Finding warmest and coldest temperatures

Question

A penguin is walking on an ice ball of radius 1 m centered at the origin.

Let $T$ be the temperature at a point $(x,y,z)$ on the sphere and it is given by $T(x,y,z) = x^2 +4y^2 +9z^2 +4xy +6xz +12yz$ °F.

What are the warmest and coldest temperatures that the penguin encounters?

I found the eigenvector of the quadratic form and I do not know how to go on to the next step.

Answer 1

Guide:

• For a positive semidefinite matrix, $v^TAv \geq 0$, hence if you can attain this value, you have obtain the smallest value.

• Also prove that we have $v^TAv$ to be at most the largest eigenvalue where $v^Tv=1$.

To see this, consider the langrangian $v^TAv-\lambda (v^Tv-1)$ and differentiate it and equate it to $0$.

Answer 2

Eigenvalues and quadratic forms only complicate the matter here.

Since $T(x,y,z)=(x+2y+3z)^2$, the coldest parts of the sphere are where $x+2y+3z=0$; $T$ is 0 °F at these points. Likewise, the warmest parts are the points of tangency of the sphere and planes parallel to $x+2y+3z=0$ – where the line defined by the vector $(1,2,3)$ intersects the sphere. These points are $\frac{(1,2,3)}{\Vert1,2,3\Vert}=\frac1{\sqrt{14}}(1,2,3)$ and its negation; $T$ is 14 °F.