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The plane $ x + y + 2z = 4$ intersects the paraboloid $z=x^2+y^2$ in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.

From this, I thought that $x^2+y^2+z^2$ was the distance equation that I needed to minimize, and $x+y+2z=4$ and $z=x^2+y^2$ were the two constraints. I rearranged the second constraint equation to get $z-x^2-y^2=0$.

So I set it up as follows, ($a$ is supposed to be lambda and $b$ is supposed to be mu):

$2x = a(1) + b(-2x)$

$2y = a(1) + b(-2y)$

$2z = a(2) + b(1)$.

From this, I then got:

$2xyz = a(yz) + b(-2xyz)$

$2xyz = a(xz) + b(-2xyz)$

$2xyz = a(2xy) + b(xy)$

This gave me that $yz - 2xyz = xz-2xyz$. Therefore, $yz = xz$ and therefore $x=y$.

Plugging this into the plane equation, I got that $2x +2z$ $=$ $4$, and therefore $z$ $=$ $2-x$. This gave the values of $x=2$, $y=2$ and $z=0$ and therefore the point $(2, 2, 0)$. But this point is not right for either the minimum or maximum distance.

If anyone knows where I may have gone wrong and can provide feedback, I would greatly appreciate it!

Here, the two constraints are $g (x, y, z) = x+y+2z=4$ and $h (x, y, z) = x^2 + y^2 − z$.

Any critical point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that $(x, y, z)$ is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from $(x, y, z)$ to the origin is given by $sqrt{(x-0)^2+(y-0)^2+(z-0)^2}$.This expression
(and its partial derivatives) would be cumbersome to work with, so we will find the the extrema
of the square of the distance. Thus, our objective function is
$$f (x, y, z) = x^2 + y^2 + z^2$$and$$∇f = langle2x, 2y, 2zrangle$$
$$lambda∇g = langleλ, λ, 2λrangle$$
$$mu∇h = langle2mu x, 2mu y, −mu rangle $$
Thus the system we need to solve for $(x, y, z)$ is
$$2x=lambda+2mu xtag1$$
$$2y=lambda+2mu ytag2$$
$$2z=2lambda-mu xtag3$$
$$x+y+2z=4tag4$$
$$x^2+y^2-z=0tag5$$
Subtracting $(2)$ from $(1)$ and factoring gives
$$2(x-y)=2mu(x-y)$$
so $mu=1$ whenever $xneq y$. Substituting $mu=1$ into $(1)$ gives us $lambda=0$ and substituting $mu=1$ and $lambda=0$ into $(3)$ gives us $2z=-1$ and thus $z=-dfrac12$ into $(4)$ and $(5)$ gives us $$x+y-5=0$$
$$x^2+y^2+dfrac12=0$$
However, $x^2+y^2+dfrac12=0$ has no solution. Thus we must have $x=y$

Since we now know $x=y$, $(4)$ and $(5)$ become
$$2x+2z=4$$
$$2x^2-z=0$$so
$$z=2-x$$
$$z=2x^2$$
Combining these together gives us $$2x^2=2-x$$, so $2x^2+x-2=0$.

Now solve the above equation for $x$.

Then substitute to get the critical points $(x_1,y_1,z_1)$ and $x_2,y_2,z_2$.

When you substitute them into the objective functions then you get $f(x_1,y_1,z_1)=?$ and $f(x_2,y_2,z_2)=?$

Then compare both $sqrt{f(x_1,y_1,z_1)=?}$ and $sqrt{f(x_2,y_2,z_2)=?}$ to find minimum and maximum distance.