A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner 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 formed by cutting squares from the corners of a square piece of tin, we will follow these steps:

### Step 1: Define the Variables
Let \( x \) be the side length of the square cut from each corner of the tin. The original side length of the square piece of tin is 18 cm.

### 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 = \( 18 - 2x \)
- Width = \( 18 - 2x \)
- Height = \( x \)

### Step 3: Write the Volume Function
The volume \( V \) of the box can be expressed as:
\[ V = \text{Length} \times \text{Width} \times \text{Height} = (18 - 2x)(18 - 2x)(x) \]
This simplifies to:
\[ V = (18 - 2x)^2 \cdot x \]

### Step 4: Expand the Volume Function
Expanding the volume function:
\[ V = (18 - 2x)(18 - 2x) \cdot x = (324 - 72x + 4x^2) \cdot x = 4x^3 - 72x^2 + 324x \]

### Step 5: Find the First Derivative
To find the maximum volume, we need to take the derivative of \( V \) with respect to \( x \):
\[ \frac{dV}{dx} = 12x^2 - 144x + 324 \]

### Step 6: Set the First Derivative to Zero
Setting the first derivative equal to zero to find critical points:
\[ 12x^2 - 144x + 324 = 0 \]
Dividing the entire equation by 12:
\[ x^2 - 12x + 27 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -12, c = 27 \):
\[ x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 27}}{2 \cdot 1} = \frac{12 \pm \sqrt{144 - 108}}{2} = \frac{12 \pm \sqrt{36}}{2} = \frac{12 \pm 6}{2} \]
This gives us:
\[ x = \frac{18}{2} = 9 \quad \text{and} \quad x = \frac{6}{2} = 3 \]

### Step 8: Determine Valid Solutions
Since \( x = 9 \) would result in a base area of \( 0 \) (as \( 18 - 2 \cdot 9 = 0 \)), we discard \( x = 9 \). Thus, the only valid solution is:
\[ x = 3 \]

### Step 9: Verify Maximum Volume
To confirm that this value of \( x \) gives a maximum volume, we can check the second derivative:
\[ \frac{d^2V}{dx^2} = 24x - 144 \]
Substituting \( x = 3 \):
\[ \frac{d^2V}{dx^2} = 24(3) - 144 = 72 - 144 = -72 < 0 \]
This indicates that the volume is maximized at \( x = 3 \).

### Step 10: Calculate Maximum Volume
Now, substituting \( x = 3 \) back into the volume equation:
\[ V = 3(18 - 2 \cdot 3)^2 = 3(18 - 6)^2 = 3(12)^2 = 3 \cdot 144 = 432 \text{ cm}^3 \]

### Final Answer
The side of the square to be cut off for maximum volume is \( 3 \) cm, and the maximum volume of the box is \( 432 \) cm³.

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