Exercise 38

A rectangular box with length $x$, width $y$, and height $z$ is being built. The box is positioned so that one corner is stationed at the origin and the box lies in the first octant where $x$, $y$, and $z$ are all positive. There is an added constraint on how the box is constructed: it must fit underneath the plane with equation $x + 2y + 3z = 6$. In fact, we will assume that the corner of the box “opposite” the origin must actually lie on this plane. The basic problem is to find the maximum volume of the box.

1. Sketch the plane $x + 2y + 3z = 6$, as well as a picture of a potential box. Label everything appropriately.

2. Explain how you can use the fact that one corner of the box lies on the plane to write the volume of the box as a function of $x$ and $y$ only. Do so, and clearly show the formula you find for $V(x,y)$.

3. Find all critical points of $V$. (Note that when finding the critical points, it is essential that you factor first to make the algebra easier.)

4. Without considering the current applied nature of the function $V$, classify each critical point you found above as a local maximum, local minimum, or saddle point of $V$.

5. Determine the maximum volume of the box, justifying your answer completely with an appropriate discussion of the critical points of the function.

6. Now suppose that we instead stipulated that, while the vertex of the box opposite the origin still had to lie on the plane, we were only going to permit the sides of the box, $x$ and $y$, to have values in a specified range (given below). That is, we now want to find the maximum value of $V$ on the closed, bounded region $$\begin{gathered} \frac{1}{2} \le x \le 1, \ \ 1 \le y \le 2. \end{gathered}$$ Find the maximum volume of the box under this condition, justifying your answer fully.