Let's go through each part of the problem and fill in the blanks step-by-step.

Problem: A rectangular box without a lid is made from $108 \, \text{m}^2$ of cardboard. We need to find the maximum volume of such a box.

Given:

• Let $x$, $y$, and $z$ be the length, width, and height of the box.
• The volume of the box, $V$, is given by: $V = xyz$

1. Expressing $V$ in terms of $x$ and $y$: Since the box has no lid, the surface area consists of four sides and the bottom, leading to the equation: $2xz + 2yz + xy = 108$ The missing blank should be: $2yz$

2. Solving for $z$: Rearranging the surface area equation to solve for $z$: $z = \frac{108 - xy}{2(x + y)}$

3. Expressing $V(x, y)$: Substitute $z = \frac{108 - xy}{2(x + y)}$ into $V = xyz$: $V(x, y) = xy \cdot \frac{108 - xy}{2(x + y)} = \frac{xy(108 - xy)}{2(x + y)}$

4. Partial Derivatives: To find the critical points, we calculate the partial derivatives $\frac{\partial V}{\partial x}$ and $\frac{\partial V}{\partial y}$, then set them to zero.

5. Finding Critical Points: Solve the equations to find $x$ and $y$ such that both partial derivatives are zero.

Would you like a detailed explanation of each partial derivative calculation or further steps on solving for $x$ and $y$?

Related Questions:

1. How does the constraint on surface area help determine $z$ in terms of $x$ and $y$?
2. Why do we use partial derivatives to find the maximum volume?
3. What is the significance of setting the partial derivatives to zero?
4. How would the solution change if the box had a lid?
5. Can we verify the maximum volume using the Second Derivative Test?

Tip:

When maximizing functions with constraints, try to reduce the number of variables by using the constraint equation.