Question

6. Find the absolute maximum and minimum values of fon D, where f(x,y)=x² – 2xy +2y and D={(x,y)|0SX 33,05ys2.

Answer 1

Given function f(ny) = x²-axy tay where D= {(2,4) losa<3,084{2} first Partialy diff. w. h. t.'&' fx Now Partially cliff.co.kit. 'y' fy= –2x+2 - 2x - 24 Now, 2(2-4)=0 and 2(x+1)=0 so, 1 and y = +1 critical point (ny) = (+1+1) Here the gilen domain is a rectangle c.e. 6,2) (3,2). LA L2 Now, the absolute Maximum or minimum values could be any one of the four end points that all have on rectangle it could be the Point on the dine segment L, L2, L3 & Lu. OR it could be on critica point (1,1) here critical point lies on rectangle. Now, we make a table for all your points of rectangle and critical pornt. f(ay) These values find further وهو لر در سلم ) در لر و منہ o- &-0

x² for Li: y=o f(0,0) = 22 - 2x (0) +2(0) f 001,0) Between osaa3, x? is increasing function so, o is #i oguees Minimum value 3 grees Manimum value f(0,0) = 0 f(3,0) = 9 x=3 f(3,4) = 3-20)(9) +2(y) - g-64 +20y - 9- 4y Between osy<2 (44) is Decreasing function o grius Maximum value 2 gives minimum value f(3,0) = = 9 f (3,2)= 9-4(2) L3: Y=2 & osn23 $(3,2)= x?-22 (2) +2(2) x² - 4x + 4 It is a parabolic feniction f(0,2)= 4 f (3,2) = 1 for Minimum value of f(1/2) = 22-44 +4 22-4 = 2(7-3) & f'(0, 2) f(3,2) = 22- 4(2) +4 - 4-8 + 4

for La : x=0& osys 2 f(0, y) = 0 - 2004+24 - 24 ay is Increasing o is Manimum 2 is Minimum f(0,0)=0 f(0,2)= 4 for ciritical point &(1,1) I 2+2 Now from Table of o-cubs Minimum 9 abs Manimun, /و تر له 0 - - 2 1 abs Minimum so, Absolute Maxima Point :'(3,0)) and value g' and Absolute minima Point 6,0), (2, 2) and value is 'o'.