## anonymous 5 years ago 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?

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(!) $y^2+4z^2=12.$