Determine which points on the surface z² = xy + 1 are nearest to the origin of the plane.
$ xy + 1 \ge 0$ therefore the points nearest to O(0,0,0) are $z = 0, y = \frac{-1}{x}$ and $x\rightarrow0$.
Is this mathematically correct?
Is there a way to prove this using concepts from multivariable calculus?
On
Since you asked for a solution via multivariable calculus, Dr. Graubner’s answer is a good approach. For this problem, I think direct substitution for $z^2$ in the distance formula is much more straightforward then introducing Lagrange multipliers.
As an alternative approach, this can be solved using well-known properties of quadric surfaces: This is a nondegenerate quadric centered at the origin, so the nearest points will be along the principal axes with the greatest positive eigenvalues. Rewriting the equation of the surface as $z^2-xy=1$ gives $$Q=\begin{bmatrix}0&-\frac12&0\\-\frac12&0&0\\0&0&1\end{bmatrix}$$ for the matrix of the quadratic form. One can find the principal axes by inspection: they are $(0,0,1)^T$, $(1,1,0)^T$ and $(1,-1,0)^T$, with eigenvalues $1$, $-\frac12$ and $\frac12$, respectively (so that the surface is a hyperboloid of one sheet). The nearest points to the origin are therefore $\pm\frac1{\sqrt1}{(0,0,1)^T\over\|(0,0,1)^T\|}=\pm(0,0,1).$
the distance to the origin is given by $$d=\sqrt{x^2+y^2+z^2}=\sqrt{x^2+y^2+xy+1}$$ this is a function in $x,y$,differentiate with respect to $x,y$