To solve this problem, we need to minimize the surface area of the open box, subject to the constraint that its volume is fixed at 6912 cm³. Let's break it down step by step:

(a) Finding the function to minimize (Surface Area)

The surface area $A$ of an open box with width $x$, length $y$, and height $z$ can be expressed as:

$A(x, y, z) = \text{Area of the bottom} + \text{Area of the sides}$

• The bottom has area $xy$ (length × width).
• The box has 2 sides with area $xz$ (width × height) and 2 sides with area $yz$ (length × height).

Thus, the surface area is given by: $A(x, y, z) = xy + 2xz + 2yz$

So, the function to minimize is:

$f(x, y, z) = xy + 2xz + 2yz$

(b) Finding the constraint (Volume)

The volume $V$ of the box is given by the product of its width, length, and height: $V = x \cdot y \cdot z$

We are told that the volume is 6912 cm³, so the constraint is: $x \cdot y \cdot z = 6912$

Thus, the constraint function $g(x, y, z)$ is:

$g(x, y, z) = x \cdot y \cdot z - 6912 = 0$

Summary:

• Function to minimize (Surface Area):
  $f(x, y, z) = xy + 2xz + 2yz$

• Constraint (Volume):
  $g(x, y, z) = x \cdot y \cdot z - 6912 = 0$

Would you like me to continue solving for the dimensions that minimize the surface area? Let me know if you'd like additional details or explanations.

Further Questions:

1. How would you use the method of Lagrange multipliers to solve this problem?
2. What would happen to the optimal dimensions if the volume were increased to 8000 cm³?
3. Can we solve this problem using partial derivatives and optimization techniques?
4. How would you approach the problem if the box had a top?
5. How does changing the shape of the box (e.g., making it more cubic) affect the material used?

Tip: The method of Lagrange multipliers is powerful for optimizing functions subject to constraints, especially when dealing with surface areas and volumes.