An open box is to be made of square sheet of tin with side 20 cm, by cutting off small squares from each corner and foding the flaps. Find the side of small square, which is to be cut off, so that volume of box is maximum.

Answer

To solve the problem of maximizing the volume of an open box made from a square sheet of tin with side 20 cm, we will follow these steps:

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

### Step 2: Determine the dimensions of the box
After cutting out squares of side \( x \) from each corner and folding the flaps, the dimensions of the box will be:
- Length = \( 20 - 2x \)
- Width = \( 20 - 2x \)
- Height = \( x \)

### Step 3: Write the volume function
The volume \( V \) of the box can be expressed as:
\[ V = \text{Length} \times \text{Width} \times \text{Height} = (20 - 2x)(20 - 2x)(x) \]
This simplifies to:
\[ V = (20 - 2x)^2 \cdot x \]

### Step 4: Expand the volume function
Expanding \( V \):
\[ V = (20 - 2x)(20 - 2x) \cdot x = (400 - 80x + 4x^2) \cdot x \]
\[ V = 400x - 80x^2 + 4x^3 \]

### Step 5: Differentiate the volume function
To find the maximum volume, we differentiate \( V \) with respect to \( x \):
\[ \frac{dV}{dx} = 400 - 160x + 12x^2 \]

### Step 6: Set the derivative to zero
Setting the derivative equal to zero to find critical points:
\[ 12x^2 - 160x + 400 = 0 \]

### Step 7: Simplify the equation
Dividing the entire equation by 4:
\[ 3x^2 - 40x + 100 = 0 \]

### Step 8: Use the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3 \), \( b = -40 \), and \( c = 100 \).
\[ x = \frac{40 \pm \sqrt{(-40)^2 - 4 \cdot 3 \cdot 100}}{2 \cdot 3} \]
\[ x = \frac{40 \pm \sqrt{1600 - 1200}}{6} \]
\[ x = \frac{40 \pm \sqrt{400}}{6} \]
\[ x = \frac{40 \pm 20}{6} \]

### Step 9: Calculate the values of \( x \)
Calculating the two possible values:
1. \( x = \frac{60}{6} = 10 \)
2. \( x = \frac{20}{6} = \frac{10}{3} \)

### Step 10: Determine which value is valid
Since \( x = 10 \) would result in a box with no base (as \( 20 - 2x = 0 \)), we discard this solution. Therefore, the valid solution is:
\[ x = \frac{10}{3} \text{ cm} \]

### Step 11: Verify if it is a maximum
To confirm that this value of \( x \) gives a maximum volume, we can check the second derivative:
\[ \frac{d^2V}{dx^2} = 24x - 160 \]
Substituting \( x = \frac{10}{3} \):
\[ \frac{d^2V}{dx^2} = 24 \cdot \frac{10}{3} - 160 = 80 - 160 = -80 \]
Since the second derivative is negative, this confirms that \( x = \frac{10}{3} \) is indeed a maximum.

### Final Answer
The side of the small square that should be cut off to maximize the volume of the box is:
\[ \boxed{\frac{10}{3} \text{ cm}} \]