Question 1149596
Among all pairs of numbers (x,y) such that 6x+2y=38, find the pair for which the
sum of squares, x²+y², is minimum.  Write your answers as fractions reduced to
lowest terms.
<pre>
Let S = x² + y²

To reduce the number of unknowns from 3 to 2, we solve 6x+2y=38 for y:
                                                          2y=38-6x
                                                           y=19-3x
Substituting 19-3x for y in S = x² + y²

S  = x² + (19-3x)²

Take the derivative:

S' = 2x + 2(19-3x)(-3)
S' = 2x - 6(19-3x)

Set the derivative equal to 0:

    2x - 6(19 - 3x) = 0
    2x -  114 + 18x = 0
          20x - 114 = 0
                20x = 114
                  x = 114/10
                  x = 57/10

Check to see if it's a minimum.  We take the second derivative

S" = 2 - 6(-3)
S" = 2 + 18
s" = 20

S" is not 0, so the second derivative test succeeds. It's positive 
which means that it's concave upward there, which means that it is
a minimum.

Now we find the y-value by substituting in y=19-3x

y=19-3x
y=19-3(57/10)
y=190/10-171/10
y=19/10

Answer: (57/10, 19/10)

Edwin</pre>