## Problem 1Find the shortest distance from the point (x0,y0,z0) to the plane Ax+By+Cz+D=0.## Solution 1The 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 previous way
of doing it. Do both methods give the same answer?
## Problem 2Find 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:

## Solution 2The 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. |

Link to the commands in this file

Carlos Rodriguez <carlos@math.albany.edu> Last modified: Tue Oct 22 09:11:02 EDT 1996