Question

2. Let f(x,y) = 2x2 - 6xy + 3y2 be a function defined on xy-plane (a) Find first and second partial derivatives of (b) Determine the local extreme points off (max., min., saddle points) if there are any. (c) Find the absolute max. and absolute min. values of f over the closed region bounded by the lines x= 1, y = 0, and y = x

Answer 1

f(,y)= 222 63y + 3y? fx = 6x² by foxx = 128 ar fyy = 6 st fey = –6, =s. & critical points Put for=0 & fy=o = 6x²-6y=0 -6x+6y=0 =) y = x² x=y = x=x2 x(x-1)=0 x= 0,1

2 Point (o,o),(,1). Now callishy. ret - s² = 72x - 36 at (oo) sut-s2 = 0-36=-36co - (0,0) Saddle point p at (1,1) set - s² = 72-36 = 36 70 8 = 12 20 =) (1,1) is a point of minima. minimum value +(1:1) = 2-6+3 = -1)

c). Absolute maxima or minima inn x= 1, y = 0 , y = x 1 16,1) We dont have any critical point in this region. So check only boundry. aty=0 f(x,y) = 2x² OLASI - Emin=0 fman-2 for lyzo

at x=1, f(y) = 2-6y + 3y2 'I' (y) = -6 +by = y=1 for f (y) = 0 = (1,1) is a point already (1,1) is a point of minima Now at & y=x. fox) = 2x² - 6x²+3x² f(x) = 2x² – 3x² f(x) = 6x² - 6x , put of Casco Point 6x (at) to be H=0,1 Point (ool & Coll. = absolute maxima = 2 at(1,0) absolute minima = -1 at (1,1)