Find a point on a plane that is closest to the origin

Question

Find the closest point to the origin on the plane which is tangent to $f(x,y)=x^2 e^y$ at point $P(0.9,0.1)$.

Answer 1

Here's the figure with the function in yellow, the tangent plane (to be computed, below) in blue, and $P$ in red. Note the non-uniform scaling in different axes:

The function is: $f(x,y) = x^2 e^y$

The derivatives with respect to $x$ and to $y$, are:

$2 x e^y$ and $x^2 e^y$, respectively, which when evaluated at $P$ have values:

$1.98931$ and $0.895188$.

Thus the tangent plane has the form:

$z = 1.98931 x + 0.895188 y + c$, where $c$ must be found so that the plane goes through $P$: $c = -0.984709$.

Now we want to find the point on this plane $z = 1.98931 x + 0.895188 y -0.984709$ that is closest to the origin. Compute the (square) of that distance:

$d^2 = x^2 + y^2 + (1.98931 x + 0.895188 y -0.984709)^2$.

Take the derivatives with respect to $x$ and $y$, set them to zero and find:

$x^* = 0.340158$ and $y^* = 0.153071$.

Note the uniform scaling in this figure, and the confirmation that the green separation line to the origin is perpendicular to the tangent plane.