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

Read more about volumes at:

link

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 .

step-by-step explanation:

3024 rhe base is what up in worry about

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3