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=6.

Let (x, y, z) be a point on the plane in the first octant. The box formed by this point has volume xyz, and you want to maximize this subject to the equation of the plane.

Use the method of Lagrange multipliers: the Lagrangian is

L(x, y, z) = xyz - λ (x + 2y + 3z - 6)

Find its critical points:

∂L/∂x = yz - λ = 0

∂L/∂y = xz - 2λ = 0

∂L/∂z = xy - 3λ = 0

∂L/∂λ = -(x + 2y + 3z - 6) = 0

Solving the first three equations for λ gives

λ = yz = xz/2 = xy/3

Solve these equations for y and z :

• yz = xz/2   =>   y = x/2   =>   2y = x

• yz = xy/3   =>   z = x/3   =>   3z = x

Substitute these solutions into the last equation and solve for x, then again for y and z :

x + 2y + 3z - 6 = 3x - 6 = 0   =>   3x = 6   =>   x = 2, y = 1, z = 2/3

At this critical point, the maximum volume is

xyz = 2*1*2/3 = 4/3

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.