# Thread: Stationary points of two variables function. Advance section

1. ## Stationary points of two variables function. Advance section

Help, please, I have to solve this difficult task. I tried to decide, but I understand that I will not be able to solve this task.

Find the stationary points of this function
z(x, y) = (x^3)*y – x*(y^3) – 4*sqrt(3)*x – 4y.

2. Originally Posted by Dext91
Help, please, I have to solve this difficult task. I tried to decide, but I understand that I will not be able to solve this task.

Find the stationary points of this function
z(x, y) = (x^3)*y – x*(y^3) – 4*sqrt(3)*x – 4y.
So $\displaystyle z = {x^3}y - x{y^3} - 4\sqrt 3 x - 4y$

$\displaystyle \frac{{\partial z}}{{\partial x}} = 3{x^2}y - {y^3} - 4\sqrt 3$ and $\displaystyle \frac{{\partial z}}{{\partial y}} = {x^3} - 3x{y^2} - 4.$

So, we got the following system of equations

$\displaystyle \left\{ \begin{gathered}{x^3} - 3x{y^2} - 4 = 0, \hfill \\3{x^2}y - {y^3} - 4\sqrt 3 = 0; \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}{x^3} - 3x{y^2} = 4, \hfill \\3{x^2}y - {y^3} = 4\sqrt 3 . \hfill \\ \end{gathered} \right.$

Now, we multiply the second equation of the system to imaginary unit $\displaystyle i = \sqrt { - 1}$ and subtract the second equation from the first equation.

$\displaystyle {x^3} - 3i{x^2}y - 3x{y^2} - i{y^3} = 4 - i4\sqrt 3 \Leftrightarrow$

$\displaystyle \Leftrightarrow {x^3} - 3{x^2}\left( {iy} \right) + 3x{\left( {iy} \right)^2} - {\left( {iy} \right)^3} = 4 - i4\sqrt 3 \Leftrightarrow$

$\displaystyle \Leftrightarrow {\left( {x - iy} \right)^3} = 4 - i4\sqrt 3 \Leftrightarrow$

$\displaystyle \Leftrightarrow {\left( {x - iy} \right)^3} = 8\left( {\cos \frac{\pi }{3} + i\sin \frac{\pi }{3}} \right) \Leftrightarrow$

$\displaystyle \Leftrightarrow x - iy = 2\left( {\cos \left( {\frac{\pi }{9} + \frac{{2\pi k}}{3}} \right) + i\sin \left( {\frac{\pi }{9} + \frac{{2\pi k}}{3}} \right)} \right),{\text{ }}k = 0,1,2 \Leftrightarrow$

$\displaystyle \Leftrightarrow \left[ \begin{gathered}x - iy = 2\cos \frac{\pi } {9} + 2i\sin \frac{\pi }{9}, \hfill \\x - iy = 2\cos \frac{{5\pi }}{9} + 2i\sin \frac{{5\pi }}{9}, \hfill \\x - iy = 2\cos \frac{{11\pi }}{9} + 2i\sin \frac{{11\pi }}{9}. \hfill \\ \end{gathered} \right.$$\displaystyle {\text{ }} \Leftrightarrow \left[ \begin{gathered}{x_1} = 2\cos \frac{\pi }{9},{\text{ }}{y_1} = - 2\sin \frac{\pi }{9}, \hfill \\{x_2} = 2\cos \frac{{5\pi }}{9},{\text{ }}{y_2} = - 2\sin \frac{{5\pi }}{9}, \hfill \\{x_3} = 2\cos \frac{{11\pi }}{9},{\text{ }}{y_3} = - 2\sin \frac{{11\pi }}{9}. \hfill \\ \end{gathered} \right.$

We got three pairs of real roots of the system.
Now you need to calculate the three values of the function $\displaystyle z$, i.e.

$\displaystyle {z_1} = f\left( {2\cos \frac{\pi }{9}, - 2\sin \frac{\pi }{9}} \right),$

$\displaystyle {z_2} = f\left( {2\cos \frac{{5\pi }}{9}, - 2\sin \frac{{5\pi }}{9}} \right),$

$\displaystyle {z_3} = f\left( {2\cos \frac{{11\pi }}{9}, - 2\sin \frac{{11\pi }}{9}} \right).$

3. Thank's Demath!
But we can not solve this problem without using complex numbers?

How can algebraic system equations have roots in the transcendental form of sine and cosines?