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# How do you find critical points of multivariable function $f(x,y) = {x^3} + xy - {y^3}?$

Last updated date: 17th Jun 2024
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Hint:First of all we will take the given expression and take differentiation with respect to “x” and “y” one by one and then find the factors and place the values of one variable to get the value for the other variable.

Complete step by step solution:
Take the given expression:
$f(x,y) = {x^3} + xy - {y^3}$
Differentiate the above expression with respect to “x”
$\dfrac{d}{{dx}}f(x,y) = \dfrac{d}{{dx}}({x^3} + xy - {y^3})$
Apply differentiation to all the terms on the right hand side of the equation inside the bracket.
${f_x} = \dfrac{d}{{dx}}({x^3}) + \dfrac{d}{{dx}}(xy) - \dfrac{d}{{dx}}({y^3})$
Apply the formula in the above equation and place the value in it.
${f_x} = 3{x^2} + y$
To find the critical value in the above equation equal to zero.
$0 = 3{x^2} + y$ …. (A)
Similarly take derivative in the given expression with respect to “y”
$\dfrac{d}{{dy}}f(x,y) = \dfrac{d}{{dy}}({x^3} + xy - {y^3})$
Apply differentiation to all the terms on the right hand side of the equation inside the bracket.
${f_y} = \dfrac{d}{{dy}}({x^3}) + \dfrac{d}{{dy}}(xy) - \dfrac{d}{{dy}}({y^3})$
Apply the formula in the above equation and place the value in it.
${f_y} = x - 3{y^2}$
To find the critical value in the above equation equal to zero.
$0 = x - 3{y^2}$ …. (B)
Take equation (A)
$y = - 3{x^2}$
Place the above value in the equation (B)
$x - 3{( - 3{x^2})^2} = 0$
Simplify the above equation-
$x - 27{x^4} = 0$
Finding the factors of the above equation.
$\Rightarrow x(1 - 27{x^3}) = 0$
Implies
$x = 0$ …. (C)
or $1 - 27{x^3} = 0$
Simplify the above equation:
$1 = 27{x^3}$
When term multiplicative on one side is moved to the opposite side, then it goes to the
denominator.
$\Rightarrow {x^3} = \dfrac{1}{{27}}$
Take a cube-root on both the sides of the equation.
$\Rightarrow x = \dfrac{1}{3}$ …. (D)
Using the equation (C) and equation (A)
$0 = 3{x^2} + y$
Make “y” the subject, when any term is moved from one side to another the sign of the term also changes.
$y = - 3{x^2}$
Place $x = 0$
$\Rightarrow y = - 3(0)$
Zero multiplied with anything gives zero.
$\Rightarrow y = 0$ … (E)
Now, Place $x = \dfrac{1}{3}$ in equation (A)
$\Rightarrow y = - \dfrac{1}{3}$
Hence, the critical points are: $(0,0)$ and $\left( {\dfrac{1}{3}, - \dfrac{1}{3}} \right)$

Note: Be careful about the sign convention when doing simplification. When you move any term from one side to another then the sign of the term also changes. Positive terms become negative and vice-versa.