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

Answer

To solve the problem of maximizing the volume of a box made from a square piece of tin, we will follow these steps:

### Step 1: Define the variables
Let the side length of the square piece of tin be \( L = 24 \) cm. Let \( x \) be the side length of the square cut from each corner.

### Step 2: Determine the dimensions of the box
After cutting out squares of side \( x \) from each corner and folding up the sides, the dimensions of the box will be:
- Length = \( L - 2x = 24 - 2x \)
- Width = \( L - 2x = 24 - 2x \)
- Height = \( x \)

### Step 3: Write the volume formula
The volume \( V \) of the box can be expressed as:
\[ V = \text{Length} \times \text{Width} \times \text{Height} = (24 - 2x)(24 - 2x)(x) = (24 - 2x)^2 \cdot x \]

### Step 4: Expand the volume formula
Expanding the volume formula:
\[ V = (24 - 2x)^2 \cdot x = (576 - 96x + 4x^2) \cdot x = 576x - 96x^2 + 4x^3 \]

### Step 5: Differentiate the volume function
To find the maximum volume, we need to differentiate \( V \) with respect to \( x \):
\[ \frac{dV}{dx} = 576 - 192x + 12x^2 \]

### Step 6: Set the derivative to zero
Setting the derivative equal to zero to find critical points:
\[ 12x^2 - 192x + 576 = 0 \]
Dividing the entire equation by 12:
\[ x^2 - 16x + 48 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -16, c = 48 \):
\[ x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 48}}{2 \cdot 1} = \frac{16 \pm \sqrt{256 - 192}}{2} = \frac{16 \pm \sqrt{64}}{2} = \frac{16 \pm 8}{2} \]
Thus, we get:
\[ x = \frac{24}{2} = 12 \quad \text{or} \quad x = \frac{8}{2} = 4 \]

### Step 8: Determine which critical point gives maximum volume
To find out which critical point gives the maximum volume, we can check the second derivative:
\[ \frac{d^2V}{dx^2} = 24x - 192 \]
Evaluating at \( x = 4 \):
\[ \frac{d^2V}{dx^2} = 24(4) - 192 = 96 - 192 = -96 \quad (\text{less than 0, so maximum}) \]
Evaluating at \( x = 12 \):
\[ \frac{d^2V}{dx^2} = 24(12) - 192 = 288 - 192 = 96 \quad (\text{greater than 0, so minimum}) \]

### Step 9: Calculate the maximum volume
Now, substituting \( x = 4 \) back into the volume formula:
\[ V = (24 - 2(4))^2 \cdot 4 = (24 - 8)^2 \cdot 4 = (16)^2 \cdot 4 = 256 \cdot 4 = 1024 \text{ cm}^3 \]

### Conclusion
The side of the square to be cut off for maximum volume is \( x = 4 \) cm, and the maximum volume of the box is \( 1024 \) cm³.

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