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Math Help - Stationary points of two variables function. Advance section

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Dext91 View Post
    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 z = {x^3}y - x{y^3} - 4\sqrt 3 x - 4y

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

    So, we got the following system of equations

    \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 i = \sqrt { - 1} and subtract the second equation from the first equation.

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

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

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

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

    \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

    \Leftrightarrow \left[ \begin{gathered}x - iy = 2\cos \frac{\pi }<br />
{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. {\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 z, i.e.

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

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

    {z_3} = f\left( {2\cos \frac{{11\pi }}{9}, - 2\sin \frac{{11\pi }}{9}} \right).<br />
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  3. #3
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    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?
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