# Computing Extrema with Maple

### Problem 1

Find the shortest distance from the point (x0,y0,z0) to the plane Ax+By+Cz+D=0.

### Solution 1

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 previous way of doing it. Do both methods give the same answer?

### Problem 2

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:

### Solution 2

```
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>