14.7 Maximum and Minimum Values

Size: px

Start display at page:

Download "14.7 Maximum and Minimum Values"

Shavonne Pierce
5 years ago
Views:

1 CHAPTER 14. PARTIAL DERIVATIVES Maximum and Minimum Values Definition. Let f(x, y) be a function. f has a local max at (a, b) iff(a, b) (a, b). f(x, y) for all (x, y) near f has a local min at (a, b) iff(a, b) apple f(x, y) for all (x, y) near (a, b). f has an absolute max at (a, b) iff(a, b) f(x, y) for all (x, y) in the domain. f has an absolute min at (a, b) iff(a, b) apple f(x, y) for all (x, y) in the domain. The number f(a, b) is called the maximum value or minimum value. Theorem. Let f have a local maximum or minimum at (a, b). Then each partial derivative, f x (a, b) and f y (a, b), either equals 0 or does not exist. In particular, if the partial derivatives exist, then max means f(a, b) f(x, y), min means f(a, b) apple f(x, y). Local means (x, y) nearby, absolute means all (x, y) in domain. Local max/min implies rf = 0 (or DNE). rf(a, b) =0. Proof. We can reduce this statement to Calculus I by taking the traces. For instance, the y = b trace gives a function of just one variable, f(x, b). This function has a local max or min at x = a, and so f x (x, b) equals 0 or does not exist at x = a. Definition. We say ha, bi is a critical point of f(x, y) ifrf(a, b) =0 or at least one of the partial derivatives is undefined. Example 1. Find the critical points of f(x, y) =x 2 + y 2 them as local max/min/neither. 4x +8y + 10 and identify f x =2x 4=0 2x 4=0 x =2 f y =2y +8 2y +8=0 y = 4 ha, bi = h2, We will learn a version of the 2nd derivative test below to figure out if this point is a max/min/neither. But in this case we can figure it out with some algebra. Specifically, we ll complete the square: f(x, y) =(x 2 4x + 4) + (y 2 +8y + 16) f(x, y) = 4i =(x 2) 2 +(y + 4) positive numbers 10 for all x, y f(2, 4) = 10 is abs/local min

2 CHAPTER 14. PARTIAL DERIVATIVES 116 Definition (2nd Derivative Test). Suppose (a, b) is a critical point of f with f x (a, b) =f y (a, b) = 0 and suppose all the 2nd partial derivatives are continuous at and near (a, b). Let D = D(a, b) =f xx (a, b) f yy (a, b) 2 f xy (a, b) Let D = f xx f yy (f xy ) 2. (1) D>0, f xx < 0, min, (2) D>0, f xx > 0, max, (3) D<0, neither, (4) D = 0, try something else. 1. If D>0 and f xx (a, b) > 0thenf(a, b) is a local min. 2. If D>0 and f xx (a, b) < 0thenf(a, b) is a local max. 3. If D < 0thenf(a, b) is neither a max nor min and (a, b) isa saddle point. 4. If D = 0, then the 2nd derivative test is gives you no information. 2 Comments. 1. We can write D in other ways. Note that f xy (a, b) = fxy (a, b) f yx (ab) = 2 f yx (a, b) so we can use either fxy or f yx or both to calculate D. 2. In conditions (1) and (2), instead of f xx (a, b) > 0 or f xx (a, b) < 0, we can use f yy (a, b) > 0 or f yy (a, b) < 0 without changing any other details of the test. In e ect, the condition D >0 means that the information we get from f xx and from f yy will be the same (i.e. if D>0thenf xx < 0 () f yy < 0 and f xx > 0 () f yy > 0). 3. We can remember D using matrices: it is the determinant of the matrix of second partial derivatives: D = f xx(a, b) f xy (a, b) f yx (a, b) f yy (a, b) Furthermore, the matrix of all second partial derivatives is called the Hessian matrix. Below in a few examples we will occasionally use the notation r 2 f(x, y) for the matrix of second order partial derivatives. Example 2. Apply the second derivative test to to identify the local/max/min/neither of f(x, y) =x 2 + y 2 4x +8y f x =2x 4 f y =2y +8 (2, 4) = critical pt f xx =2 f yy =2 f yx = f xy =0 This is where we ended on Thursday, February 28 D = =4 D>0 and f xx (2, 4) = 2 > 0so(2, 4) is a local min. Example 3. Find the critical points and identify them as local max/min/neither for

