Question

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.`?

Answer 1

`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`