A square piece of tin of side 24 cm is to be made into a box without top by cutting a square from each corner and foding 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 this maximum volume.

Answer

To find the side of the square that should be cut off from each corner of a square piece of tin of side 24 cm in order to maximize the volume of the resulting box, we can follow these steps:

### Step 1: Define the variables
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 = \( 24 - 2x \)
- Width = \( 24 - 2x \)
- Height = \( x \)

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

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

### Step 5: Find the derivative of the volume function
To maximize the volume, we need to find the derivative of \( V \):
\[ \frac{dV}{dx} = 576 - 192x + 12x^2 \]

### 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, \( a = 1, b = -16, c = 48 \):
\[ x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 48}}{2 \cdot 1} \]
\[ x = \frac{16 \pm \sqrt{256 - 192}}{2} \]
\[ x = \frac{16 \pm \sqrt{64}}{2} \]
\[ x = \frac{16 \pm 8}{2} \]
This gives us two potential solutions:
\[ x = \frac{24}{2} = 12 \quad \text{and} \quad x = \frac{8}{2} = 4 \]

### Step 8: Determine which value maximizes the volume
Since \( x \) must be less than half of the side of the square (i.e., \( x < 12 \)), we consider \( x = 4 \).

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

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

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