LESSON 23: EXTREMA OF FUNCTIONS OF 2 VARIABLES OCTOBER 25, 2017
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1 LESSON : EXTREMA OF FUNCTIONS OF VARIABLES OCTOBER 5, 017 Just like with functions of a single variable, we want to find the minima (plural of minimum) and maxima (plural of maximum) of functions of several variables. We call the minima and maxima the extrema (plural of extremum). Definition 1. A local (or relative) minimum/maximum point is the point (x, y) that makes the function the smallest/largest in some area. A global (or absolute) minimum/maximum point is the point (x, y) that make the function the smallest/largest on the whole graph. Note. The extrema, i.e. the minima and maxima, of a function are the function values, e.g., if (, ) is a minima point of the function f(x, y), then the minima of f is the value f(, ). Ex 1. Figure 1. Functions of a Single Variable A difference between functions of a single variable and functions of several variables is that functions of several variables have what are called saddle points. Figure. Functions of a Several Variables Definition. The points (x 0, y 0 ) such that f x (x 0, y 0 ) = 0 = f y (x 0, y 0 ) are called critical points. 1
2 MATH 1600 Definition 4. We call the function is called the discriminant. MEMORIZE this formula. D(x, y) = f xx (x, y)f yy (x, y) (f xy (x, y)) Second Derivative Test: Suppose (x 0, y 0 ) is a critical point of f. If (1) D(x 0, y 0 ) > 0 and f xx (x 0, y 0 ) < 0, (x 0, y 0 ) is a local maximum point () D(x 0, y 0 ) > 0 and f xx (x 0, y 0 ) > 0, (x 0, y 0 ) is a local minimum point () D(x 0, y 0 ) < 0, then (x 0, y 0 ) is a saddle point (4) D(x 0, y 0 ) = 0, the test is inconclusive (i.e., this test doesn t give you any information) MEMORIZE this test. Examples. 1. Find and classify the critical points of f(x, y) = x + y y x. Solution: We apply the following steps. Critical points are points (x 0, y 0 ) that make both f x and f y equal to 0. Write Hence, and f x (x, y) = x 1 = (x 1)(x + 1) f y (x, y) = y 1 = (y 1)(y + 1) if f x = 0 then (x 1)(x + 1) = 0 x = ±1 iff y = 0 then (y 1)(y + 1) = 0 y = ±1. So, setting both f x and f y equal to zero, our critical points are (x 0, y 0 ) = (1, 1), (1, 1), ( 1, 1), ( 1, 1). Step : Find second derivatives Write Step : Find discriminant f xx = x, f yy = y, and f xy = 0. Our formula for the discriminant is D = f xx f yy (f xy ).
3 MATH 1600 So, D(x, y) = (x) (y) ( }{{}}{{}}{{} 0 ) = 4xy. f xx f yy f xy We go through each critical point and apply the test. Critical Point D(x 0, y 0 ) f xx (x 0, y 0 ) Classification (1, 1) 4(1)(1) = 4 > 0 (1) = > 0 local min (1, 1) 4(1)( 1) = 4 < 0 saddle point ( 1, 1) 4( 1)(1) = 4 < 0 saddle point ( 1, 1) 4( 1)( 1) = 4 > 0 ( 1) < 0 local max. Find and classify the critical points of g(u, v) = u v uv v. Solution: Again, we go through our steps We have g u = uv v = v(u 1) g v = u u v Recall that our critical points are the (u 0, v 0 ) that make both g u and g v equal to 0. We write 0 = g u = v(u 1). Here, we have a choice: either v = 0 or u = 1. Moreover, we have 0 = g v = u u v v = u u. At this point, it is not clear what our points (u 0, v 0 ) should be. This is where we break it into cases: Case 1. v = 0 If v = 0, then v = u u becomes 0 = u u = u(u 1). So u = 0, 1. This means that two of our critical points are (0 u, 0 v ) and (1 u, 0 ). v (Because we were given g(u, v), the order is going to be (u, v).)
4 4 MATH 1600 Case. u = 1 If u = 1, then v = u u becomes which implies v = ( ) 1 1 = = 1 4, v = 1 8. Hence, our last critical point is, 1. 8 Putting this all together, our critical points are (0, 0), (1, 0), and, 1. 8 Step : Find second derivatives We have g uu = v, g vv =, and g uv = u 1. Step : Find discriminant The formula for the discriminant is given by which becomes D = g uu g vv (g uv ) D(u, v) = (v) ( ) (u }{{}}{{}}{{ 1 } ) = 4v (u 1). g uu g vv g uv Write D, 1 8 g uu, 1 8 ( = 4 1 ) ( 8 ( = 1 ) = ( ) ) 1 1 = 1 D(0, 0) = 4(0) ((0) 1) = 1 D(0, 1) = 4(1) ((1) 1) = 4 1 = 5
5 MATH So we get Critical Point D(u 0, v 0 ) g xx (u 0, v 0 ) Classification, > 0 1 < 0 4 local max (0, 0) 1 < 0 saddle point (0, 1) 5 < 0 saddle point. Find the local minima and maxima of f(x, y) = x + y x + y. Solution: Observe that this question is different than the previous examples. Previously, we were asked to classify the critical points, but now we are asked to find the actual function values at the critical points. Our process, fortunately, doesn t change too much. We still need to find and classify the critical points but then we need to plug them back into f(x, y) to determine the function value. We have f x = x and f y = y +. Thus, 0 = f x = x x = 1 and 0 = f y = y + y = 1. This means our critical point is (1, 1). Step : Find second derivatives Step : Find discriminant f xx =, f yy =, f xy = 0 D(x, y) = f xx f yy (f xy ) = ()() (0) = 4 Since D(1, 1) = 4 > 0 and f xx (1, 1) = > 0, we have a local min at (1, 1) and no local max. Step 5: Determine function values By the previous steps, we know that f(1, 1) = (1) + ( 1) (1) + ( 1) = = is a local minimum of f(x, y). 4. Find the local minima and maxima of g(x, y) = x + xy y.
6 6 MATH 1600 Solution: We go through our steps and then plug our points into the function. g x = x + y and g y = x y So 0 = g x = x + y x = y and 0 = g y = x y x = y. Since we have that y = x, we can substitute this into y = x which means (x) = x 4x = x x = 0 or x = 4. Then we break this down into cases. Case 1. x = 0 If x = 0, then y = (0) = 0. Hence, one critical point is (0, 0). Case. x = 4 If x = 4, then y = (4) = 8. Thus, another critical point is (4, 8). Putting this together, our critical points are (0, 0) and (4, 8). Step : Find second derivatives g xx = x, g yy = 1, and g xy = Step : Find discriminant D(x, y) = g xx g yy (g xy ) = ( x)( 1) () = x 4 We write D(0, 0) = (0) 4 = 4 D(4, 8) = (4) 4 = 8 4 = 4 g xx (4, 8) = (4) = 8 Hence, Critical Point D(x 0, y 0 ) g xx (x 0, y 0 ) Classification (0, 0) 4 < 0 saddle point (4, 8) 4 > 0 8 < 0 local max
7 MATH Step 5: We only have one local maximum which is g(4, 8) = (4) = 64 = + (4)(8) (8)
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