Volume in first octant cut off by a plane

[Frillth]

Homework Statement

Find the equation of the plane through the point [1,2,2] that cuts off the smallest possible volume in the first octant.

Homework Equations

Volume of a pyramid = 1/3Ah

The Attempt at a Solution

The plane is going to cut out a pyramid with the x-, y-, and z-intercepts, so let x, y, and z be the intercepts. Then V = 1/6xyz. But since the plane must go through [1,2,2] and three points define a plane, we can write one of x, y, and z in terms of the other two. Any plane passing through intercepts x, y, and z has a general point [a,b,c] so that:
a/x + b/y + c/z = 1
Since [1,2,2] is on the plane, plug that in for [a,b,c]:
1/x + 2/y + 2/z = 1
Solve for x (just because I'm guessing that it would be the easiest):
x = -1/(2y + 2z - 1)
Now plug that into the volume formula:
V = -yz/(12y + 12z - 6)

Is this right so far? If not, what did I do wrong? If so, how can I continue?

 

[Dick]

I think it's ok so far. Now take the partial derivatives dV/dy and dV/dz and set them both to zero. That's two equations in two unknowns.

 

[Frillth]

All right, so from there I take the partials and I get:

dV/dy = (-12z^2 + 6z)/(12y + 12z - 6)^2
dV/dz = (-12y^2 + 6y)/(12y + 12z - 6)^2

Setting these equal to zero, I get:

y = z = 0 and y = z = 1/2

However, I know that y and z can't be 0, and if y and z are 1/2, then by my earlier result:

x = -1/(1 + 1 - 1) = -1

I can't have x = -1, because the plane has to intersect the positive x-axis or else the volume will be infinite. What did I do wrong here?

 

[Dick]

You have a mistake solving for x in 1/x + 2/y + 2/z = 1. 1/x=1-(2/y+2/z). That's
x=1/(1-(2/y+2/z)). Not the same as your expression. I missed it. Sorry.