3 CHAPTER 14. PARTIAL DERIVATIVES 117 the function f(x, y) =x 4 y 4. f x =4x 3 f y = 4y 3 (0, 0) = critical pt apple r 2 12x 2 0 f = 0 12y 2 D(0, 0) = =0. Since D = 0, the Second Derivative Test doesn t help. To finish, we look at traces y = 0 trace: f(x, 0) = x 4 x = 0 trace: f(0,y)= y 4 x = 0 is local min y = 0 is local max (0, 0) is saddle point Absolute Maximum and Minimum Comments. Recall the Extreme Value Theorem of Calculus I, also known as the Absolute Max/Min Test. It said two things. (1) Every continuous function defined on an interval [a, b] always has an absolute maximum and an absolute minimum. (2) To find the absolute max/min you (i) find the critical points, (ii) evaluate the function at each critical point and each endpoint, (iii) identify the largest/smallest values calculated in step (ii) as the absolute max/mins. Theorem (Extreme Value Theorem). Let f be a continuous function defined on a closed and bounded domain D R 2. Then f has an absolute maximum an absolute minimum on D. Tofindthemdothe following: 1. Find the critical points of f in D, 2. Evaluate f at each critical point, 3. Find the absolute maximum and minimum values of f on the boundary of D (usually using Calculus I methods, or Lagrange Multipliers in the next section), 4. Identify the largest/smallest values calculated in steps (2) and (3) as the absolute max/mins. Given f contin, D closed bounded. (1) Find crit. pts in D, (2) Evaluate f, (3) Find max/mins on boundary of D, (4) Compare values. Example 4. Let f(x, y) =xy 2, D = {x 0, y 0, x 2 + y 2 apple 3}. Find the absolute maximum and minimum values of f on D. Let s start by identifying D a little better. It is part of a circle with radius p 3, in Quadrant I:

4 CHAPTER 14. PARTIAL DERIVATIVES 118 This domain is bounded, because it does not extend infinitely far in any direction, and it is closed, because it includes its boundary (since the definition had x 0 not just x>0, and similarly y 0 not just y>0 and x 2 + y 2 apple 3 and not just x 2 + y 2 < 3). Now we find the critical points: f x = y 2 y 2 =0 y =0 f y =2xy 2xy =0 x = 0 or y =0 In other words the critical points are all along the x-axis. Since this is also the boundary of the domain D, we ll find the max/min values of f on these critical points in steps (2) and (3). We start by breaking the boundary of D down into three pieces: boundary of D = D 1 [ D 2 [ D 3 D 1 = x-axis from 0 to p 3 D 2 = y-axis from 0 to p 3 D 3 = Quadrant I part of circle x 2 + y 2 =3 Now we figure out the absolute max/min of f on each of these sets: f on D 1 : f(x, 0) = 0 f on D 2 : f(0,y)=0 f on D 3 : f(x, y) =? x 2 + y 2 =3 y = p 3 x 2 f(x, y) =f(x, p 3 x 2 ) = x(3 x 2 ) Now we use Calculus I with the function g(x) =x(3 x 2 ) on the domain [0, p 3]. g(x) =3x x 3, [0, p 3] g 0 (x) =3 3x 2 3 3x 2 =0 x = ±1

5 CHAPTER 14. PARTIAL DERIVATIVES 119 x =1, since x = g(0) = 0 g(1) = 3(1) 1 3 =2 g( p 3) = 3 p 3 ( p 3) 3 = p 3 3 ( p 3) 2 =0 1 is not in the domain So we see that g has absolute min at x = 0 or x = p 3, with value 0, and absolute max at x = 1 with value 2. Translating this back to f we get: y = p 3 x 2 x =0: f(x, y) = 0 everywhere on D 1 x =1: y = p y = p 2 x = p q 3: y = 3 ( p 3) 2 y =0 Abs Max f(1, p 2) = 2 f(x, y) = 0 everywhere on D 2 Abs Min f(x, y) = 0 everywhere on D 1 [ D 2 This is where we ended on Friday, March 1 Extra Example This extra example is very similar to some homework problems, and should be studied like the preceeding examples. Example 5. [Stewart 6e, #32] Find the absolute max and min of the following function on the indicated domain: We find the critical points where f(x, y) =4x +6y x 2 y 2 D = {(x, y) 0 apple x apple 4, 0 apple y apple 5} h 2x +4, 2y +6i = h0, 0i rf = h 2x +4, 2y +6i hx, yi = h2, 3i f(2, 3) = 13 Now we break the boundary of the domain into 4 pieces: Boundary of D = D 1 [ D 2 [ D 3 [ D 4 D 1 : y =0, x is in [0, 4]

