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

1. ## 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

\frac{1}{x} - \frac{K_3}{K^3x^2}=\frac{2}{x}\nonumber

\implies x=-\frac{1}{K_3}

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