A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum ?

Options

A. `x=18`
B. `x=24`
C. `x=5`
D. None of these

Explanation

To solve the problem of maximizing the volume of a box made from a rectangular sheet of tin by cutting squares from each corner, we will follow these steps:

### Step 1: Define the dimensions of the box
Let the side length of the square cut from each corner be \( x \) cm. The original dimensions of the rectangular sheet are 45 cm by 24 cm. After cutting out squares from each corner, the dimensions of the base of the box will be:
- Length: \( 45 - 2x \)
- Width: \( 24 - 2x \)
- Height: \( x \)

### Step 2: Write the volume function
The volume \( V \) of the box can be expressed as:
\[ V = \text{Length} \times \text{Width} \times \text{Height} = (45 - 2x)(24 - 2x)(x) \]
Expanding this, we first calculate \( (45 - 2x)(24 - 2x) \):
\[ (45 - 2x)(24 - 2x) = 1080 - 90x - 48x + 4x^2 = 4x^2 - 138x + 1080 \]
Now, substituting back into the volume equation:
\[ V = x(4x^2 - 138x + 1080) = 4x^3 - 138x^2 + 1080x \]

### Step 3: Find the derivative of the volume function
To find the value of \( x \) that maximizes the volume, we need to take the derivative of \( V \) with respect to \( x \):
\[ \frac{dV}{dx} = \frac{d}{dx}(4x^3 - 138x^2 + 1080x) = 12x^2 - 276x + 1080 \]

### Step 4: Set the derivative equal to zero
To find the critical points, set the derivative equal to zero:
\[ 12x^2 - 276x + 1080 = 0 \]
Dividing the entire equation by 12 simplifies it:
\[ x^2 - 23x + 90 = 0 \]

### Step 5: Factor the quadratic equation
Next, we will factor the quadratic:
\[ (x - 18)(x - 5) = 0 \]
Thus, the solutions are:
\[ x = 18 \quad \text{or} \quad x = 5 \]

### Step 6: Determine the feasible solution
Since \( x \) represents the side length of the square cut from each corner, it must be less than half of the shorter side of the rectangle. The shorter side is 24 cm, so:
\[ x < \frac{24}{2} = 12 \]
Thus, \( x = 18 \) is not a feasible solution. The only feasible solution is:
\[ x = 5 \]

### Conclusion
The side of the square to be cut off to maximize the volume of the box is \( \boxed{5} \) cm.

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