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. |