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

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1