# Maximizing a volume

• Jul 8th 2009, 01:01 PM
arbolis
Maximizing a volume
I find the following problem very hard!
A rectangular box has 3 faces in the planes given by $x=0$, $z=0$ and $y=0$. The vertex that is not in the latter planes is in the plane $4x+3y+z=36$. Determine the dimensions that maximizes the volume of the box.
I tried to visualize the box by drawing a sketch and I think I couldn't. So I'm stuck on giving an expression of the volume of the box. I'd like some help for this task.
Precisely I've drew a rectangular parallelepiped in the xyz plane but I don't see how it could have a maximum volume so I think I drew it wrongly.
• Jul 8th 2009, 01:57 PM
halbard
We'll assume that the box is suppose to lie in the first octant. This seems like a reasonable starting point.

If the vertex in the plane $4x+3y+z=36$ has coordinates $(x,y,z)$ then the volume of the box is $xyz$.

So the problem boils down to maximising the quantity $xyz$ subject to the conditions that $x$, $y$, $z$ are positive and $4x+3y+z=36$.

Now try the AM-GM inequality, namely $\frac{4x+3y+z}3\geq\root 3\of{4x.3y.z}$ with equality iff $4x=3y=z$.

You should now be able to deduce that the maximum volume is 144. Happy hunting.
• Jul 8th 2009, 07:08 PM
arbolis
Ok thanks a lot. I'll try it.
Err... I would have never thought to suppose that the box has to lie in the first octant. I hope the question will be clearer in the exam.
• Jul 10th 2009, 01:52 PM
arbolis
Instead of using the inequality, I used Lagrange's multipliers method. I found $x=3$, $y=4$ and $z=12$, hence $V=144$ as you pointed out.
I still have some difficulties to see why we can suppose the vertex to have the coordinates $(x,y,z)$. I don't know if I could solve a similar problem. (Hopefully yes).