A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top by cutting off squares from the corners and folding up the flaps. What should be the side of the square in order the volume of the box is maximum.

Answer

To find the side of the square that should be cut from each corner of the rectangular sheet in order to maximize the volume of the box, we will follow these steps:

### Step 1: Define the dimensions of the box
Let the side of the square cut from each corner be \( x \) cm.

- The length of the box after cutting the squares will be:
\[ L = 45 - 2x \]
- The width of the box will be:
\[ W = 24 - 2x \]
- The height of the box will be:
\[ H = x \]

### Step 2: Write the volume formula
The volume \( V \) of the box can be expressed as:
\[ V = L \times W \times H = (45 - 2x)(24 - 2x)(x) \]

### Step 3: Expand the volume expression
Now, we will expand the volume expression:
\[ V = (45 - 2x)(24 - 2x)x \]
First, expand \( (45 - 2x)(24 - 2x) \):
\[ = 45 \times 24 - 90x + 4x^2 = 1080 - 90x + 4x^2 \]
Now multiply by \( x \):
\[ V = x(1080 - 90x + 4x^2) = 1080x - 90x^2 + 4x^3 \]

### Step 4: Differentiate the volume function
To find the maximum volume, we need to differentiate \( V \) with respect to \( x \):
\[ \frac{dV}{dx} = 1080 - 180x + 12x^2 \]

### Step 5: Set the derivative to zero
Setting the derivative equal to zero to find critical points:
\[ 12x^2 - 180x + 1080 = 0 \]
Dividing the entire equation by 12:
\[ x^2 - 15x + 90 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -15, c = 90 \):
\[ x = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 1 \cdot 90}}{2 \cdot 1} \]
\[ x = \frac{15 \pm \sqrt{225 - 360}}{2} \]
\[ x = \frac{15 \pm \sqrt{-135}}{2} \]
Since the discriminant is negative, we need to check the critical points from the factorization:
\[ 12(x - 5)(x - 18) = 0 \]
This gives us \( x = 5 \) and \( x = 18 \).

### Step 7: Determine the maximum volume
To determine which value gives a maximum volume, we can check the second derivative:
\[ \frac{d^2V}{dx^2} = 24x - 180 \]
Evaluating at \( x = 5 \):
\[ \frac{d^2V}{dx^2} = 24(5) - 180 = 120 - 180 = -60 < 0 \]
This indicates a maximum at \( x = 5 \).

### Conclusion
The side of the square that should be cut off from each corner to maximize the volume of the box is:
\[ \boxed{5 \text{ cm}} \]