6 CHAPTER 14. PARTIAL DERIVATIVES 120 D 2 : x =4, y is in [0, 5] D 3 : y =5, x is in [0, 4] D 4 : x =0, y is in [0, 5] Now we have a series of Calculus I problems: D 1 : find max/min of g 1 (x) =f(x, 0) = 4x x 2, on [0, 4] D 2 : find max/min of g 2 (y) =f(4,y) = y 16 y 2 =6y y 2 on [0, 5] D 3 : find max/min of g 3 (x) =f(x, 5) = 4x x on [0, 4] D 4 : find max/min of g 4 (x) =f(0,y)=6y y 2 on [0, 5] Calc I on D 1 : g 1 (x) =4x x 2, on [0, 4] g1(x) 0 = 2x +4 2x +4=0) x =2 g 1 (0) = 0 g 1 (2) = 4 g 1 (4) = 0 Calc I on D 2 : g 2 (y) =6y y 2 on [0, 5] g2(y) 0 = 2y +6 2y +6=0) y =3 g 2 (0) = 0 g 2 (3) = 18 9=9 g 2 (5) = = 5 Calc I on D 3 g 3 (x) =4x x on [0, 4] g3(x) 0 = 2x +4 2x +4=0) x =2 g 3 (0) = 5 g 3 (2) = 8 4+5=9 g 3 (4) = = 5 Calc I on D 4 : g 4 (x) =6y y 2 on [0, 5] g4(x) 0 = 2y +6 2y +6=0) y =3 g 4 (0) = 0 g 4 (3) = 18 9=9

7 CHAPTER 14. PARTIAL DERIVATIVES 121 g 4 (5) = = 5 After solving the above Calc I problems, and comparing values, you should get the following: f(x, y) : f(2, 3) = 13 g 1 (x) : g 1 (0) = 0, g 1 (2) = 4 g 1 (4) = 0 g 2 (y) : g 2 (0) = 0, g 2 (3) = 9, g 2 (5) = 5 g 3 (x) : g 3 (0) = 5, g 3 (2) = 9, g 3 (4) = 5 g 4 (y) : g 4 (0) = 0, g 4 (3) = 9, g 4 (5) = 5 From the above numbers, we simply want to identify the largest and smallest: Extra Optional Material abs max at (2, 3), value = 13 abs min at (4, 0) and (0, 0), value = 0 Everything in this subsection is kind of interesting, but does not need to be studied, and is not included on the homework or tests. Example 6. [The Ackley Function] The Ackley function was invented as a good test case for numerical algorithms that are designed to find maximums and minimums. We give a simplified version of it here (it can be generalized to higher dimensions, and various coe cients can be tweaked to change the relative size of the mins and maxs). f(x, y) = 10e 5p x 2 +y 2 e cos(2 x)+cos(2 y). Use a computer to find approximate values of the critical points, and the max and mins, on the domain 3 apple x apple 3 and 3 apple y apple 3, and identify them as local/global max/min. Comments. The second derivative test given above is the analogue in two variables of of the second derivative test from Calculus I. There is also a first derivative test analogue. Here s one way to state it. Theorem. Suppose f is continuous at (a, b) and di erentiable near (a, b). 1. If there exists an open disk around (a, b) such that for all (x, y) in that disk we have rf(x, y) hx a, y bi > 0then(a, b) is a local minimum. 2. If there exists an open disk around (a, b) such that for all (x, y) in that disk we have rf(x, y) hx a, y bi < 0then(a, b) is a local maximum. 3. If for all open disks around (a, b) we have some (x, y) inthediskwithrf(x, y) hx a, y bi < 0 and some (x, y) inthediskwithrf(x, y) hx a, y bi > 0, then (a, b) is neither a local max nor local min. (See by Michael W. Botsko in The American Mathematical Monthly, vol. 93, 1986, no. 7, pages ) Comments. 1. The conditions above are more easily understood as directional derivatives. The expression rf(x, y) hx a, y bi > 0 is equivalent to Duf(a + hu 1,b+ hu 2 ) > 0 where x = a + hu 1 and y = b + hu 2. Thus, if we ignore all the complicated parts of the statements about open disks, the previous theorem

