Question

Find the point (x,y) of x2+14xy+49y2=100 that is closest to the origin and lies in the first quadrant.

Answer 1

Notice that

$x^2+14xy+49y^2=(x+7y)^2$

so the constraint is a set of two lines,

$(x+7y)^2=100\implies\begin{cases}x+7y=10\\x+7y=10\end{cases}$

and only the first line passes through the first quadrant.

The distance between any point $(x,y)$ in the plane is $\sqrt{x^2+y^2}$, but we know that $\sqrt{f(x,y)}$ and $f(x,y)$ share the same critical points, so we need only worry about minimizing $x^2+y^2$. The Lagrangian for this problem is then

$L(x,y,\lambda)=x^2+y^2+\lambda(x+7y-10)$

with partial derivatives (set equal to 0)

$L_x=2x+\lambda=0$

$L_y=2y+7\lambda=0$

$L_\lambda=x+7y-10=0$

We have

$L_y-7L_x=2y-14x=0\implies y=7x$

which tells us that

$x+7y-10=0\iff x+49x=10\implies x=\dfrac15\implies y=\dfrac75$

so that $\left(\dfrac15,\dfrac75\right)$ is a critical point. The Hessian for the target function $x^2+y^2$ is

$H(x,y)=\begin{bmatrix}2&0\\0&2\end{bmatrix}$

which is positive definite for all $x,y$, so the critical point is the site of a minimum. The minimum distance itself (which we don't seem to care about for this problem, but we might as well state it) is $\sqrt{\left(\dfrac15\right)^2+\left(\dfrac75\right)^2}=2$.