A plane,sphere and line all in an inequality

Question

$x+y+z=4$

$x^2 +y^2+z^2=6$

If $x$,$y$ and $z$ are real numbers that satisfy the equation above, and that they lie in the range of $[a,b]$,then find values of both a and b.

My attempt:-

Instead of using Lagrange Multipliers ,I tried to approach this problem using Geometric interpretation.

The solution set, being the intersection of a sphere and a plane, is a non-horizontal circle with a unique highest as well as lowest point. Since the solution set is symmetric in $x$ and $y$(WLOG),we must have $x=y$,at the point of maxima or minima.

Now,this question boils down to 2 variables and 2 equations.

$2x+z=4$

$2x^2 + z^2=6$

Solving these by plugging in $z=4-2x$ in the third equation,we get the value of $z=2$ or $2/3.$ So $a=2/3$ and $b=2$,but the answer is marked incorrect.

Question:-What is wrong with the above solution?

Answer 1

I used a different method, and found that one kind of extremal point has two coordinates equal to $1$ and the other equal to $2$; the other kind of extremal point has two coordinates equal to $5/3$, the other equal to $2/3$. That is, I found complete agreement with your results.

EDIT: In response to your request, I’ll show you how I did it. First notice that you’re asking for the projection on any one of the axes of the full figure, which of course is a circle. I choose to project onto the $y$-axis, in other words to ask the range of $y$-values for the points of the circle. Now I make the substitution $z=4-x-y$ into the second equation, and get $$ x^2 + xy + y^2 - 4x - 4y=-5\,. $$ Then I just did an implicit differentiation to find where the tangents are horizontal. The result is $\frac{dy}{dx}=-(2x+y-4)\big/(2y+x-4)$, so I want the intersection of the line $y=4-2x$ with the ellipse, which I hope you know has major axis of slope $-1$, minor of slope $1$, not that this is significant. At any rate, when you substitute this $y$ for the $y$ in the equation of the ellipse, you get $3x^2-8x+5=0$, factoring as $(3x-5)(x-1)$. Your point of maximum $y$-value is $(1,2,1)$ and your point of minimum $y$-value is $(5/3,2/3,5/3)$.

Answer 2

What is wrong with the above proof?

There seems nothing wrong, after all. Effectively, you cross your objects with plane $x-y=0$ and find your point there.

Alternate geometric way is to notice that in point of maximal (minimal) x tangent to the circle has direction $(0, 1, -1)$, so orthogonal circle radius has direction $(2,-1,-1)$; centre of circle is at $(4/3, 4/3, 4/3)$ and radius is $\sqrt{2 \over 3}$. After all this, we get $a = {2 \over 3}, b = 2$ (fixed mistake with initial calculations where sphere radius 6 was assumed).

Considering these two ways yield the same result, I don't think it's wrong. Are you sure the statement about your answer being incorrect is flawless?

Answer 3

Here is yet another method to get the answer. By a well known inequality, $$2(x^2+y^2)\ge (x+y)^2 \iff 2(6-z^2)\ge (4-z)^2 \iff (3z-2)(z-2)\le 0$$