Find Max/Min of 2x + 6y + 10z Bounded by x^2 + y^2 + z^2 = 35

[tnutty]

Homework Statement

Find mix/max

2x + 6y + 10Z bounded by x^2 + y^2 + z^2 = 35

After finding the partial derivative and letting it equal to the bounding equation * C

2 = C * 2x
6 = C * 2y
10 = C * 2z

x^2 + y^2 + z^2 = 35

x = 1/C
y = 3C
z = 5C

inputting this into the equation of sphere

(1/C)^2 + (3C)^2 + (5C)^2 = 35

=

1/C^2 + 9C^2 + 25C^2 =35

=

1/C^2 + 34C^2 = 35

now I tried a couple of different things and I think the simplest for I got it down to was
this :
1 = (35 - 34C^2) C^2

It looks like C = plus/minus 1 from observation, but that's not deductive enough, and way
to show it with math, and not just plugging it in please.

From there If I find the constant C the I can try to find other variables , x, y, z

 

[lurflurf]

2 = C * 2x
6 = C * 2y
10 = C * 2z
so
x = 1/C
y = 3/C
z = 5/C
Then you will get a surprisingly nice answer, too nice in fact.
if 1 = (35 - 34C^2) C^2 would have resulted without a mistake it is a quartic in quadratic form 34x^2-35x+1=(34x-1)(x-1)=0 where x=C^2