An open topped box is to be constructed by removing equal squares from each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the sides. Find the volume of the largest such box.

Options

A. `200/3 m^3`
B. `250/3 m^3`
C. `100/3 m^3`
D. `350/3 m^3`

Explanation

To find the volume of the largest open-topped box that can be constructed by removing equal squares from each corner of a 3m by 8m rectangular sheet of aluminum, we can follow these steps:

### Step 1: Define the variables
Let \( x \) be the length of the side of the square cut from each corner. After cutting out the squares, the dimensions of the box will be:
- Length = \( 8 - 2x \)
- Width = \( 3 - 2x \)
- Height = \( x \)

### Step 2: Write the volume formula
The volume \( V \) of the box can be expressed as:
\[ V = \text{Length} \times \text{Width} \times \text{Height} = x(8 - 2x)(3 - 2x) \]

### Step 3: Expand the volume expression
Now, we need to expand the expression for volume:
\[ V = x(8 - 2x)(3 - 2x) \]
First, expand \( (8 - 2x)(3 - 2x) \):
\[ (8 - 2x)(3 - 2x) = 24 - 16x + 6x - 4x^2 = 24 - 10x - 4x^2 \]
Now, substituting back into the volume equation:
\[ V = x(24 - 10x - 4x^2) = 24x - 10x^2 - 4x^3 \]

### Step 4: Differentiate the volume function
To find the maximum volume, we differentiate \( V \) with respect to \( x \):
\[ \frac{dV}{dx} = 24 - 20x - 12x^2 \]

### Step 5: Set the derivative to zero
To find critical points, set the derivative equal to zero:
\[ 24 - 20x - 12x^2 = 0 \]
Rearranging gives:
\[ 12x^2 + 20x - 24 = 0 \]
Dividing the entire equation by 4:
\[ 3x^2 + 5x - 6 = 0 \]

### Step 6: Factor the quadratic equation
Next, we factor the quadratic:
\[ (3x - 2)(x + 3) = 0 \]
Thus, the solutions are:
\[ x = \frac{2}{3} \quad \text{and} \quad x = -3 \]
Since \( x \) must be positive, we take \( x = \frac{2}{3} \).

### Step 7: Determine if it is a maximum
To confirm that this critical point is a maximum, we can use the second derivative test:
\[ \frac{d^2V}{dx^2} = -20 - 24x \]
Evaluating at \( x = \frac{2}{3} \):
\[ \frac{d^2V}{dx^2} = -20 - 24 \left(\frac{2}{3}\right) = -20 - 16 = -36 \]
Since this is negative, \( x = \frac{2}{3} \) is indeed a maximum.

### Step 8: Calculate the maximum volume
Now, substitute \( x = \frac{2}{3} \) back into the volume formula:
\[ V = \left(\frac{2}{3}\right) \left(8 - 2 \cdot \frac{2}{3}\right) \left(3 - 2 \cdot \frac{2}{3}\right) \]
Calculating the dimensions:
\[ = \left(\frac{2}{3}\right) \left(8 - \frac{4}{3}\right) \left(3 - \frac{4}{3}\right) \]
\[ = \left(\frac{2}{3}\right) \left(\frac{24}{3} - \frac{4}{3}\right) \left(\frac{9}{3} - \frac{4}{3}\right) \]
\[ = \left(\frac{2}{3}\right) \left(\frac{20}{3}\right) \left(\frac{5}{3}\right) \]
\[ = \frac{2 \cdot 20 \cdot 5}{27} = \frac{200}{27} \text{ cubic meters} \]

### Final Answer
Thus, the maximum volume of the box is:
\[ \frac{200}{27} \text{ cubic meters} \]