The problem is about a closed cuboid box with the following dimensions and information:

• Width $x$ cm
• Length $y$ cm
• Height $5$ cm (given)
• Volume of the box is $40 \, \text{cm}^3$
• External surface area of the box is $76 \, \text{cm}^2$

We are asked to form two equations using this information and solve for $x$ and $y$.

Step 1: Volume Equation

The volume $V$ of a cuboid is given by: $V = x \times y \times 5$ Since the volume is $40 \, \text{cm}^3$, we have: $x \times y \times 5 = 40$ $x \times y = \frac{40}{5} = 8 \quad \text{(Equation 1)}$

Step 2: Surface Area Equation

The surface area $S$ of a cuboid is given by: $S = 2(xy + xh + yh)$ Substituting $h = 5$ and the given surface area $S = 76 \, \text{cm}^2$: $76 = 2(xy + 5x + 5y)$ $38 = xy + 5x + 5y$ Using $xy = 8$ from Equation 1, we can substitute $xy = 8$ into this: $38 = 8 + 5x + 5y$ $30 = 5x + 5y$ $6 = x + y \quad \text{(Equation 2)}$

Step 3: Solving the System of Equations

Now, we have two equations:

1. $xy = 8$
2. $x + y = 6$

To solve this system: From Equation 2, solve for $y$ in terms of $x$: $y = 6 - x$ Substitute this into Equation 1: $x(6 - x) = 8$ $6x - x^2 = 8$ $x^2 - 6x + 8 = 0$ Solve this quadratic equation using the quadratic formula: $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(8)}}{2(1)}$ $x = \frac{6 \pm \sqrt{36 - 32}}{2}$ $x = \frac{6 \pm \sqrt{4}}{2}$ $x = \frac{6 \pm 2}{2}$ So, the two possible values for $x$ are: $x = \frac{6 + 2}{2} = 4 \quad \text{or} \quad x = \frac{6 - 2}{2} = 2$

Step 4: Find Corresponding Values of $y$

For $x = 4$: $y = 6 - 4 = 2$

For $x = 2$: $y = 6 - 2 = 4$

Final Solution:

The two possible sets of dimensions are:

1. $x = 4 \, \text{cm}, y = 2 \, \text{cm}, h = 5 \, \text{cm}$
2. $x = 2 \, \text{cm}, y = 4 \, \text{cm}, h = 5 \, \text{cm}$

Would you like further details or explanations on any part of this process?

Related Questions:

1. How is the surface area of a cuboid calculated for different dimensions?
2. What is the general formula for the volume of a cuboid?
3. How can the quadratic formula be derived and applied to different types of equations?
4. What are the practical applications of cuboids in real-world packaging or construction?
5. How would the process change if the height of the box was unknown?

Tip:

When solving systems of equations, always check for possible simplifications early, like isolating one variable, which can make the overall process easier.