solution Let f(, ), g(, ) 1 and h(, ). We find the critical points: a) f (, ), f (, ) Thus the critical point is (0, 0). The function f has a local minimum at (0, 0) because and are both nonnegative, ielding 0. b) g(, ), g(, ) Thus the critical point is (0, 0). The function g has a local maimum at (0, 0) because z 1 and and are both nonnegative, so the largest value z occurs at (0, 0). c) h (, ), h (, ) Thus the critical point is (0, 0). The function h has neither a local maimum nor a local minimum at (0, 0). h is minimum on the -ais (where = 0) and a maimum on the -ais (where = 0). Such point is called a saddle point. 6
Note In general, a surface z f(, ) has a saddle point at (a, b) if there are two distinct vertical planes through this point such that the trace of the surface in one of the planes has a local maimum at (a, b) and the trace in the other has a local minimum at (a, b). Eample.0 (c) illustrates the fact that f( a, b ) 0 and f ( a, b ) 0 does not guarantee that there is a local etremum at (a, b). The net theorem gives a criterion for deciding what is happening at a critical point. This theorem is analogous to the Second Derivative Test for functions of one variable. Theorem.11 Second-Partials Test Let f(, ) have a critical point at (a, b) and assume that f has continuous second-order partial derivatives in a disk centered at (a, b). Let D f ( a, b) f ( a, b) [ f ( a, b )] 7
(i) If D > 0 and f( a, b ) 0, then f has a local minimum at (a, b). (ii) If D > 0 and f( a, b ) 0, then f has a local maimum at (a, b). (iii) If D < 0, then f has a saddle point at (a, b). (iv) If D = 0, then no conclusion can be drawn. 8
Remark The epression f f f is called the discriminant or Hessian of f. It is sometimes easier to remember it in the determinant form, f f f If the discriminant is positive at the point (a, b), then the surface curves the same wa in all directions: downwards if f 0, giving rise to a local maimum upwards if f( a, b ) 0, giving a local minimum. If the discriminant is negative at (a, b), then the surface curves up in some directions and down in others, so we have a saddle point. f f f f Illustration Finding relative etrema using first partial derivative using second partial derivative 9
Eample.34 Locate all local etrema and saddle points of f(, ) 1. Solution First determine f and f : f(, ) and f (, ). Secondl, solve the equations, f = 0 and f = 0 for and : 0 and 0 So the onl critical point is at (0, 0). Thirdl, evaluate f, f and f at the critical point. f(, ), f(, ) 0 and f(, ) At the point (0, 0), f (0,0), f (0,0) 0 and f (0,0) Compute D: 10
D 0 0 4 Since D = 4 > 0 and f ( 0, 0) < 0, the second partials test tell us that a local maimum occurs at (0, 0). In other words, the point (0, 0, 1) is a local maimum, with f having a corresponding maimum value of 1. Eample.35 Locate all local etrema and saddle points of 3 3 f (, ) 8 4. Prompts/Questions What are the critical points? o How are the calculated? How do ou classif these points? o Can ou use the Second Derivative Test? Solution f 4 4, f 4 3 Find the critical points, solve 11
4 4 0 (1) 4 3 0 () From Eqn. (1), () to find 4 3( ) 0 If = 0, then = 0 If =, then = 4 0,. Substitute this into Eqn. So the critical points are (0, 0), (, 4). Find f, f and f and compute D: D f(, ) 48, f(, ) 4 and f(, ) 6. f f 48 4 f f 4 6 88 576 Evaluate D at the critical points: At (0, 0), D = 576 < 0, so there is a saddle point at (0, 0). At (, 4), D = 88()(4)-576 = 178 > 0 1
and f (, 4) = 48() = 96 > 0. So there is a local minimum at (, 4). Thus f has a saddle point (0, 0, 0) and local minimum (, 4, 64). Eample.36 Find the local etreme values of the function. (i) (ii) f (, ) 4 3 3 h(, ) Prompts/Questions What are the critical points? Can ou use the second partials test? o What do ou do when the test fails? How does the function behave near the critical points? 13
Solution (i) The partial derivatives of f are 3 f 4. 4 f, Solving f 0 and f 0 simultaneousl, we note that the critical points occurs whenever 0 or 0. That is ever point on the or ais is a critical point. So, the critical points are ( o, ) and (0, ). Using the Second Derivative Test: 4 3 8 D 3 8 1 4 64 6 6 6 40 For an critical point ( 0,0) or (0, 0), the second partials test fails. Let s analse the function. Observed that f(, ) 0 for ever critical point (either 0 or 4 0 or both. Since f(, ) 0 when 0 and 0, it follows that each critical point must be a local minimum. 14
The graph of f is shown below. z Graph of f(, ) 4 15
(ii) h (, ) 3, h (, ) 3. Solving the equations h = 0 and h = 0 simultaneousl, we obtain (0, 0) as the onl critical point. The second partials test fails here. Wh? Let us eamine the traces on the coordinate planes finish it off Graph of 3 3 h(, ) (, ) h has neither kind of local etremum nor a saddle point at (0, 0). 16
Reflection What can ou sa about the partial derivatives of a differentiable function at a local (relative) maimum or minimum?... How do ou find the points for where a local (relative) maimum or minimum might be located? Saddle point?... What is the second derivative test? What do ou do if the second partials test is inconclusive?... 17
