If p and q are two quantities such that `p^(2) +q^(2) =1`, then maximum value of p + q is

Options

A. `(1)/(2)`
B. `(1)/(sqrt(2))`
C. `sqrt(2)`
D. 2

Answer

To find the maximum value of \( p + q \) given that \( p^2 + q^2 = 1 \), we can use the Cauchy-Schwarz inequality or the method of Lagrange multipliers. Here’s a step-by-step solution using the Cauchy-Schwarz inequality:

### Step 1: Understand the Given Condition
We start with the equation:
\[ p^2 + q^2 = 1 \]
This represents a circle of radius 1 in the \( p-q \) plane.

### Step 2: Apply the Cauchy-Schwarz Inequality
According to the Cauchy-Schwarz inequality:
\[ (p + q)^2 \leq (1^2 + 1^2)(p^2 + q^2) \]
Here, we take \( a_1 = p \), \( a_2 = q \), \( b_1 = 1 \), and \( b_2 = 1 \).

### Step 3: Substitute the Known Values
Substituting \( p^2 + q^2 = 1 \) into the inequality gives:
\[ (p + q)^2 \leq (1 + 1)(1) = 2 \]

### Step 4: Take the Square Root
Taking the square root of both sides, we find:
\[ p + q \leq \sqrt{2} \]

### Step 5: Determine When Equality Holds
Equality in the Cauchy-Schwarz inequality holds when:
\[ \frac{p}{1} = \frac{q}{1} \implies p = q \]
Substituting \( p = q \) into the equation \( p^2 + q^2 = 1 \):
\[ 2p^2 = 1 \implies p^2 = \frac{1}{2} \implies p = \frac{1}{\sqrt{2}}, q = \frac{1}{\sqrt{2}} \]

### Step 6: Calculate the Maximum Value
Thus, the maximum value of \( p + q \) is:
\[ p + q = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \]

### Conclusion
The maximum value of \( p + q \) is:
\[ \boxed{\sqrt{2}} \]

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