Question

Find the stationary points of u=x^3+y^3-3xy​

Answer 1

$\frac{du}{dy} = 3 {y}^{2} - 3x$

$\frac{du}{dy} = 3 {x}^{2} - 3y$

• considering above equation as (1) and (2),
• from equation (1),

$3 {y}^{2} - 3x = 0$

$3 {y}^{2} = 3x$

${y}^{2} = x$

putting above values in equation (2),

$3 {x}^{2} - 3y = 0$

$3 {y}^{4} - 3y = 0$

$3y(y {}^{3} - 1) = 0$

${y}^{3} - 1 = 0$

$y = 1$

putting value of y in equation of X,

$x = 1$

thus, the stationary points are ( 1 , 1 )