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To solve this problem, let's break it down step by step:

We are tasked with creating an open-top box by cutting equal-sized squares (of side \( x \)) from each corner of a rectangular sheet of metal and folding up the sides. The dimensions of the sheet are given as:

- Length \( l = 200 \, \text{mm} \)
- Width \( w = 150 \, \text{mm} \)

### Step 1: Express the Volume of the Box

The volume \( V \) of the open-top box is given by:

\[
V = \text{length} \times \text{width} \times \text{height}
\]

The dimensions of the base of the box, after cutting out squares of side \( x \) from each corner, will be:

- New length of the box = \( l - 2x \)
- New width of the box = \( w - 2x \)
- The height of the box is \( x \) (since this is the height formed by folding up the sides).

Thus, the volume of the box is:

\[
V(x) = (200 - 2x)(150 - 2x)x
\]

Expanding this expression:

\[
V(x) = x(200 - 2x)(150 - 2x)
\]

\[
V(x) = x[(200 \times 150) - 2 \times 200x - 2 \times 150x + 4x^2]
\]

\[
V(x) = x[30000 - 700x + 4x^2]
\]

\[
V(x) = 30000x - 700x^2 + 4x^3
\]

### Step 2: Find the Value of \( x \) that Maximizes the Volume

To maximize the volume, we need to take the derivative of \( V(x) \) with respect to \( x \), and set it equal to zero:

\[
\frac{dV}{dx} = 30000 - 1400x + 12x^2
\]

Set \( \frac{dV}{dx} = 0 \):

\[
12x^2 - 1400x + 30000 = 0
\]

This is a quadratic equation. We can solve it using the quadratic formula:

\[
x = \frac{-(-1400) \pm \sqrt{(-1400)^2 - 4 \times 12 \times 30000}}{2 \times 12}
\]

\[
x = \frac{1400 \pm \sqrt{1960000 - 1440000}}{24}
\]

\[
x = \frac{1400 \pm \sqrt{520000}}{24}
\]

\[
x = \frac{1400 \pm 721.11}{24}
\]

Thus, we have two possible solutions for \( x \):

\[
x = \frac{1400 + 721.11}{24} = \frac{2121.11}{24} \approx 88.38 \, \text{mm}
\]

\[
x = \frac{1400 - 721.11}{24} = \frac{678.89}{24} \approx 28.29 \, \text{mm}
\]

Since \( x = 88.38 \, \text{mm} \) would result in negative dimensions for the base, we discard this solution. Hence, the valid value for \( x \) is:

\[
x \approx 28.29 \, \text{mm}
\]

### Step 3: Calculate the Maximum Volume

Substitute \( x = 28.29 \, \text{mm} \) into the volume equation to find the maximum volume:

\[
V(28.29) = 30000(28.29) - 700(28.29)^2 + 4(28.29)^3
\]

Calculating each term:

\[
V(28.29) = 30000 \times 28.29 - 700 \times (28.29)^2 + 4 \times (28.29)^3
\]
\[
V(28.29) \approx 848700 - 55983.64 + 9025.82
\]
\[
V(28.29) \approx 803742.18 \, \text{mm}^3
\]

So, the maximum volume is approximately **803742 mm³**.

### Step 4: Comment on the Value Obtained

The value obtained represents the maximum volume that can be achieved for the given dimensions of the metal sheet. It demonstrates how optimizing the cut size maximizes the volume, and any deviation from this optimal cut would result in a smaller volume.

---

Would you like further details or clarifications? 

Here are some related questions:

1. How would the maximum volume change if the width \( w \) was 180 mm instead of 150 mm?
2. Can you derive the formula for the maximum volume for a general \( l \) and \( w \)?
3. What is the relationship between the dimensions of the sheet and the height of the box that produces maximum volume?
4. How does the shape of the base change as \( x \) approaches the limits of its range?
5. What is the maximum possible height of the box before it no longer forms a box?

**Tip:** In optimization problems, critical points (found by setting the derivative equal to zero) often indicate maximum or minimum values, but remember to check which by using the second derivative or the context of the problem.

You plan to make a simple, open-topped box from a piece of sheet metal by cutting a square of equal size from each corner and folding up the sides. Given l = 200mm and w = 150mm, calculate: a) The value of x which will give the maximum volume, b) The maximum volume of the box, c) Comment on the value obtained in part b.

Learn how to maximize the volume of an open-topped box created by cutting squares from the corners of a metal sheet. This detailed solution involves optimization, quadratic equations, and differentiation, making it ideal for high school students (Grades 10-12).

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