To solve this problem, let's set up an equation based on the information given.

Problem Breakdown

1. The initial dimensions of the metal sheet are:

   • Length = 30 cm
   • Width = 18 cm

2. Squares of side length $x$ are cut from each corner.

3. After cutting the squares and folding the sides upwards, the dimensions of the open box formed are:

   • New Length = $30 - 2x$
   • New Width = $18 - 2x$
   • Height = $x$

Step 1: Express the Volume of the Box

The volume $V$ of the box is given by: $V = \text{length} \times \text{width} \times \text{height}$ So, $V = (30 - 2x)(18 - 2x)x$

Step 2: Expand the Equation

Let's expand $(30 - 2x)(18 - 2x)x$: $V = (540 - 60x - 36x + 4x^2)x$ Simplifying, $V = 4x^3 - 96x^2 + 540x$

Step 3: Set Up the Equation

Since we're only asked to determine the values of $x$ (not the volume), we set the equation $V = 0$ (considering the boundary conditions): $4x^3 - 96x^2 + 540x = 0$ Dividing through by $x$ (assuming $x \neq 0$): $4x^2 - 96x + 540 = 0$

Step 4: Simplify and Use the Quadratic Formula

Now, divide by 4 to simplify: $x^2 - 24x + 135 = 0$

This is a quadratic equation in the form $ax^2 + bx + c = 0$, where:

• $a = 1$
• $b = -24$
• $c = 135$

Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute the values of $a$, $b$, and $c$: $x = \frac{-(-24) \pm \sqrt{(-24)^2 - 4 \cdot 1 \cdot 135}}{2 \cdot 1}$ $x = \frac{24 \pm \sqrt{576 - 540}}{2}$ $x = \frac{24 \pm \sqrt{36}}{2}$ $x = \frac{24 \pm 6}{2}$

This gives us two solutions:

1. $x = \frac{24 + 6}{2} = 15$
2. $x = \frac{24 - 6}{2} = 9$

Solution

The two values of $x$ are: $x = 15 \text{ cm} \quad \text{and} \quad x = 9 \text{ cm}$

Would you like more details on any part of the solution, or have questions?

Further Related Questions

1. What is the volume of the box when $x = 9$ cm?
2. What is the volume of the box when $x = 15$ cm?
3. Why are only certain values of $x$ practical for this open box problem?
4. What happens to the dimensions of the box if we increase the initial size of the metal sheet?
5. How does changing $x$ affect the overall volume of the box?

Tip

Always ensure that the values you solve for are within practical limits for the problem. For example, here $x$ must be less than half of the smallest dimension (18 cm), so values of $x$ greater than 9 cm would not make sense geometrically.