Question 2. Find the maximum value of the function f(x, y) = x + y subject to the con- straint x2 + y = 2.Solutions Given: F(oly) = alty subject to the constraint g(x,y ) = x ty - 2 By method of Lagrange multiplier. The lagrange function is Lay, V= traystage L(xy) = xty + 7( 2 2+ y)- 21 -Q Now we will find out the partial dessivatives of L(xing) with respect for, X and: y = d ty - 2 3 2 Now solve these 3 derivatives equal to zero them find out values of day and ). - - 0 = ) d = 3 a4 1/3 3 OL =>Now from egine and 2 and 4 in put values egin + 14 we will get . - 2 = 0 = 2 413 2 4 1 2 On taking whole cube an both side we will get. 4 X 3 3 3 = = 12 4 12 + 6 1679 8 - 3 + 3 12 8 14 256 64 76 = 5121 - 1 - 192x*+ 2472 25619 -64 16 Xo 2 = 512 10 - 1 - 1927 + 2472 A = 2048 1 6 - 4 - 468 1 + 9672 07 204870 - 7687 + 967 2 - 12- 4=0 204816 - 7687 5There is no method to give solution to oth degree polynomial equation Hence graphing calculator and find out that only real solution of egin (5) are 1 3x = - Q4/0478 and 1 = 0. 410478 Hence for 1= -0. 410478 + 2x0, 4 10478 2 = 1.218 and Y = ( xux 0. 4 1047 8 ) y = 0. 84765 Hence , point is ( 1. 218 , 0. 84765 ) Now for 1= 0. 410478 2 = 2 x 0, 4 10478 2 5 - 1- 218 and y = - (yxo. 41 0478 y = - 0. 84765 Hence point is ( - 1.218, -0, 84765) Now , f ( - 1. 218, -0. 84765) = - 2.06565 f ( 1. 218, 0. 84765 ) = 206565 (maximum value