Question

Im not sure how to do this one:
The temperature on the surface given by x^2 + y^2 + 4z^2 = 12 is given by T(x, y, z) = y^2 + 4z^2. What is the maximum temperature on this surface?

Answer 1

You should use Lagrange multipliers to find the maximum temperature on the surface (which is an ellipsoid).
The function is
\[T(x,y,z)=y^2+4z^2.\]
The constraint is
\[g(x,y,z)=x^2+y^2+4z^2=12.\]
The auxiliary function is
\[\Lambda(x,y,z,\lambda)=T(x,y,z)+\lambda [g(x,y,z)-12]=\]
\[=y^2+4z^2+\lambda[x^2+y^2+4z^2-12].\]
Then you have to solve the following system of linear equations:
\[\begin{array}{rcl}
\frac{\partial\Lambda}{\partial x}=2x&=&0\\
\frac{\partial\Lambda}{\partial y}=2y+2\lambda y&=&0\\
\frac{\partial\Lambda}{\partial z}=8z+8\lambda z&=&0\\
\frac{\partial\Lambda}{\partial\lambda}=x^2+y^2+4z^2-12&=&0\\
\end{array}\]
The solution is
\[x=0,\quad \lambda=-1,\quad\textrm{ and you'll get: }y^2+4z^2=12.\]
Since
\[T(x,y,z)=y^2+4z^2\]
therefore
\[T(0,y,z)=12,\quad \textrm{ if } y=\pm\sqrt{12-4z^2},\quad -\sqrt{3}\leq z\leq \sqrt{3}.\]
So the maximum temperature on the ellipsoid is 12 and the critical point are on the ellipse(!)
\[y^2+4z^2=12.\]