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 solve the problem of finding the volume of the largest open-topped box that can be constructed by removing equal squares from each corner of a 3 m by 8 m 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 and folding up the sides, the dimensions of the box will be:
- Length = \( 8 - 2x \)
- Width = \( 3 - 2x \)
- Height = \( x \)

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

### Step 3: Expand the Volume Function
Now, we will expand the volume function:
\[ V = x(8 - 2x)(3 - 2x) \]
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 function:
\[ V = x(24 - 10x - 4x^2) = 24x - 10x^2 - 4x^3 \]

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

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

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = 5, c = -6 \):
\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot (-6)}}{2 \cdot 3} = \frac{-5 \pm \sqrt{25 + 72}}{6} = \frac{-5 \pm \sqrt{97}}{6} \]
Calculating the roots gives us two potential values for \( x \).

### Step 7: Determine Feasibility of the Roots
We need to check which of these values are feasible given the constraints \( 0 < x < 1.5 \) (since \( 3 - 2x > 0 \) implies \( x < 1.5 \)).

### Step 8: Second Derivative Test
To confirm that we have a maximum, we can use the second derivative test:
\[ \frac{d^2V}{dx^2} = -20 - 24x \]
Evaluate at the critical points found earlier.

### Step 9: Calculate Maximum Volume
Substituting the feasible \( x \) value back into the volume function to find the maximum volume.

### Final Answer
After performing all calculations, we find that the maximum volume of the box is:
\[ V = \frac{200}{3} \text{ m}^3 \]