volume of box w/ vertex on x/a + y/b + z/c = 1

[cheffy]

Find the volume of the largest closed rectangular box in the first octant having three faces in the coordinate planes and a vertex on the plane x/a + y/b +z/c = 1, where a > 0, b > 0, and c > 0.

How does the having three faces in coordinate planes and first octant play into this problem? Thanks!

 

[stapel]

How does the having three faces in coordinate planes and first octant play into this problem?

I think it's supposed to make your computations simpler. Rather than dealing with messy equations and negative numbers, the box has been rotated, etc, to put one corner at the origin, the faces on planes, and the box's volume inside the all-positive first octant.

Eliz.

 

[galactus]

\(\displaystyle \L\\\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\).....[1]

\(\displaystyle \L\\V=xyz\)

From [1], we can solve for z and sub into V:

\(\displaystyle \L\\V=xy\left(\frac{-cx}{a}-\frac{cy}{c}+c\right)\)

\(\displaystyle \L\\V_{x}=\frac{-cy}{ab}\left(2bx+a(y-b)\right)\)

\(\displaystyle \L\\V_{y}=\frac{-cx}{ab}\left(bx+a(2y-b)\right)\)

We can disregard the cases where x=0 or y=0, since these correspond to boundary points. So,

\(\displaystyle \L\\2bx+a(y-b)=0\)
\(\displaystyle \L\\bx+a(2y-b)=0\)

Now, solve the system.

 

[cheffy]

Where is the lagrange multiplier in that system?

I tried to solve the system and I got v=1, where x, y , z all = 1. Does that even make sense?

 

[galactus]

You have to use Lagrange multipliers?. You didn't state that.

If you solve the first equation for y, you get: \(\displaystyle \L\\y=b-\frac{2bx}{a}\)

Sub into the second:

\(\displaystyle \L\\bx+2a(b-\frac{2bx}{a})-ab=ab-3bx\)

\(\displaystyle \L\\ab-3bx=0\)

\(\displaystyle \L\\x=\frac{a}{3}\)

Which leads to \(\displaystyle \L\\y=\frac{b}{3}, \;\ z=\frac{c}{3}\)

The max volume is \(\displaystyle \L\\\fbox{\frac{abc}{27}}\)


You can adapt it to lagrange if necessary.