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Thread: Constrained optimization with a tricky function (long post)

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    Constrained optimization with a tricky function (long post)

    I want to find the minimum of a function T inside a region of interest bounded by the given constraint curves:


    \begin{eqnarray}
    T(x,y) = K_0 + \frac{K_3}{y}+\frac{y}{x} \nonumber\\
    x>0\nonumber\\
    y>0\nonumber\\
    y^2\leq K_1\nonumber\\
    y^4+x^2\leq K_2\nonumber\\
    \frac{y^2}{x}\leq K_3 \nonumber\\
    K_1 > 0\nonumber\\
    K_2 > 0\nonumber\\
    K_3 > 0\nonumber\\
    \end{eqnarray}


    Observation #1:


    \begin{eqnarray}
    \nabla T=\bigg<-\frac{y}{x^2}, \frac{1}{x}-\frac{K_3}{y^2}\bigg>
    \end{eqnarray}


    I can see that there are no critical points inside the region of interest. If there is a minimum, it'll be on the boundary.


    Approach #1:


    Use the Lagrangian multiplier method to find the minimum on boundary.


    \begin{eqnarray}
    G_1(x,y) = y^2 - K_1 = 0 \nonumber\\
    G_2(x,y) = y^4+x^2 - K_2 = 0 \nonumber\\
    G_2(x,y) = \frac{y^2}{x} - K_3 = 0\nonumber
    \end{eqnarray}
    Define the auxiliary Lagrangian function
    \begin{eqnarray}
    L(x,y,\lambda_1,\lambda_2,\lambda_3) = T(x,y) + \lambda_1 G_1(x,y) + \lambda_2 G_2(x,y) + \lambda_3 G_3(x,y)\nonumber\\
    = T_0 + \frac{K_3}{y}+\frac{y}{x} + \lambda_1(y^2-K_1) + \lambda_2(y^4+x^2-K_2) + \lambda_3\bigg(\frac{y^2}{x} - K_3\bigg)
    \end{eqnarray}
    Then solve the following system of equations
    \begin{eqnarray}
    L_x(x,y,\lambda_1,\lambda_2,\lambda_3) = 2\lambda_2 x-\frac{y}{x^2}-\frac{\lambda_3 y^2}{x^2} = 0\\
    L_y(x,y,\lambda_1,\lambda_2,\lambda_3) = \frac{1}{x}-\frac{K_3}{y^2}+2\lambda_1 y + \frac{2 \lambda_3 y}{x} + 4 \lambda_2 y^3 = 0\\
    \label{Llam1}L_{\lambda_1}(x,y,\lambda_1,\lambda_2 ,\lambda_3) = -K_1 + y^2 = 0\\
    L_{\lambda_2}(x,y,\lambda_1,\lambda_2,\lambda_3) = -K_2 + x^2 + y^4 = 0\\
    L_{\lambda_3}(x,y,\lambda_1,\lambda_2,\lambda_3) = -K_3 + \frac{y^2}{x}=0
    \end{eqnarray}


    I feel that this is too much to solve by hand. With my limited knowledge of Mathematica, I cannot get anything other than a numerical solution for some hand-pick values for the constants.


    Is it possible to simplify the problem?


    Observation #2:


    A closer look at $G_1$


    \begin{eqnarray}
    \nabla G_1 = \big<0, 2y\big>\nonumber
    \end{eqnarray}
    Then
    \begin{eqnarray}
    \label{nablaG1} \nabla T = \lambda \nabla G_1
    \end{eqnarray}
    \begin{cases}
    \frac{1}{x}-\frac{K_3}{y^2} = 2\lambda y\\
    -\frac{y}{x^2} = 0\\
    y = K_1
    \end{cases}


    Since $y >0$, $-\frac{y}{x^2} \neq 0$ and so I do not expect to find the minimum of T along this boundary.

    A similar conclusion can be reached for $G_3$.

    \begin{eqnarray}
    \label{nablaG3}\nabla T = \lambda \nabla G_3
    \end{eqnarray}
    \begin{cases}
    -\frac{y}{x^2} = -\lambda\frac{y^2}{x^2}\\
    \frac{1}{x}-\frac{K_3}{y^2} = \lambda \frac{2y}{x}\\
    \frac{y^2}{x} = K_3
    \end{cases}


    Based on the above
    \begin{eqnarray}
    \lambda=\frac{1}{y}\nonumber\\
    y^2=K_3x\nonumber
    \end{eqnarray}
    Substituting the above expressions into the second sub-equation above we get
    \begin{equation}
    \frac{1}{x} - \frac{K_3}{K^3x^2}=\frac{2}{x}\nonumber
    \end{equation}
    \begin{equation}
    \implies x=-\frac{1}{K_3}
    \end{equation}

    Since $x = > 0, K_3 > 0$, we see that $x=-\frac{1}{K_3}$ is not possible and so I don't expect to find the minimum of T on this boundary either.

    Are these observations valid so far?

    On the other hand, using a similar approach I see that it should be possible to find the minimum of T on the $G_2$ boundary (verified in software).

    Now, in the presence of all three constraints, only the values of T on a specific segment of $G_2$ are admissible.

    Is there a way to simplify the problem and obtain a nice human-readable solution? Or is there another technique that could be used?

    Attached is an example plot showing the constraint boundaries and the gradient vector field of T for some hand-picked values of $K_1,K_2,K_3$.

    Thanks

    Constrained optimization with a tricky function (long post)-funct.png
    Last edited by jumbo1985; Oct 5th 2017 at 02:25 PM.
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