Example 4

Find the point on the intersection of the plane z= 3 x+ 4 y+ 6  and the paraboloid    which is closest to the origin.  

(commands used to generate figure above)

Solution:  

The equation of the plane and the equation of the paraboloid are the constraints.   Our objective is to minimize the square of the distance from a point ( x,y,z ) to the origin.  Thus, our constrained optimization problem is of the form

Minimize

subject to  3 x+ 4 y+ 6  - z  = 0    and     

Thus, the Lagrangian is as follows:

>	Lagrangian:=x^2+y^2+z^2+lambda[1]*(3*x+4*y+6-z)+lambda[2]*(x^2+y^2-z);

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Next we set the gradient equal to 0 and solve:

>	grad(Lagrangian,[x,y,z,lambda[1],lambda[2]]); convert(%,'set'): sols:=[solve(%,{x,y,z,lambda[1],lambda[2]})];

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There are two solutions in this case.

>	solution1:=sols[2]; solution2:=sols[3];

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Let's now evaluate f  at each of these points:

>	seq(print("At ",sols[j],":  f(x,y,z) = ",subs(sols[j],x^2+y^2+z^2)),j=2..3);

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Clearly, the closer of these two points is

( )

which is shown as a black point in the plot below:

>	p1:=plot3d(3*x+4*y+6,x=-2..2,y=-x-2..-x+2,color=red,grid=[5,5]): p2:=plot3d(x^2+y^2,x=-2..2,y=-sqrt(4-x^2)..sqrt(4-x^2)): p3:=sphere([-3/5,-4/5,1],0.1,color=black,grid=[10,10]): display(p1,p2,p3,view=[-2..2,-2..2,0..7],axes=normal);

>