
Maximizing a volume
I find the following problem very hard!
A rectangular box has 3 faces in the planes given by $\displaystyle x=0$, $\displaystyle z=0$ and $\displaystyle y=0$. The vertex that is not in the latter planes is in the plane $\displaystyle 4x+3y+z=36$. 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.

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 $\displaystyle 4x+3y+z=36$ has coordinates $\displaystyle (x,y,z)$ then the volume of the box is $\displaystyle xyz$.
So the problem boils down to maximising the quantity $\displaystyle xyz$ subject to the conditions that $\displaystyle x$, $\displaystyle y$, $\displaystyle z$ are positive and $\displaystyle 4x+3y+z=36$.
Now try the AMGM inequality, namely $\displaystyle \frac{4x+3y+z}3\geq\root 3\of{4x.3y.z}$ with equality iff $\displaystyle 4x=3y=z$.
You should now be able to deduce that the maximum volume is 144. Happy hunting.

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.

Instead of using the inequality, I used Lagrange's multipliers method. I found $\displaystyle x=3$, $\displaystyle y=4$ and $\displaystyle z=12$, hence $\displaystyle V=144$ as you pointed out.
I still have some difficulties to see why we can suppose the vertex to have the coordinates $\displaystyle (x,y,z)$. I don't know if I could solve a similar problem. (Hopefully yes).