To solve this problem, we aim to maximize the volume of a rectangular box under the constraint that its total surface area is $64 \, \text{cm}^2$.

Step 1: Variables and Formulas

Let the dimensions of the box be:

• Length: $x$,
• Width: $y$,
• Height: $z$.

Volume:

The volume $V$ of the box is: $V = x \cdot y \cdot z$

Surface Area:

The total surface area $S$ is given as: $S = 2(xy + xz + yz) = 64 \, \text{cm}^2$

We aim to maximize $V$ subject to $S = 64$.

Step 2: Using Constraints

From the surface area formula: $xy + xz + yz = 32$

Rewriting $z$ in terms of $x$ and $y$: $z = \frac{32 - xy}{x + y}$

Step 3: Substituting $z$ into Volume

Substitute $z = \frac{32 - xy}{x + y}$ into $V$: $V = x \cdot y \cdot \frac{32 - xy}{x + y}$

Simplify: $V = \frac{32xy - (xy)^2}{x + y}$

Step 4: Optimization

To find the maximum $V$, we take partial derivatives with respect to $x$ and $y$, and solve for the critical points. This process involves solving: $\frac{\partial V}{\partial x} = 0 \quad \text{and} \quad \frac{\partial V}{\partial y} = 0$

For simplicity, a symmetric box ($x = y = z$) often maximizes the volume. Assuming this symmetry: $x = y = z$

From the surface area constraint: $2(x^2 + x^2 + x^2) = 64$ $6x^2 = 64 \quad \Rightarrow \quad x^2 = \frac{64}{6} \quad \Rightarrow \quad x = \sqrt{\frac{32}{3}} \approx 3.27 \, \text{cm}$

Thus: $x = y = z \approx 3.27 \, \text{cm}$

Final Dimensions

The box with the maximum volume is approximately: $x = y = z \approx 3.27 \, \text{cm}$

Would you like detailed steps for the optimization or solving the derivatives?

Related Questions:

1. How is the formula for surface area derived for a rectangular box?
2. Why does symmetry (equal dimensions) maximize the volume for this problem?
3. Can this problem be solved using Lagrange multipliers? How?
4. What happens to the maximum volume if the surface area constraint changes?
5. What is the geometric interpretation of maximizing the volume under a constraint?

Tip:

When optimizing under constraints, always check if symmetry simplifies the problem—it often reduces the complexity significantly!