Question

Divide the number 20 into two parts such that the sum of their squares is minimum.

Hint:

When minimizing the sum of squares, take the derivative of the function and solve for the critical point.

Answer 1

Step 1: Set up the variables.
Let the two parts into which the number 20 is divided be \( x \) and \( 20 - x \). The sum of the squares of these parts is given by: \[ S(x) = x^2 + (20 - x)^2 \]

Step 2: Simplify the expression.
Expanding the expression: \[ S(x) = x^2 + (400 - 40x + x^2) = 2x^2 - 40x + 400 \]

Step 3: Minimize the function.
To minimize \( S(x) \), we take the derivative with respect to \( x \) and set it equal to zero: \[ \frac{dS}{dx} = 4x - 40 \] Setting the derivative equal to zero: \[ 4x - 40 = 0 $\Rightarrow$ x = 10 \]

Step 4: Verify that this is a minimum.
Taking the second derivative: \[ \frac{d^2S}{dx^2} = 4 \] Since the second derivative is positive, the function has a minimum at \( x = 10 \).

Step 5: Conclude.
Thus, the number 20 should be divided into two parts of 10 and 10 to minimize the sum of their squares.

Final Answer: \[ \boxed{10 \text{ and } 10} \]