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2. (a) Let f be defined for all (x, y) by fƒ (x, y) = x'+y' -3xy. Show that (0, 0) and (1, 1) are the only stationary points, and compute the quadratic form in (9) for f at the stationary points. (b) Check the definiteness of the quadratic form at the stationary points.

2. (a) Let f be defined for all (x, y) by fƒ (x, y) = x'+y' -3xy. Show that (0, 0) and (1, 1) are the only stationary points, and compute the quadratic form in (9) for f at the stationary points. (b) Check the definiteness of the quadratic form at the stationary points.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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SECTION 3.3/ EQUALITY CONSTRAINTS: THE LAGRANGE PROBLEM 115 ROB MS FOR SECTION 3 2 SM 1. The function f(x1, x2, x3) = x} + x3 + 3x3 Extx + Ex1x7+ 7x]x – defined on R has ohly one stationary point. Show that it is a local minimum point. 2. (a) Let f be defined for all (x, y) by ƒ (x, y) = x+y³-3xy. Show that (0, 0) and (1, 1) are the only stationary points, and compute the quadratic form in (9) for ƒ at the stationary points. (b) Check the definiteness of the quadratic form at the stationary points. (c) Classify the stationary points by using (1)(3). SM 3. Classify the stationary points of (a) f(x, y, z) = x² +x?y + y²z+z? – 4z (b) f(x1, x2, x3, X4) = 20x2 + 48x3 + 6x4 + 8x,.x2 – 4x} – 12x3 – x; – 4x3 HARDER PROBLEMS 4. Suppose f(x, y) has only one stationary point (x*, y*) which is a local minimum point. Is (x*, y*) necessarily a global minimum point? It may be surprising that the answer is no. Prove this by examining the function defined for all (x, y) by f (x, y) = (1 +y)³x²+y². (Hint: Look at f(x, –2) as x → .) 3.3 Equality Constraints: The Lagrange Problem A general optimization problem with equality constraints is of the form max (min) f (x1, ..., Xn) subject to 1q = ("x'· · 'Ix)18 (u > w) (1) "q = ("x' · 'x) 48 where the b; are constants. We assume that m < n because otherwise there are usually no degrees of freedom. In vector formulation, the problem is 12:5 4/18
Transcribed Image Text:SECTION 3.3/ EQUALITY CONSTRAINTS: THE LAGRANGE PROBLEM 115 ROB MS FOR SECTION 3 2 SM 1. The function f(x1, x2, x3) = x} + x3 + 3x3 Extx + Ex1x7+ 7x]x – defined on R has ohly one stationary point. Show that it is a local minimum point. 2. (a) Let f be defined for all (x, y) by ƒ (x, y) = x+y³-3xy. Show that (0, 0) and (1, 1) are the only stationary points, and compute the quadratic form in (9) for ƒ at the stationary points. (b) Check the definiteness of the quadratic form at the stationary points. (c) Classify the stationary points by using (1)(3). SM 3. Classify the stationary points of (a) f(x, y, z) = x² +x?y + y²z+z? – 4z (b) f(x1, x2, x3, X4) = 20x2 + 48x3 + 6x4 + 8x,.x2 – 4x} – 12x3 – x; – 4x3 HARDER PROBLEMS 4. Suppose f(x, y) has only one stationary point (x*, y*) which is a local minimum point. Is (x*, y*) necessarily a global minimum point? It may be surprising that the answer is no. Prove this by examining the function defined for all (x, y) by f (x, y) = (1 +y)³x²+y². (Hint: Look at f(x, –2) as x → .) 3.3 Equality Constraints: The Lagrange Problem A general optimization problem with equality constraints is of the form max (min) f (x1, ..., Xn) subject to 1q = ("x'· · 'Ix)18 (u > w) (1) "q = ("x' · 'x) 48 where the b; are constants. We assume that m < n because otherwise there are usually no degrees of freedom. In vector formulation, the problem is 12:5 4/18
oy%20Knut%20Sydsaeter,Peter%20Hammond,Atle%20Seierstad, Arne%20Strom%20(z-lib.org).pdf We next study conditions that allow the stationary points of a function of n variables to be classified as local maximum points, local minimum points, and saddle points. First recall the second-order conditions for local extreme points for functions of two variables. If f (x, y) is a C2 function with (x*, y*) as an interior stationary point, then 12) | SH(**, y*) f**, y*)| | *, y°) f*, y" | fi(*, y") > 0 & >0 = local min. at (x*, y*) (1) | K**, y") f(**, y") | |f(**, y) f*, y*) | | S(**, y") fi(**, y*) | |**, y*) f**, y") | >0 = local max. at (x*, y*) (2) <0 = (x*, y*) is a saddle point (3) In order to generalize these results to functions of n variables, we need to consider the n leading principal minors of the Hessian matrix f"(x) = (f#(x))nxn: (x)f (x) f (x) (x) S (x) f (x) k = 1,..., n = (x)a (t) (x) S THEOREM 3 2.1 (LOCAL SECOND-ORDER CONDITIONS). (x)l (x) Suppose that f (x) = ƒ(x1, .., Xn) is defined on a set S in R" and that x* is an interior stationary point. Assume also that f is C² in an open ball around x*. Let Dr(x) be defined by (4). Then: (a) Dr(x*) > 0, k = 1,...,n x* is a local minimum point. (b) (-1)* DE(x*) > 0, k = 1,..., n x* is a local maximum point. (c) Dn(x*) #0 and neither (a) nor (b) is satisfied x* is a saddle point. Show 12:51 PM 4/18/202
Transcribed Image Text:oy%20Knut%20Sydsaeter,Peter%20Hammond,Atle%20Seierstad, Arne%20Strom%20(z-lib.org).pdf We next study conditions that allow the stationary points of a function of n variables to be classified as local maximum points, local minimum points, and saddle points. First recall the second-order conditions for local extreme points for functions of two variables. If f (x, y) is a C2 function with (x*, y*) as an interior stationary point, then 12) | SH(**, y*) f**, y*)| | *, y°) f*, y" | fi(*, y") > 0 & >0 = local min. at (x*, y*) (1) | K**, y") f(**, y") | |f(**, y) f*, y*) | | S(**, y") fi(**, y*) | |**, y*) f**, y") | >0 = local max. at (x*, y*) (2) <0 = (x*, y*) is a saddle point (3) In order to generalize these results to functions of n variables, we need to consider the n leading principal minors of the Hessian matrix f"(x) = (f#(x))nxn: (x)f (x) f (x) (x) S (x) f (x) k = 1,..., n = (x)a (t) (x) S THEOREM 3 2.1 (LOCAL SECOND-ORDER CONDITIONS). (x)l (x) Suppose that f (x) = ƒ(x1, .., Xn) is defined on a set S in R" and that x* is an interior stationary point. Assume also that f is C² in an open ball around x*. Let Dr(x) be defined by (4). Then: (a) Dr(x*) > 0, k = 1,...,n x* is a local minimum point. (b) (-1)* DE(x*) > 0, k = 1,..., n x* is a local maximum point. (c) Dn(x*) #0 and neither (a) nor (b) is satisfied x* is a saddle point. Show 12:51 PM 4/18/202
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