# The minimum distance from the origin to the intersection of x^2+y^2= z with the plane x+y+z=12 ?

Dec 15, 2017

The coordinates of the intersection point, closest to origin are $\left(x , y , z\right) = \left(2 , 2 , 8\right)$, which is at a distance of $6 \setminus \sqrt{2}$ units from the coordinate origin.

#### Explanation:

Distance on parabola: The distance of any arbitrary point on the parabola $z = {x}^{2} + {y}^{2}$ is
$S \left(x , y\right) = \setminus \sqrt{{x}^{2} + {y}^{2} + {z}^{2} \left(x , y\right)} = \setminus \sqrt{{x}^{2} + {y}^{2} + {\left({x}^{2} + {y}^{2}\right)}^{2}}$
$\setminus q \quad \setminus q \quad \setminus q \quad = \setminus \sqrt{\left({x}^{2} + {y}^{2}\right) \left(1 + {x}^{2} + {y}^{2}\right)}$ ...... (Eq 1)

This is the function in whose extremum we are interested in

Constraint Condition: But we do not consider all the points on the parabola but only those points that lie on the intersection with the plane $x + y + z = 12$.

Plane: $\setminus \quad z = 12 - \left(x + y\right)$
Parabola: $\setminus \quad z = {x}^{2} + {y}^{2}$

To find the equation representing the locus of intersection points, eliminate $z$ in the parabola equation using its value on the plane equation,

(x^2+y^2)=12-(x+y);
${x}^{2} + {y}^{2} + \left(x + y\right) - 12 = 0$ ......(Eq 2)

So $x$ and $y$ cannot take arbitrary values but are subjected to a further constraint given by Eq 2. The constraint function is then $\setminus \phi \left(x , y\right) = {x}^{2} + {y}^{2} + \left(x + y\right) - 12$;

Lagrange Multiplier Technique: When you want to extremize a function $S \left(x , y\right)$ subject to the constraint $\setminus \phi \left(x , y\right) = 0$, define a new function $\setminus L \left(x , y\right) = S \left(x , y\right) - \setminus \lambda \setminus \phi \left(x , y\right)$. This new function $L \left(x , y\right)$ is the Lagrangian and $\setminus \lambda$ is the Lagrangian Multiplier. Extremizing the Lagrangian $L \left(x , y\right)$ is equivalent to extremizing $S \left(x , y\right)$, subject to the constraint $\setminus \phi \left(x , y\right) = 0$.

Applying this technique to our problem -

Step 1: Construct the Lagrangian function and calculate its gradient,

L(x,y) = S(x,y) - \lambda\phi(x,y);

S(x,y) = \sqrt{(x^2+y^2)(1+x^2+y^2)};
\phi(x,y) = (x^2+y^2)+(x+y)-12;

\frac{\delS}{\delx} = \frac{x{1+2(x^2+y^2)}}{\sqrt{(x^2+y^2)(1+x^2+y^2)}};

\frac{\delS}{\delx} = \frac{y{1+2(x^2+y^2)}}{\sqrt{(x^2+y^2)(1+x^2+y^2)}};

\frac{\del\phi}{\delx} = (2x+1); \qquad \qquad \frac{\del\phi}{\delx} = (2x+1);

\gradL(x,y) = \frac{\delL}{\delx}\hati + \frac{\delL}{\dely}\hatj;

\frac{\delL}{\delx} = \frac{\delS}{\delx} - \lambda\frac{\del\phi}{\delx}; \qquad \qquad \frac{\delL}{\dely} = \frac{\delS}{\dely} - \lambda\frac{\del\phi}{\dely};

Step 2: The extremum points of the Lagrangian is found by setting its gradient to zero and solve the set of equations that each component of the gradient yields,

\gradL(x,y) = 0; \qquad \frac{\delL}{\delx}=0; \qquad \qquad \frac{\delL}{\dely}=0

Step 3: Solve these equations to find the coordinates of the extremum point.

