Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 2y + 3z = 15.

The volume of a box is the amount of space in it.

The volume of the largest rectangular box is 125/6

The vertex of the plane is given as:

Subtract 15 from both sides

Using Lagrange multipliers, we have:

Substitute

Differentiate

Equate the above derivatives o 0

Make the subject in

Substitute in and

Factor out z and y

Divide both sides by z

Make x the subject

Substitute and in

Divide by 3

Substitute in   and

The volume of the largest rectangular box is then calculated as:

So, we have:

Hence, the volume of the largest rectangular box is 125/6

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Such a box has one vertex at the origin (0, 0, 0), and the vertex opposite this one is affixed to the plane with coordinates . The volume of the box is then , which we want to maximize subject to the constraint . Of course, we don't want a degenerate box, so we assume each of is positive.

We can use Lagrange multipliers - the Lagrangian is

with partial derivatives (set equal to 0)

Then

So the largest volume that can be attained is .

answer for #1 is c

answer for #2 is d

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