8 CHAPTER 14. PARTIAL DERIVATIVES 122 basically says: if all the directional derivatives right around (a, b) are positive, then it s a local min, if all the directional derivatives right around (a, b) are negative, then it s a local max, and if some are positive and some are negative then it s neither. 2. In Calculus practice problems it s rare to find a problem where we need the full strength of part (3). In almost textbook problems instead of needing to look at all directional derivatives to find some that are positive and some that are negative. Really, we almost always get away with just needing to look at the directional derivatives just with respect to the directions h1, 0i and h0, 1i. This means we end up showing that the x-trace has a local max/min, and the y-trace has a local max/min, but one is a max and the other is a min. This is basically what happens in the example f(x, y) =x 4 y 4 above (which we repeat below using the First Derivative Test). Example 7. The article by Botsko mentioned above has this example: Use the first derivative test to show that the critical point (1, 2) of f(x, y) =x 2 +2xy +3y 2 +2x + 10y +9 We take the dot product approach from the previous theorem rf(x, y) hx 1, y+2i = h2x +2y +2, 2x +6y + 10i hx 1, y+2i =(2x +2y + 2)(x 1) + (2x +6y + 10)(y + 2). We can see now why the first derivative approach may not be easy to sort out: how can we hope to show that the above formula is > 0 or < 0? Well, remember it s only meant to be positive or negative for (x, y) su ciently close to (1, 2). Su ciently close means that both x 1 and y + 2 should be close to 0. Since x 1 and y + 2 are the crucial quantities, it would be easier to write the whole formula in terms of them. In other words, change variables. Using these variables we have let w = x 1 let v = y +2 then 2x +2y +2=2w +2v and 2x +6y + 10 = 2w +6v. (2x +2y + 2)(x 1) + (2x +6y + 10)(y + 2) = (2w +2v)w +(2w +6v)v = 2(w 2 + wv + wv +3v 2 ) =2 (w + v) 2 +2v 2 > 0 for all w and v Therefore, (1, 2) is a local minimum. The reader may see that we were a bit lucky here: the formula simplified rather well with the substitutions we made. But it s not so easy in general to factor a quadratic in two variables. Example 8. Apply the first derivative test to identify the behavior of f(x, y) =x 2 + y 2 4x +8y + 10 at the point (2, 4). rf(x, y) hx 2, y+4i = h2x 4, 2y +8i hx 2, y+4i

9 CHAPTER 14. PARTIAL DERIVATIVES 123 Therefore (2, 4) is a local minimum. =(2x 4)(x 2) + (2y + 8)(y + 4) = 2(x 2)(x 2) + 2(y + 4)(y + 4) = 2(x 2) 2 + 2(y + 4) 2 > 0 for all x and y. Example 9. Apply the first derivative test to the function and the critical point (0, 0). f(x, y) =x 4 y 4 rf(x, y) hx 0, y 0i = 4x 3, 4y 3 hx, yi =4x 4 4y 4 For any open disk around (0, 0) we can find an (x, y) value in the disk that makes this positive (for example x in the disk and y = 0) and we can find another (x, y) value that makes this negative (for example x = 0 and y in the disk). Therefore, this is a saddle point. Example 10. Apply the first derivative test to the function and the critical point (0, 0). Therefore (0, 0) is a local maximum. f(x, y) = x 4 y 4 rf(x, y) hx 0, y 0i = 4x 3, 4y 3 hx, yi = 4x 4 4y 4 < 0 for all x and y. Example 11. From the article by Botsko listed above we have more examples: (a) Apply the 1st derivative test to f(x, y) =1 x 2/3 y 4/5 at the critical point (0, 0). Show also that the second derivative test fails. (b) Apply the 1st derivative test to f(x, y) = 25 + (x y) 4 +(x 1) 4 at the critical point (1, 1). Show also that the second derivative test fails. (a) rf(x, y) hx 0, y 0i = 2 3 x 1/3, = 2 3 x2/3 4 5 y4/5 < 0 for all x and y. Therefore (0, 0) is a local max. Note that 2 2 r 2 6 f(x, y) = 49 x 4/ x 6/5 4 5 y 1/5 hx, yi

10 CHAPTER 14. PARTIAL DERIVATIVES 124 Since r 2 f(0, 0) is undefined, we cannot apply the second derivative test. (b) rf(x, y) hx 0, y 0i = 4(x y) 3 + 4(x 1) 3, 4(x y) 3 hx 1, y 1i = 4(x y) 3 + 4(x 1) 3 (x 1) 4(x y) 3 (y 1). Factor out the 4 (and remove it entirely), and then apply the substitution w = x 1 v = y 1 and note that x y = w v (x y) 3 +(x 1) 3 (x 1) (x y) 3 (y 1) = (w v) 3 + w 3 w (w v) 3 v =(w v) 3 w + w 4 (w v) 3 v =(w v) 3 w v + w 4 =(w v) 4 + w 4 > 0 for all w and v. Therefore (1, 1) is a local min. Note that apple r 2 12(x y) (x 1) 2 12(x y) 2 f = 12(x y) 2 12(x y) 2 apple r f(1, 1) = 0 0 Therefore D = 0 and we can t apply the second derivative test.

11.7 Maximum and Minimum Values

Arkansas Tech University MATH 2934: Calculus III Dr. Marcel B Finan 11.7 Maximum and Minimum Values Just like functions of a single variable, functions of several variables can have local and global extrema,