Finding the extrema; an optimization problem

Question:
Find the extrema of $x^2+2y^2+3z^2$ when $3x+2y+z=-1$.

Answer:
From the condition, we can solve for $z$.
\[ z = -(3x+2y+1)\]
Thus, we can define a function.
\[ F(x,y) = x^2+2y^2+3(-(3x+2y+1)) \]
It is supposed to find the extrema of this function. Take the partial derivatives.
\[ \frac{\partial F(x,y)}{\partial x} = 56x+36y+18,  \\
  \frac{\partial F(x,y)}{\partial y} = 36x+28y+12,   \\
   \frac{\partial^2 F(x,y)}{\partial x^2} = 56,  \\
    \frac{\partial^2 F(x,y)}{\partial y^2} = 28,  \\
      \frac{\partial^2 F(x,y)}{\partial x \partial y} = 36  \]
The value of the extremum is found when $\frac{\partial F(x,y)}{\partial x} = 0$ and $\frac{\partial F(x,y)}{\partial y} = 0$. Namely,
\[ x = -\frac{9}{34}  \\
   y = -\frac{3}{34}  \]
In order to find if this is maximum or minimum, calculate $D = F_{xx}F_{yy}-F^2_{xy}$. Then, use the following condition:
If $D>0$ and $F_{xx}>0$, then it is a minimum.
If $D>0$ and $F_{xx}<0$, then it is a maximum.
If $D<0$, then it is not a extremum.
If $D=0$, then it cannot be determined in this method.
For this problem, $D = 272>0$ and $F_{xx}=56>0$, so $F(-9/34,-3/34) = 3/34$ is the minimum.

Powered by Hirophysics.com