Final answer:
The maximum volume of the rectangular parallelepiped in the 1st octant, with one vertex on the plane 2x + y + 4z = 12, is found by expressing the volume in terms of x, derivating it, and finding the critical points that yield the maximum volume.
Step-by-step explanation:
The maximum possible volume of a rectangular parallelepiped, which lies in the 1st octant with three sides on the coordinate planes and one vertex on the plane 2x + y + 4z = 12, can be found using optimization techniques in calculus.
For a point (x,y,z) on this plane, the parallelepiped's volume V can be expressed as V = xyz.
Since the point lies on the given plane, we have a constraint 2x + y + 4z = 12.
To find the maximum volume, we express y and z in terms of x: y = 12 - 2x - 4z.
Then, we can solve for z in terms of x by setting the derivative of the volume function in terms of x to zero and finding the critical points.
After calculating, we substitute the x-value back into the constraints to find the corresponding y and z values. The product of these values gives the maximum volume of the parallelepiped.