Grade 9 Reading Comprehension Worksheets | Reading comprehension ... - Free Printable

Based on the text provided in the image, here is a clear breakdown and solution to the problem.

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The Problem (Restated)

You are given a large rectangular sheet of cardboard. The goal is to create an open-top box by cutting out identical squares from each of the four corners and then folding up the sides.

The dimensions of the original cardboard sheet are:
- Length = 30 inches
- Width = 20 inches

Let the side length of each square cut from the corners be `x` inches.

After cutting and folding, you will have a box with:
- Height = x inches
- Length = (30 - 2x) inches (because you remove `x` from both ends)
- Width = (20 - 2x) inches (same reason)

The task is to find the value of `x` that maximizes the volume of the box.

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Step 1: Write the Volume Formula

Volume of a rectangular box = Length × Width × Height

So, the volume `V` as a function of `x` is:

> V(x) = x * (30 - 2x) * (20 - 2x)

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Step 2: Expand the Volume Function

First, multiply the two binomials:

(30 - 2x)(20 - 2x) =
= 30*20 - 30*2x - 2x*20 + 2x*2x
= 600 - 60x - 40x + 4x²
= 600 - 100x + 4x²

Now multiply by `x`:

> V(x) = x*(600 - 100x + 4x²)
> V(x) = 600x - 100x² + 4x³

Or, written in standard polynomial form:

> V(x) = 4x³ - 100x² + 600x

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Step 3: Find the Maximum Volume

To find the maximum volume, we need to find the critical points of `V(x)` by taking its derivative and setting it equal to zero.

#### First Derivative:

V'(x) = d/dx [4x³ - 100x² + 600x]
> V'(x) = 12x² - 200x + 600

#### Set V'(x) = 0:

12x² - 200x + 600 = 0

We can simplify this equation by dividing all terms by 4:

> 3x² - 50x + 150 = 0

#### Solve the Quadratic Equation:

Use the quadratic formula:

> x = [50 ± √( (-50)² - 4*3*150 )] / (2*3)
> x = [50 ± √(2500 - 1800)] / 6
> x = [50 ± √700] / 6

Simplify √700:

√700 = √(100 * 7) = 10√7 ≈ 10 * 2.64575 ≈ 26.4575

So:

> x ≈ [50 ± 26.4575] / 6

This gives two solutions:

1. x₁ ≈ (50 + 26.4575)/6 ≈ 76.4575/6 ≈ 12.74 inches
2. x₂ ≈ (50 - 26.4575)/6 ≈ 23.5425/6 ≈ 3.92 inches

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Step 4: Determine Which Value is Valid and Maximizes Volume

We must consider the physical constraints of the problem.

The width of the cardboard is 20 inches. Since we cut `x` from both sides, the remaining width is `20 - 2x`. This must be positive:

> 20 - 2x > 0 → x < 10

Similarly, for length: `30 - 2x > 0 → x < 15`

So the stricter constraint is x < 10.

Therefore, x ≈ 12.74 is invalid — it would result in negative width.

Only x ≈ 3.92 inches is physically possible.

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Step 5: Confirm It’s a Maximum

We can use the second derivative test.

V'(x) = 12x² - 200x + 600
V''(x) = 24x - 200

At x ≈ 3.92:

V''(3.92) ≈ 24*(3.92) - 200 ≈ 94.08 - 200 = -105.92 < 0

Since the second derivative is negative, the function is concave down at this point → local maximum.

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✔ Final Answer

To maximize the volume of the box, you should cut squares with side length approximately:

> x ≈ 3.92 inches

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Bonus: What is the Maximum Volume?

Plug x ≈ 3.92 back into the volume formula:

V(x) = x(30 - 2x)(20 - 2x)

≈ 3.92 * (30 - 7.84) * (20 - 7.84)
≈ 3.92 * 22.16 * 12.16

First, 22.16 * 12.16 ≈ 269.4656
Then, 3.92 * 269.4656 ≈ 1056.3 cubic inches

So the maximum volume is approximately 1056.3 in³.

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📌 Summary

- Optimal cut size: ~3.92 inches per corner
- Resulting box dimensions:
- Height: 3.92 in
- Length: 30 - 2*3.92 = 22.16 in
- Width: 20 - 2*3.92 = 12.16 in
- Maximum volume: ~1056.3 cubic inches

This is a classic optimization problem in calculus, often called the “box with maximum volume” problem. The key steps are: write the volume function, take its derivative, find critical points, and check which one makes sense physically and gives a maximum.