TITLE: Computing Extrema with Maple
# <h3>Problem 1</h3>
  Find the shortest distance from the point (x0,y0,z0) to the 
  plane Ax+By+Cz+D=0.
  <hr>
  <h3>Solution 1</h3>
  The square of the distance from the given point to an arbitrary point on the
  plane is:
> z := solve(A*x+B*y+C*z+D,z); SqrDist := (x-x0)^2+(y-y0)^2+(z-z0)^2;

                                     A x + B y + D
                              z := - -------------
                                           C

                              2           2   /  A x + B y + D     \2
           SqrDist := (x - x0)  + (y - y0)  + |- ------------- - z0|
                                              \        C           /

# this is assuming that C is not zero. We now minimize the above
  expression as a function of x, and y. To do this we look first
  for the point where the partials are both zero i.e.
> solve({diff(SqrDist,x), diff(SqrDist,y)},{x,y});

                            2                            2
                      - x0 C  + A D + A z0 C + B A y0 - B  x0
               {x = - ---------------------------------------,
                                     2    2    2
                                    A  + C  + B

                               2    2
                         - y0 C  - A  y0 + A B x0 + B D + B z0 C
                   y = - ---------------------------------------}
                                        2    2    2
                                       A  + C  + B

# You may want to compare this result with the 
  <a href="../distance-to-point-dir/ptoplane-maple.html">previous way
  of doing it.</a> Do both methods give the same answer?
  
  <h3>Problem 2</h3>
  Find the global maximum and minimum values of the function f
  on the closed triangular region with vertices (-1,1),(2,1)
  and (-1,-2). Where,
> f := (x,y) -> x^2 + 2*x*y + 3*y^2:
# <hr>
  <h3>Solution 2</h3>
  The gradient of f is,
> gf := grad(f(x,y),[x,y]);

                         gf := [ 2 x + 2 y, 2 x + 6 y ]

# and the only point where gf is zero is:
> solve({gf[1],gf[2]},{x,y});

                                 {y = 0, x = 0}

# and the origin is inside the triangular region so it is
  its global min. Clearly the function has no global max since
  we can always increase the value of f by increasing x and or y.
>