Use Lagrange multipliers to find the minimum and maximum val

Use Lagrange multipliers to find the minimum and maximum values of

x^(2)*y+x+y subject to xy=4

x² y + x + y = ?

xy = 4

x²y + x + y -(xy) = 0 where is a scalar

take partial derivative wrt x:

2xy + 1 - y = 0 = > y = 9/

x = 4/9

take partial derivative wrt y

x² + 1 - x = 0

16² / 81 + 1 - (4/9)=0

20/81 ² = 1

= ±9/20

x = (4/9)(9/20) = 2/5 , y= 25

when = -9/20

x = -2/5 , -25

maximum value of f(x) = 5(2/5) + 25 = 45

minimum value = 5(-2/5) -25 = -45 answer !!!!

ut value here in this equation