Use Lagrange multipliers to find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the given plane. x 3y 4z

I assume the given plane has equation

x + 3y + 4z = d

for some constant d ≠ 0.

You want to maximize the function V(x, y, z) = xyz subject to the above constraint. The Lagrangian would be

L(x, y, z, λ) = xyz - λ (x + 3y + 4z - d)

Find its critical points:

∂L/∂x = yz - λ = 0

∂L/∂y = xz - 3λ = 0

∂L/∂z = xy - 4λ = 0

∂L/∂λ = x + 3y + 4z - d = 0

The first three equations tell you that λ = yz = xz/3 = xy/4. Then, for instance,

yz = xz/3   →   3yz - xz = z (3y - x) = 0   →   z = 0   or   3y - x = 0

If z = 0, we get zero volume which is not useful. So 3y = x. Similarly, you would find x = 4z and 3y = 4z, so a critical point occurs when x = 3y = 4z.

In the fourth equation, we get upon substituting

x + 3y + 4z = 3x = d   →   x = d/3, y = d/9, z = d/12

Then the largest volume of this box would be

V(d/3, d/9, d/12) = d ³ / 324

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2

b

step-by-step explanation: