How to find minimum and maximum value of x, if x+y+z=4 and $x^2 + y^2 + z^2 = 6$?

Question

I just know that putting y=z, we will get 2 values of x. One will be the minimum and one will be the maximum. What is the logic behind it?

Answer 1

Hint: Prof vector had it right except for a small factor missing. Use $y^2+z^2 \ge \frac12 (y+z)^2$ and build a quadratic inequality in $x$.

You should get $\frac23 \le x \le 2$.

Answer 2

Hint:

$x+y+z = 4$ is a plane which has equal intercepts of $4$ on coordinate axes and $x^2 + y^2 +z^2 = 6$ is a sphere centred at $(0,0)$ with radius $\sqrt{6}$

Answer 3

$$6-x^2=y^2+z^2\ge2\,\left(\frac{y+z}2\right)^2=2\,(2-x/2)^2\tag1$$ implies $$\frac32x^2-4x+2\le0,$$ i.e. $$\frac23\le x\le2.$$ Equality in (1) means $y=z$.

Answer 4

$z=4-x-y$

$x^2+y^2+(4-x-y)^2-6=0$

Differentiate wrt $x$ and $y$

$2x-2(4-x-y)=0$

$2y-2(4-x-y)=0$

Gives $x=y=4/3$

Answer 5

The second condition gives $$x^2+(y+z)^2-2yz=6$$ or $$x^2+(4-x)^2-2yz=6$$ or $$yz=x^2-4x+5,$$ which with $y+z=4-x$ gives $$(4-x)^2-4(x^2-4x+5)\geq0$$ or $$3x^2-8x+4\leq0$$ or $$\frac{2}{3}\leq x\leq2.$$ The equality occurs for $y=z$ because $y$ and $z$ are roots of the equation $$t^2-(4-x)t+x^2-4x+5=0$$ and the equality in $\Delta\geq0$ happens for $y=z$.

Id est, $$\max_{x+y+z=4,x^2+y^2+z^2=6}x=2$$ and $$\min_{x+y+z=4,x^2+y^2+z^2=6}x=\frac{2}{3}.$$ Done!

Answer 6

One easy way to see the symmetry is to write down the KKT conditions for the optimization problem using Lagrange multipliers for the constraints. If you write down the KKT conditions you can easily see that at the optimality $y=z$. Using this fact you can compute the two answers for x which one of them minimizes the objective and the other one maximizes it.

Answer 7

I think , i have a simple solution to this problem , though it is not an elegant one . We have $x^2+y^2+(4-x-y)^2-6=0$ by eliminating z from the 2 given equations . And from this condition ($x^2+y^2+(4-x-y)^2-6=0$) we have to find maximum and minimum value of x . In other words we have to find range of x for this function $x^2+y^2+(4-x-y)^2-6=0$ to exist (this is an ellipse , not a function though but a relation) . One simple thing we can do is to expand this expression to get $2x^2+2y^2+10+2xy-8x-8y=0$. This is a quadratic in y . x and y can only take real values , hence for y to be real , discriminant of this quadtratic equation in y must be greater than or equal to 0 . So solve this inequality to get range of x between 2/3 and 2