Find the volume of the largest rectangular box in the first octant with three fa

Question

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. I'm not sure what the question is asking, or how to set up this problem. Please Show Steps.

Explanation / Answer

We want to maximize the volume V = xyz subject to the constraint g = x + 2y + 3z = 6. We need to solve grad(V) = t * grad(g) ==> (yz, xz, xy) = t * (1,2,3) Thus, yz = t, xz = 2t, and xy = 3t ==> xyz = xt = 2yt = 3zt Adding these equations yields 3xyz = (x + 2y + 3z)t = 6t ==> xyz = 2t. Thus, 2t = xt = 2yt = 3zt. (t can't be 0, since that would force x,y,z = 0, which is impractical since x,y,z > 0 in this word problem.) Therefore, 2 = x = 2y = 3z ==> x = 2, y = 1, and z = 2/3. So, the maximal volume is V = xyz = 4/3.

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