The sum of squares of two parts of a number 100 is minimum, then two parts are

Options

A. 50,50
B. 25,75
C. 40,60
D. 30,70

Answer

To solve the problem of finding two parts of the number 100 such that the sum of their squares is minimized, we can follow these steps:

### Step 1: Define the Variables
Let the two parts be \( A \) and \( B \). According to the problem, we have:
\[ A + B = 100 \]

### Step 2: Express One Variable in Terms of the Other
From the equation above, we can express \( B \) in terms of \( A \):
\[ B = 100 - A \]

### Step 3: Write the Function to Minimize
We need to minimize the sum of the squares of these two parts:
\[ S = A^2 + B^2 \]
Substituting \( B \) from Step 2:
\[ S = A^2 + (100 - A)^2 \]

### Step 4: Expand the Equation
Now, we expand the equation:
\[ S = A^2 + (10000 - 200A + A^2) = 2A^2 - 200A + 10000 \]

### Step 5: Differentiate the Function
Next, we differentiate \( S \) with respect to \( A \):
\[ \frac{dS}{dA} = 4A - 200 \]

### Step 6: Set the Derivative to Zero
To find the minimum, we set the derivative equal to zero:
\[ 4A - 200 = 0 \]
Solving for \( A \):
\[ 4A = 200 \implies A = 50 \]

### Step 7: Find the Value of \( B \)
Now, we can find \( B \) using the equation from Step 2:
\[ B = 100 - A = 100 - 50 = 50 \]

### Step 8: Verify Minimum Condition
To confirm that this is a minimum, we can check the second derivative:
\[ \frac{d^2S}{dA^2} = 4 \]
Since \( \frac{d^2S}{dA^2} > 0 \), this indicates that \( S \) has a minimum at \( A = 50 \).

### Final Answer
Thus, the two parts are:
\[ A = 50 \quad \text{and} \quad B = 50 \]

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