Explanation
Concept:
A critical point of a function of f(x) is a value x0 in the domain of f where it is not differentiable or its derivative is 0 (f′(x0)=0).
To find the nature of the critical points, we apply the second derivative test.
I. If f′′(x) is less than 0, then the given function is said to be maxima.
II. If f′′(x) is greater than 0, then the function is said to be minima.
Calculation:
Here, f(x, y) = x² - 2x + 2y² + 4y - 2
Partial derivatives,
f′(x)=2x−2 and f′(y)=4y+4
Now, for critical points, f′(x)=0
⇒ 2x−2=0
⇒ x=1,
Also, f′(y)=0
⇒ 4y+4=0
⇒ y=−1,
So, critical points (1,−1)
f''(x) = 2 > 0 and f''(y) = 4 > 0
So, at (1,−1) minimum
Hence, option (4) is correct.