Question

If \[ x+y=k,\quad x\gt 0,\quad y\gt 0, \] then \(x^2+y^2\) is minimum, if

• A. \(x\gt y\)
• B. \(x\lt y\)
• C. \(x=y\)
• D. \(x=2y\)

Hint:

For positive numbers with fixed sum, \[ x+y=k, \] the quantity \[ x^2+y^2 \] is minimum when the numbers are equal: \[ x=y=\frac{k}{2}. \] This follows directly from AM-GM or differentiation.

Answer 1

The Correct Option is C

Step 1: Rewrite \(x^2+y^2\) using the constraint.
\(x^2+y^2 = (x+y)^2-2xy = k^2-2xy\). To minimise \(x^2+y^2\), maximise \(xy\).

Step 2: Apply AM-GM.
By AM-GM: \(\dfrac{x+y}{2} \geq \sqrt{xy} \Rightarrow xy \leq \left(\dfrac{k}{2}\right)^2 = \dfrac{k^2}{4}\), with equality when \(x=y\).

Step 3: Conclude.
\(x^2+y^2\) is minimum when \(x=y=k/2\).
\[ \boxed{x = y} \]