Distance on parabola: The distance of any arbitrary point on the parabola z=x^2+y^2 is
S(x,y) = \sqrt{x^2+y^2+z^2(x,y)}=\sqrt{x^2+y^2+(x^2+y^2)^2}
\qquad \qquad \qquad = \sqrt{(x^2+y^2)(1+x^2+y^2)} ...... (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: \quad z = 12 - (x+y)
Parabola: \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+(x+y)-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 \phi(x,y) = x^2 + y^2 + (x+y) - 12;
Lagrange Multiplier Technique: When you want to extremize a function S(x,y) subject to the constraint \phi(x,y)=0, define a new function \L(x,y) = S(x,y) - \lambda\phi(x,y). This new function L(x,y) is the Lagrangian and \lambda is the Lagrangian Multiplier. Extremizing the Lagrangian L(x,y) is equivalent to extremizing S(x,y), subject to the constraint \phi(x,y)=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};
\frac{x{1+2(x^2+y^2)}}{\sqrt{1+(x^2+y^2)}} = \lambda(2x+1)
\lambda = (\frac{x}{2x+1})(\frac{1+2(x^2+y^2)}{\sqrt{1+x^2+y^2}}) ...... (Eq 3)
Y component:
\frac{\delS}{\delx} - \lambda\frac{\del\phi}{\delx}=0; \qquad \rightarrow \qquad \frac{\delS}{\delx} = \lambda\frac{\del\phi}{\delx};
\frac{y{1+2(x^2+y^2)}}{\sqrt{1+(x^2+y^2)}} = \lambda(2y+1)
\lambda = (\frac{y}{2y+1})(\frac{1+2(x^2+y^2)}{\sqrt{1+x^2+y^2}}) ...... (Eq 4)
Eliminating \lambda between Eq 3 and Eq 4,
(\frac{x}{2x+1}) = (\frac{y}{2y+1}) \qquad \rightarrow \qquad x(2y+1)=y(2x+1)
cancel{2xy} + x = cancel{2xy} + y; \qquad x = y
Lagrangian L(x,y) 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 \phi(x,y)=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 (x,y)=(2,2) 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(x,y) shows that the extreme value is at (x,y,z)=(2,2,8).
It lies at a distance of
S(x=2,y=2) = \sqrt{(2^2+2^2)(1+2^2+2^2)} = 6\sqrt{2} units from the coordinate origin.
