Question

The maximum value of the objective function \((x+y)\) subject to the constraint \((x^2+y^2)=2\) is (in integer).

Hint:

For constrained optimization involving \(x^2+y^2\), identities like \((x+y)^2=x^2+y^2+2xy\) and inequalities such as AM-GM are very useful.

Answer 1

Correct Answer: 2

Step 1: State the problem.
Maximise $x+y$ on the circle $x^2+y^2=2$.

Step 2: Square the target.
\[ (x+y)^2=x^2+y^2+2xy=2+2xy \]

Step 3: Bound the cross term.
Since $x^2+y^2\ge2xy$ and $x^2+y^2=2$, we get $xy\le1$. So $(x+y)^2\le2+2=4$, which means $x+y\le2$.

Step 4: Check when equality holds.
Equality needs $x=y$. Then $2x^2=2$, so $x=1$ and $y=1$, which satisfies the circle.

Step 5: Read the maximum.
At $(1,1)$ the sum is $1+1=2$.
\[ \boxed{2} \]