# Calculating the minimum distance to the origin from a curve defined by $\frac{x^2}{4}+y^2+\frac{z^2}{4}=1$ and $x+y+z=1$

I want to calculate the points of the curve given by $$\frac{x^2}{4}+y^2+\frac{z^2}{4}=1,\qquad x+y+z=1$$ which are minimum and maximum distance to the origin. Using Lagrange multipliers, the maximum and minimum will be in the solutions of the system $$\begin{cases}(x,y,z)=\lambda(x/2,2y,z/2)+\mu(1,1,1)\\ \frac{x^2}{4}+y^2+\frac{z^2}{4}=1 \\ x+y+z=1\end{cases}$$

I don't understand why the gradient of the function is $$(x,y,z)$$. I thought the function we were trying to minimize or maximize here is the distance function to the origin, which should be $$\sqrt{x^2+y^2+z^2}$$?.

• Minimizing $\sqrt{x^2+y^2+z^2}$ and minimizing $\frac{1}{2}(x^2+y^2+z^2)$ are the same thing, but the gradient of the second function is simpler. – AlexanderJ93 Jan 20 at 2:16

$$1=\frac{x^2}{4}+y^2+\frac{z^2}{4}\geq y^2,$$ which gives $$0\leq y^2\leq1.$$
Thus, since $$x^2+z^2=4-4y^2,$$ we obtain: $$\sqrt{x^2+y^2+z^2}=\sqrt{4-3y^2}\leq2.$$ The equality occurs for $$y=0$$, $$x+z=1$$ and $$x^2+z^2=4,$$ which says that we got a maximal distance.
Also, $$\sqrt{x^2+y^2+z^2}=\sqrt{4-3y^2}\geq1.$$ The equality occurs for $$y=1$$ and $$x=z=0,$$ which says that we got a minimal distance.