A rectangular box is to be made form a sheet of 24 inch length and 9 inch wideth cutting out indetical squares of side length x from the four corners and turning up the sides
What is the value of x for width the vulume is maximum ?

Options

A. 1 inch
B. 1.5 inch
C. 2 inch
D. 2.5 inch

Explanation

To find the value of \( x \) for which the volume of the rectangular box is maximum, we will follow these steps:

### Step 1: Define the dimensions of the box
The original dimensions of the sheet are:
- Length = 24 inches
- Width = 9 inches

When we cut out squares of side length \( x \) from each corner, the new dimensions of the base of the box will be:
- New Length = \( 24 - 2x \)
- New Width = \( 9 - 2x \)

### Step 2: Write the volume function
The volume \( V \) of the box can be expressed as:
\[ V(x) = \text{Length} \times \text{Width} \times \text{Height} = (24 - 2x)(9 - 2x)(x) \]
Expanding this, we have:
\[ V(x) = (24 - 2x)(9 - 2x)x \]
Calculating the product:
\[ V(x) = (216 - 48x - 18x + 4x^2)x = 4x^3 - 66x^2 + 216x \]

### Step 3: Differentiate the volume function
To find the maximum volume, we need to differentiate \( V(x) \) with respect to \( x \):
\[ V'(x) = \frac{d}{dx}(4x^3 - 66x^2 + 216x) = 12x^2 - 132x + 216 \]

### Step 4: Set the derivative to zero
To find the critical points, set \( V'(x) = 0 \):
\[ 12x^2 - 132x + 216 = 0 \]
Dividing the entire equation by 12 simplifies it to:
\[ x^2 - 11x + 18 = 0 \]

### Step 5: Factor the quadratic equation
Now we factor the quadratic:
\[ (x - 9)(x - 2) = 0 \]
Thus, the solutions are:
\[ x = 9 \quad \text{or} \quad x = 2 \]

### Step 6: Determine which critical point gives maximum volume
We need to check the second derivative to determine whether these points are maxima or minima:
\[ V''(x) = \frac{d}{dx}(12x^2 - 132x + 216) = 24x - 132 \]
Now, we evaluate \( V''(x) \) at both critical points:
1. For \( x = 9 \):
\[ V''(9) = 24(9) - 132 = 216 - 132 = 84 \quad (\text{positive, indicating a local minimum}) \]
2. For \( x = 2 \):
\[ V''(2) = 24(2) - 132 = 48 - 132 = -84 \quad (\text{negative, indicating a local maximum}) \]

### Conclusion
The value of \( x \) for which the volume is maximum is:
\[ \boxed{2 \text{ inches}} \]