Let d be the least distance between a point of the ellipse (x^2/(4)+y^2=1 and a point of the line x+y=4. Find d.

The ellipse lies entirely on one side of the line (the origin side), since $x+y\le\sqrt{4+1}=\sqrt5<4$ on the ellipse. Parametrise a point as $(2\cos\theta,\sin\theta)$. Its distance from the line is $\dfrac{4-(2\cos\theta+\sin\theta)}{\sqrt2}$. Using $2\cos\theta+\sin\theta=\sqrt5\cos(\theta-\alpha)$ with $\tan\alpha=\tfrac12$, the maximum of this expression is $\sqrt5$, attained at $\theta=\alpha$. The least distance between the ellipse and the line is therefore $d=\dfrac{4-\sqrt5}{\sqrt2}\approx\dfrac{1.76393}{1.41421}\approx1.2473$, i.e. $1.25$. The nearest point on the line is the foot of the perpendicular from that point of the ellipse, which also confirms this is a minimum over both curves: no point of the line is nearer to the ellipse than this perpendicular gap.