A closed rectangular box with faces parallel to the coordinate planes... what is the max volume?

Question

A closed rectangular box with faces parallel to the coordinate planes has one bottom corner at the origin and the opposite top corner in the first octant on the plane 5x+6y+z=1. What is the maximum volume of such a box?

Answer 1

Given that one bottom corner is at the origin and the opposite top corner is at (x,y,z), the volume of the rectangular box will be simply x*y*z.

So the problem is to maximize x*y*z such that 5x+6y+z=1.

And in the first octant, we have that x > 0, y > 0, z > 0.

We can use substitution to simplify things, solving the constraint for z: z = 1 - 5x - 6y. So the new problem is to maximize x*y*(1 - 5x - 6y) = xy -5x²y-6xy².

Critical points will be where the partial derivatives with respect to x and with respect to y are zero.

d/dx = y - 10xy - 6y² = 0, or y(1-10x-6y) = 0

d/dy = x - 5x² - 12xy = 0, or x(1-5x-12y) = 0

Since x and y are positive, the terms in the parentheses must be equal to zero:

1 - 10x - 6y = 0

1 - 5x - 12y = 0

Solving this system of equations yields x = 1/15, y = 1/18, and plugging these values into our expression for z (z = 1 - 5x - 6y) yields z = 1/3. Then multiplying x*y*z gives our volume = 1/810.

Note that to be complete, you would want to check your second-order conditions to make sure this is a max and not a min or saddle point, but I'll leave that out for the sake of simplicity.