. Question 1 Find the maximum and minimum values of the function f(x, y) = x y subject to 5x2 + 3y? = 45 Maximum value: Minimum value: Enter DNE if the value does not exist. Question Help: Video Message instructor Post to forum Submit Question MacBook Pro esc C G Search or type URL $ % - .- W # 2 4 5 Q W E R TQuestion 2 Find the maximum and minimum values of the function f(x, y) = ety subject to a3 + y3 = 128 Maximum value: Minimum value: Enter DNE if the value does not exist. Question Help: Video Message instructor Post to forum Submit Question MacBook Pro esc C G Search or type URL EA % > - .- W # 4 5 Q W E RSolution-(21 Given, Play) = zy Subject to Constraint 5x2 + 3 42 = 95 Consider 91xy ) = 5 x2 + 342 - 95 Lagrange's Multiplier Equation is. L ( x, ) = Play ) + A glory ) where is lagrange Multiplier. [(xyid ) = x24 +d ( 5x2+ 3 42 - 45) Differentiating partially writ x, we get 26 = 2x4 + 1 ( 10x) 2x "Differentiating partially wirt ly', we get ay = xe2 + + ( 64 ) Critical point is given by putting 26 - 26 0 J x dy we have 2 xy + 1 1 1029) = 0. = ) 2x ( 4 + 5 1) = 0 V 4 = - 51 08 x=0. Similarly, Re2 + bly = 0 x2 = - 614If we have x-other the Constraint gives us 3 4 2 = 45 = 42 = 15 08 4 =7015 If we have y=-51 then the second equation gives us x2 + GA(- 51 ) = 0 2 2 - 30 1 2 = 0 x - +130A Putting in Constraints, we have 5 ( 30 12 ) + 3(-51) 2= 45 150 12 + 75 12 = 45 225 1 2 = 45 2 = 45 = = 225 " - e . ) A=I which gives for A = I we get y = - V5 and x = + 56 for ' = - 1 we get. y = 15 andSo, lagrange's Multiplies gives us six points to Checke : 10, JTS ) , 10,- 515), ( 56, - 5376-56155, ( - 5GIVE ) , (VGiv5 ) To find Maximum and minimum. we need to Simply plug these Six points along with the critical points in the function. flo,vs ) = 1012 x 85 = 0 + 10, -5 5) = (012 x - 55 = 0 f ( o, vis ) = (01 2 x 515 = 0 + 10, - VIS ) = 10)2 x - V15 = 0. + (56- 85 ) = f1- 56,- VS) = (+ 56)2 x(- JS )= -605 Minimum f ( 56, VS ) = $1- 56, 55)=(+ 56)2 x5S= 605 Maximum