MATH1211/10-11(1)/Tu6sol/Department of Mathematics, The University of Hong KongMATH1211 Multivariable Calculus (2010–2011, First semester) Brief Solution to Tutorial 61.(a)False. Counter-example: the constant functionf(x, y) =C. At (0,0) (or any point inR2)fhas both a local (in fact global) minimum and a local (in fact global) maximum.(b)True. Supposeais a saddle point off. Then in any neighborhood ofathere are pointsbandcsuch thatf(b)< f(a)< f(c). Hencefcould not have a local minimum/maximum ata.(c)False. Counter-example:f(x, y) = 2x2+2y2, g(x, y) =-x2-y2. Thenfhas a local minimumat (0,0) andghas a local maximum at (0,0). But (0,0) is not a saddle point off+g=x2+y2.2.To solve for critical points:(0 =fx= 6x2-6y0 =fy=-6x+ 6y=⇒(x2=yx=y=⇒(x, y) = (0,0) or (1,1).Also,fxx= 12x, fxy(=fyx) =-6, fyy= 6.At critical point (0,0),fxx= 0 andfxxfyy-f2xy=-36<0. Hence (0,0) is a saddle point.At critical point (1,1),fxx>0 andfxxfyy-f2xy= 36>0; hencefhas a local minimum at (1,1).3.Writeg(x, y, z) =x+y-zso that the constraint becomesg(x, y, z) = 1.Critical points areobtained by solving∇f=λ∇gandg(x, y, z) = 1:2x=λ2y=λ2z=-λx+y-z=1.The first three equations give
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