X component:
\quad \frac{\delS}{\delx} - \lambda\frac{\del\phi}{\delx}=0; \qquad \rightarrow \qquad \frac{\delS}{\delx} = \lambda\frac{\del\phi}{\delx};

$\setminus \frac{x \left\{1 + 2 \left({x}^{2} + {y}^{2}\right)\right\}}{\setminus \sqrt{1 + \left({x}^{2} + {y}^{2}\right)}} = \setminus \lambda \left(2 x + 1\right)$

$\setminus \lambda = \left(\setminus \frac{x}{2 x + 1}\right) \left(\setminus \frac{1 + 2 \left({x}^{2} + {y}^{2}\right)}{\setminus \sqrt{1 + {x}^{2} + {y}^{2}}}\right)$ ...... (Eq 3)

Y component:
\frac{\delS}{\delx} - \lambda\frac{\del\phi}{\delx}=0; \qquad \rightarrow \qquad \frac{\delS}{\delx} = \lambda\frac{\del\phi}{\delx};

$\setminus \frac{y \left\{1 + 2 \left({x}^{2} + {y}^{2}\right)\right\}}{\setminus \sqrt{1 + \left({x}^{2} + {y}^{2}\right)}} = \setminus \lambda \left(2 y + 1\right)$

$\setminus \lambda = \left(\setminus \frac{y}{2 y + 1}\right) \left(\setminus \frac{1 + 2 \left({x}^{2} + {y}^{2}\right)}{\setminus \sqrt{1 + {x}^{2} + {y}^{2}}}\right)$ ...... (Eq 4)

Eliminating $\setminus \lambda$ between Eq 3 and Eq 4,

$\left(\setminus \frac{x}{2 x + 1}\right) = \left(\setminus \frac{y}{2 y + 1}\right) \setminus q \quad \setminus \rightarrow \setminus q \quad x \left(2 y + 1\right) = y \left(2 x + 1\right)$

cancel{2xy} + x = cancel{2xy} + y; \qquad x = y

Lagrangian $L \left(x , y\right)$ has an extremum along the $x = y$ plane,

Step 4: Evaluate the complete set of the coordinates, using the constraint equation,

Substituting this ($x = y$) in the constraint equation $\setminus \phi \left(x , y\right) = 0$, we get

2x^2+2x-12=0; \qquad \rightarrow \qquad x^2+x-6=0

This quadratic equation has solutions : x_1=+2; \qquad x_2=-3;

${x}_{1} = + 2$ is the meaningful solution.

At $\left(x , y\right) = \left(2 , 2\right)$ the value of $z$ is found either by using the plane equation or by using the parabola equation. Alternatively, we can use both and verify that they both result in the same value.

Parabola: x^2+y^2=z; \qquad z = 2^2+2^2=8;
Plane: z = 12-(x+y)=12-(2+2)=8;

Thus the Lagrangian $L \left(x , y\right)$ shows that the extreme value is at $\left(x , y , z\right) = \left(2 , 2 , 8\right)$.

It lies at a distance of
$S \left(x = 2 , y = 2\right) = \setminus \sqrt{\left({2}^{2} + {2}^{2}\right) \left(1 + {2}^{2} + {2}^{2}\right)} = 6 \setminus \sqrt{2}$ units from the coordinate origin.

Dec 15, 2017

See below.

#### Explanation:

Calling

$C \to f \left(x , y , z\right) = {x}^{2} + {y}^{2} - z = 0$ and
$\Pi \to g \left(x , y , z\right) = x + y + z - 12 = 0$

We want the distance from $C \cap \Pi$ to the origin but

$C \cap \Pi = \left(f \circ g\right) \left(x , y\right) = \phi \left(x , y\right) = {x}^{2} + {y}^{2} + x + y - 12 = 0$

So the problem now is

$\min \delta \left(x , y\right) = {x}^{2} + {y}^{2}$

subjected to $\phi \left(x , y\right) = {x}^{2} + {y}^{2} + x + y - 12 = 0$

This problem can be handled easily with the contribution of Lagrange multipliers. Forming the lagrangian

$L \left(x , y , \lambda\right) = \delta \left(x , y\right) + \lambda \phi \left(x , y\right)$

the stationary points are the solutions for

$\left\{\begin{matrix}{L}_{x} = 2 x + \lambda \left(2 x + 1\right) = 0 \\ {L}_{y} = 2 y + \lambda \left(2 y + 1\right) = 0 \\ {L}_{\lambda} = {x}^{2} + {y}^{2} + x + y - 12 = 0\end{matrix}\right.$

The solution gives

$x = - 3 , y = - 3 , \lambda = - \frac{6}{5}$ and
$x = 2 , y = 2 , \lambda = - \frac{4}{5}$

The qualification is given by substitution so

$\delta \left(- 3 , - 3\right) = 18$ and $\delta \left(2 , 2\right) = 8$ then the solution is for

$x = 2 , y = 2$ giving a distance of $\sqrt{8} = 2 \sqrt{2}$

The $z$ coordinate is obtained as $z = 12 - 2 - 2 = 8$ so the global minimum distance is $\sqrt{{2}^{2} + {2}^{2} + {8}^{2}} = 6 \sqrt{2}$

Attached a plot showing in red the distance segment.