How do you 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+5y+8z=40?

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Answer 1 · 2016-07-14T02:04:14

.#1600/27# cubic units.

Let (a, b, c) be the vertex that is inside the octant.

Then, volume V = a b c.

The vertex is in the plane x + 5 y +8 z = 40..

So, a + 5 b + 8 c = 40. Eliminating a,

#V = V ( b, c ) = b c ( 40 - 5 b - 8 c )#

For mini/max V, the partial derivatives of V with respect to b and c

are 0.

So, #c ( 40 - 5 b - 8 c )+b c(-5) = c (40 - 10 b - c) ) =..0#.

And so, c = 0 giving the minimum V = 0 and

#40- 10 b - 8 c= (40 - 5 b - 8 c) - 5 b = a - 5 b = 0#, giving .

#b = a/5#.

Also, the partial derivative with respect to c is 0 gives

#b ( 40 - 5 b - 16 c) = b ( ( 40 - 5 b - 8 c )- 8 c) = b ( a - 8 c ) = 0#.

Here, b = 0 gives the same minimum V = 0 and #a - 8 c = 0#, giving

#c = a/8#

Using a + 5 b + 8 c = 40, 3 a = 30, and so,

a = 40/3.

correspondingly,

#b = 8/3 and c = 5/3#

Thus, this volume becomes #(40/3)(8/3)(5/3)= 1600/27 cubic

units. Minimum is 0, and so, this is the maximum.