5. A rectangular box open at the top is to have a volume of $$ 32 \mathrm{ft}^{3} $$. What must be the dimensions so that the total surface area is a minimum. Use the second derivative test. 6. Suppose that the coordinate of the corner of a rectangular parallelepiped in the first octant is $$ (x, y, z) $$. Show that the largest rectangular parallelepiped that can be inscribed in the sphere $$ x^{2}+y^{2}+z^{2}=1 $$ has a volume of $$ \frac{8}{3 \sqrt{3}} $$. Use the Lagrange multiplier method.

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**Problem 5**

Let the dimensions of the open-top box be:
- Length: $$ x $$
- Width: $$ y $$
- Height: $$ z $$

Since the box is open at the top, its total surface area is the area of the base plus the area of the four sides:
$$
S = xy + 2xz + 2yz.
$$

The volume constraint is:
$$
V = xyz = 32.
$$

**Step 1. Express $$ z $$ in terms of $$ x $$ and $$ y $$:**

$$
z = \frac{32}{xy}.
$$

**Step 2. Substitute $$ z $$ into the surface area formula:**

$$
S(x,y)= xy + 2x\left(\frac{32}{xy}\right) + 2y\left(\frac{32}{xy}\right).
$$
Simplify the side terms:
$$
2x\left(\frac{32}{xy}\right) = \frac{64}{y} \quad \text{and} \quad 2y\left(\frac{32}{xy}\right) = \frac{64}{x}.
$$
Thus,
$$
S(x,y)= xy + \frac{64}{x} + \frac{64}{y}.
$$

**Step 3. Use symmetry.**

The expression for $$ S(x,y) $$ is symmetric in $$ x $$ and $$ y $$. It is reasonable to assume the optimal solution occurs when $$ x = y = s $$. Then
$$
z = \frac{32}{s^2},
$$
and the surface area becomes:
$$
S(s) = s^2 + \frac{64}{s} + \frac{64}{s} = s^2 + \frac{128}{s}.
$$

**Step 4. Find the critical point for $$ S(s) $$.**

Differentiate $$ S(s) $$ with respect to $$ s $$:
$$
S'(s) = \frac{d}{ds}\left(s^2 + \frac{128}{s}\right) = 2s - \frac{128}{s^2}.
$$
Set $$ S'(s) = 0 $$:
$$
2s - \frac{128}{s^2} = 0 \quad \Longrightarrow \quad 2s = \frac{128}{s^2} \quad \Longrightarrow \quad 2s^3 = 128.
$$
Solve for $$ s $$:
$$
s^3 = 64 \quad \Longrightarrow \quad s = 4.
$$

**Step 5. Determine $$ z $$:**

$$
z = \frac{32}{4^2} = \frac{32}{16} = 2.
$$

**Step 6. Verify the minimum via the second derivative test:**

Differentiate $$ S'(s) $$ to get $$ S''(s) $$:
$$
S''(s) = \frac{d}{ds}\left(2s - \frac{128}{s^2}\right) = 2 + \frac{256}{s^3}.
$$
Evaluate at $$ s=4 $$:
$$
S''(4) = 2 + \frac{256}{4^3} = 2 + \frac{256}{64} = 2 + 4 = 6 > 0.
$$
Since $$ S''(4) > 0 $$, the critical point corresponds to a minimum.

**Answer for Problem 5:**  
The dimensions that minimize the total surface area are $$ x = 4 $$ ft, $$ y = 4 $$ ft, and $$ z = 2 $$ ft.

---

**Problem 6**

We wish to find the largest rectangular parallelepiped (box) that can be inscribed in the sphere
$$
x^2 + y^2 + z^2 = 1.
$$
Assume the box is centered at the origin with its faces parallel to the coordinate planes. Then one vertex in the first octant is at $$ (x, y, z) $$ with $$ x, y, z > 0 $$. The full box extends from $$ -x $$ to $$ x $$, $$ -y $$ to $$ y $$, and $$ -z $$ to $$ z $$. Hence, the dimensions of the box are:
$$
\text{Length} = 2x,\quad \text{Width} = 2y,\quad \text{Height} = 2z.
$$
The volume of the box is:
$$
V = (2x)(2y)(2z) = 8xyz.
$$

We wish to maximize $$ V = 8xyz $$ under the constraint:
$$
x^2 + y^2 + z^2 = 1.
$$

**Step 1. Set up the Lagrange multiplier formulation.**

Let
$$
f(x,y,z) = xyz \quad \text{and} \quad g(x,y,z) = x^2+y^2+z^2 - 1 = 0.
$$
We introduce a Lagrange multiplier $$ \lambda $$ and set:
$$
\nabla f = \lambda \nabla g.
$$

**Step 2. Compute the gradients.**

The gradient of $$ f $$ is:
$$
\nabla f = \left( yz,\; xz,\; xy \right).
$$
The gradient of $$ g $$ is:
$$
\nabla g = \left( 2x,\; 2y,\; 2z \right).
$$

Setting $$ \nabla f = \lambda \nabla g $$, we have:
$$
yz = \lambda 2x,\quad xz = \lambda 2y,\quad xy = \lambda 2z.
$$

**Step 3. Solve the system by symmetry.**

Assuming $$ x, y, z > 0 $$, notice that the equations are symmetric, suggesting that the maximum occurs when:
$$
x = y = z.
$$
Substitute $$ x = y = z = a $$ into the constraint:
$$
3a^2 = 1 \quad \Longrightarrow \quad a = \frac{1}{\sqrt{3}}.
$$

**Step 4. Find the maximum volume.**

Substitute $$ a = \frac{1}{\sqrt{3}} $$ into the volume expression:
$$
V = 8xyz = 8a^3 = 8\left(\frac{1}{\sqrt{3}}\right)^3.
$$
Since
$$
\left(\frac{1}{\sqrt{3}}\right)^3 = \frac{1}{3\sqrt{3}},
$$
we have:
$$
V = \frac{8}{3\sqrt{3}}.
$$

**Answer for Problem 6:**  
The largest rectangular parallelepiped that can be inscribed in the sphere $$ x^2+y^2+z^2=1 $$ has a volume of $$ \frac{8}{3\sqrt{3}} $$.
**Problem 5:** To minimize the total surface area of an open-top box with a volume of 32 cubic feet, the dimensions should be 4 feet in length, 4 feet in width, and 2 feet in height. **Problem 6:** The largest rectangular parallelepiped that fits inside the sphere $$ x^2 + y^2 + z^2 = 1 $$ has a volume of $$ \frac{8}{3\sqrt{3}} $$ cubic units.

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