GATE 2020 IN Q.31: solution and answer

Official answer: 0.5Source: Official answer key, GATE 2020 IN (IIT Delhi)

1. Use the constraint to remove a variable: $y=1-x$. On the line, $f$ becomes a function of $x$ alone: $$g(x)=x^2+(1-x{)}^2=2x^2-2x+1$$

2. Set the first derivative to zero: $g^{\prime }(x)=4x-2=0\Rightarrow x=\frac{1}{2}$, which gives $y=1-\frac{1}{2}=\frac{1}{2}$.

3. Second-derivative test: $g^{\prime \prime }(x)=4>0$, so $x=\frac{1}{2}$ is a minimum. Because $g$ is an upward-opening parabola, this is the global minimum over the whole line, not just a local one.

4. Evaluate: $f\left(\frac{1}{2},\frac{1}{2}\right)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}=0.5$.

5. Check with Lagrange multipliers. Write the constraint as $h(x,y)=x+y-1=0$. Then $\nabla f=\lambda \nabla h$ gives $2x=\lambda$ and $2y=\lambda$, so $x=y$. With $x+y=1$ this gives $x=y=\frac{1}{2}$ again.

6. Geometric check: $f=x^2+y^2$ is the squared distance from the origin. The distance from the origin to the line $x+y-1=0$ is $$\frac{|0+0-1|}{\sqrt{1^2+1^2}}=\frac{1}{\sqrt{2}}$$ and its square is $\frac{1}{2}=0.5$. All three methods give a minimum value of $0.5$.