Question:

Find the extrema of \(x^2+2y^2+3z^2\) under the condition that \(3x+2y+z=-1\).

Answer:

From the condition, solve for \(z\).

\[ z = -(3x+2y+1)\]

Thus, we can define a function of \(x\) and \(y\):

\[ F(x,y) = x^2+2y^2+3(-(3x+2y+1)) \]

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.

This is the extrema of multi-dimensional functions.