.6 Absolute Etrema Focus of Attention Where can absolute etreme values of f(, ) occur? Under what circumstances does a function of two variables have both an absolute maimum and an absolute minimum? What is the procedure for determining absolute etrema? The onl places a function f(, ) can ever have an absolute etremum value are interior critical points boundar points of the function s domain Theorem.1 Etreme-Value Theorem If (, ) f is continuous on a closed bounded region R, then f has an absolute etremum on R. 17
Theorem.13 If f (, ) has an absolute etremum at an interior point of its domain, then this etremum occurs at a critical point. Note Absolute etremum is also known as global etremum. Finding Absolute Etrema Given a function f that is continuous on a closed, bounded region R: Step 1: Find all critical points of f in the interior of R. Step : Find all boundar points at which the absolute etrema can occur (critical points, endpoints, etc.) Step 3: Evaluate (, ) f at the points obtained in the preceding steps. 18
The largest of these values is the absolute maimum and the smallest is the absolute minimum. 19
Illustration Finding absolute etrema on closed and bounded region Critical points & boundar points Absolute etreme values smallest & largest values Eample.37 Find the absolute etrema of the function f (, ) over the disk 1. Prompts/Questions Where can absolute etreme occur? o What are the critical points? o What are the boundar points? How do ou decide there is an absolute minimum? Absolute maimum? Solution Step 1: f, f 0
f 0 and f 0 for all (, ). But f and f do not eist at (0, 0). Thus (0, 0) is the onl critical point of f and it is inside the region. Step : Eamine the values of f on the boundar curve 1. Because 1 on the boundar curve, we find that f (, ) That is, for ever point on the boundar circle, the value of f is 1. Step 3: Evaluating the value of f at each of the points we have found: (1 Critical point: f(0, 0) = 0 Boundar points: f (, ) 1 We conclude that the absolute minimum value of f on R is 0 and the absolute maimum value is 1. ) 1 1
Eample.38 Find the absolute etrema of the function f (, ) 3 6 3 7 on the closed triangular region in the first quadrant bounded b the lines 0, 0, 5 5. 3 Prompts/Questions Where can absolute etreme occur? o Can ou find the points? How do ou determine the absolute maimum? Absolute minimum?
Solution The region is shown in the figure. B(0, 5) R (0, 0) A(3, 0) Critical points: f 3 6 0, 3 3 0 f (1, ) is the onl critical point in the interior of R. Boundar points: The boundar of R consists of three line segments. We take one side at a time. 3
Graph of f (, ) 3 6 3 7 4
On the segment OA, 0. The function f (, ) simplifies to a function of single variable u ( ) f (, 0) 6 7, 0 3 This function has no critical numbers because u ( ) 6 is nonzero for all. Thus the etreme values occur at the endpoints (0, 0) and (3, 0) of R. On the segment OB, 0. v( ) f (0, ) 3 7, 0 5 This function has no critical numbers because v( ) 3 is nonzero for all. Thus the etreme values occur at the endpoints (0, 0) and (0, 5) of R. Segment AB: we alread accounted the endpoints of AB, so we look at the interior points of AB.
5 With 5, we have 3 5 5 w ( ) 3 5 6 3 5 3 3 7 5 14 8, 0 3 Setting w ( ) 10 14 = 0 gives 7 5. The critical number is ( 7 5, 8 3). Evaluating the value of f for the points we have found: (0, 0) f(0, 0) = 7 (3, 0) f(3, 0) = 11 (0, 5) f(0, 5) = 8 ( 7 5, 8 3) f (7 5, 8 3) = 9/5 (1, ) f(1, ) = 1 We conclude that the absolute maimum value of f is f(0, 0) = 7 and the absolute minimum value is f(3, 0) = 11. 3
Eample.39 Find the shortest distance from the point (0, 3, 4) to the plane z 5. Solution KNOW: the distance from a point (,, z) to (0, 3, 4) is d ( 0) ( 3) ( z 4) WANT: to minimise d Let (,, z) be a point on the plane z 5. We know z 5 So d ( 3) (5 4) d Instead of d, we can minimize the epression f (, ) ( 3) (1 ) 4
Find the critical values: f f (1 ) 4 4 ( 3) 4(1 ) 4 10 10 The onl critical point is (5/6, 4/3). Also f 4, f 10, f 4, so D > 0 which means there is a local minimum at (5/6, 4/3). This local minimum must also be the absolute minimum because there must be onl one point on the plane that is closest to the given point. The shortest distance is, d Note 5 6 4 3 3 1 5 6 4 3 In general it can be difficult to show that a local etremum is also an absolute etremum. In practice, the determination is made using phsical or geometrical considerations. 0 5 6 5 0
Eample.40 Suppose we wish to construct a rectangular bo with volume 3 ft 3. Three different materials will be used in the construction. The material for the sides cost RM1 per square foot, the material for the bottom costs RM3 per square foot, and the material for the top costs RM5 per square foot. What are the dimensions of the least epensive such bo? Reflection Where do absolute etreme values of f(, ) occur? What are the conditions that guarantee a f(, ) has an absolute maimum and an absolute minimum? 6
How do ou find the absolute maimum or minimum value of a function on a closed and bounded domain? On an open or unbounded region? 7