To find the maximum volume of a rectangular box given a fixed wire length for the edges and a fixed surface area for the faces, we need to formulate the problem mathematically. Let's denote the dimensions of the rectangular box by $x$, $y$, and $z$.

Step 1: Set Up Equations

1. Perimeter Constraint (Wire Length) The total length of the wire is the sum of all the edges of the box. A rectangular box has 12 edges: $4(x + y + z) = 40 \implies x + y + z = 10 \, \text{cm}$

2. Surface Area Constraint The surface area of a rectangular box is given by: $2(xy + yz + zx) = 45 \implies xy + yz + zx = 22.5 \, \text{cm}^2$

3. Volume of the Box The volume $V$ of the box is given by: $V = xyz$

Step 2: Express One Variable in Terms of the Others

From the perimeter constraint: $z = 10 - x - y$

Substitute $z$ in the surface area equation: $xy + y(10 - x - y) + x(10 - x - y) = 22.5$ This simplifies to: $xy + 10y - xy - y^2 + 10x - x^2 - xy = 22.5$ $10x + 10y - x^2 - y^2 - 2xy = 22.5$

Step 3: Set Up the Optimization Problem

The volume $V$ becomes: $V = xy(10 - x - y)$

To maximize this volume, we need to find the critical points by taking the partial derivatives with respect to $x$ and $y$ and setting them to zero.

Step 4: Solving the Equations

Let's solve these equations using optimization techniques.

Partial Derivatives

$\frac{\partial V}{\partial x} = y(10 - x - y) - xy = 0$ $\frac{\partial V}{\partial y} = x(10 - x - y) - xy = 0$

Let's solve this system of equations step-by-step.

The solutions for the critical points $(x, y)$ are:

1. $(0, 0)$
2. $(0, 10)$
3. $\left(\frac{10}{3}, \frac{10}{3}\right)$
4. $(10, 0)$

Step 5: Determine Valid Solutions and Maximum Volume

To be valid, the dimensions $x$, $y$, and $z = 10 - x - y$ must be positive. Let's check which of these solutions are feasible and calculate the corresponding volume $V = xyz$.

1. For $(0, 0)$: $z = 10 - 0 - 0 = 10$; Volume $V = 0 \cdot 0 \cdot 10 = 0$
2. For $(0, 10)$: $z = 10 - 0 - 10 = 0$; Volume $V = 0 \cdot 10 \cdot 0 = 0$
3. For $\left(\frac{10}{3}, \frac{10}{3}\right)$: $z = 10 - \frac{10}{3} - \frac{10}{3} = \frac{10}{3}$
4. For $(10, 0)$: $z = 10 - 10 - 0 = 0$; Volume $V = 10 \cdot 0 \cdot 0 = 0$

The only feasible solution is $\left(\frac{10}{3}, \frac{10}{3}, \frac{10}{3}\right)$. Let's calculate the volume for this solution.The maximum volume of the rectangular box that can be achieved, given the constraints, is approximately 37.04 cubic centimeters.

Summary:

• Dimensions: $x = \frac{10}{3} \, \text{cm}$, $y = \frac{10}{3} \, \text{cm}$, $z = \frac{10}{3} \, \text{cm}$
• Maximum Volume: $\approx 37.04 \, \text{cm}^3$

Would you like further details on the steps, or do you have any questions?

Related Questions:

1. How can you determine the dimensions of a box that maximizes the surface area given a fixed volume?
2. What is the relationship between surface area and volume for different shapes?
3. How do you use calculus to find maximum and minimum values in optimization problems?
4. What is the importance of the Lagrange multipliers method in optimization problems with constraints?
5. How does changing the wire length or surface area affect the maximum volume of the box?

Tip: When solving optimization problems with multiple constraints, it's often helpful to express all variables in terms of a single variable to simplify the calculations.