Question

If \(x+y=k, x>0, y>0\) then \(x^2+y^2\) is minimum if

• A. \( x>y \)
• B. \( x<y \)
• C. \( x = y \)
• D. \( x = 2y \)

Hint:

To find the minimum/maximum of a function of two variables with a constraint, use the constraint to express one variable in terms of the other, then substitute into the function to be optimized.
For a quadratic \(f(t)=At^2+Bt+C\), the extremum occurs at \(t=-B/(2A)\). It's a minimum if A>0, maximum if A<0.
Alternatively, use Lagrange multipliers or geometric arguments (e.g., symmetry). Here, \(x^2+y^2\) is minimized when x and y are as "equal as possible" given \(x+y=k\).

Answer 1

The Correct Option is C

Problem: Given \( x + y = k \), minimize \( S = x^2 + y^2 \).

Step 1: Express one variable in terms of the other

\[ y = k - x \]

Substitute into \( S \):

\[ S(x) = x^2 + (k - x)^2 = x^2 + (k^2 - 2kx + x^2) = 2x^2 - 2kx + k^2 \]

This is a quadratic expression in \( x \):

\[ S(x) = 2x^2 - 2kx + k^2 \]

Since the coefficient of \( x^2 \) is positive (2), the parabola opens upward. Therefore, the minimum value occurs at the vertex.

Vertex formula:

\[ x = \frac{-(-2k)}{2 \cdot 2} = \frac{2k}{4} = \frac{k}{2} \Rightarrow y = k - x = \frac{k}{2} \]

Thus, the minimum occurs when:

\[ x = y = \frac{k}{2} \]

Optional: Use calculus to verify

\[ S'(x) = \frac{d}{dx}(2x^2 - 2kx + k^2) = 4x - 2k \Rightarrow 4x - 2k = 0 \Rightarrow x = \frac{k}{2} \] \[ S''(x) = 4 > 0 \Rightarrow \text{Minimum confirmed} \]

✅ Final Answer:

\( \boxed{x = y} \)

This matches option (c).