Let x, y, and z be real numbers such that x^2 + y^2 + z^2 = 1. Find the maximum value of 9x+12y+8z.?

1 Answer
May 21, 2017

17

Explanation:

The equation:

x^2+y^2+z^2=1

describes the unit sphere in RR^3.

The equation:

9x+12y+8z = k

describes a plane with normal vector < 9, 12, 8 >

The unit vector in the same direction is given by dividing by:

||<9, 12, 8>|| = sqrt(9^2+12^2+8^2)

color(white)(||<9, 12, 8>||) = sqrt(81+144+64)

color(white)(||<9, 12, 8>||) = sqrt(289)

color(white)(||<9, 12, 8>||) = 17

that is:

< 9/17, 12/17, 8/17 >

This normal vector will intersect the unit sphere at the point:

(9/17, 12/17, 8/17)

This point will be the intersection of the plane and the unit sphere if the plane just touches the unit sphere - that is when k takes its maximum value.

Then we find:

9x+12y+8z = 9(9/17)+12(12/17)+8(8/17)

color(white)(9x+12y+8z) = (9^2+12^2+8^2)/17 = 289/17 = 17