Letthen at (0, 0),a)xfx + yfy = 4fb)xfx - yfy = 3fc)yfx + xfy = 4fd)yfx- xfy = 2fCorrect answer is option 'D'. Can you explain this answer?

Question

Answer

For the function f(x,y) to have minimum value at (a,b)
rt – s²>0 and r>0
where, r = ^∂²f⁄_∂x², t=^∂²f⁄_∂y², s=^∂²f⁄_∂x∂y, at (x,y) => (a,b)

Answer

Understanding Minimum Value Conditions
To determine the conditions under which a function $ f(x,y) $ has a minimum value at the point $ (a,b) $, we analyze the second derivative test in multivariable calculus.

Critical Point and Hessian Matrix
- A function has a critical point at $ (a,b) $ if the first partial derivatives $ f_x(a,b) $ and $ f_y(a,b) $ are both equal to zero.
- The Hessian matrix $ H $ is defined as:
$$
H = \begin{bmatrix}
f_{xx} & f_{xy} \
f_{xy} & f_{yy}
\end{bmatrix}
$$
- Here, $ f_{xx} $, $ f_{yy} $, and $ f_{xy} $ are the second partial derivatives.

Conditions for a Minimum
- The determinant of the Hessian, $ D = f_{xx} f_{yy} - (f_{xy})^2 $, plays a crucial role in identifying the nature of the critical point.
- The conditions for $ (a,b) $ to be a local minimum are:
- $ D > 0 $ (which corresponds to $ rt - s^2 > 0 $)
- $ f_{xx} > 0 $ (which corresponds to $ r > 0 $)

Conclusion
- Therefore, for $ f(x,y) $ to have a minimum at $ (a,b) $, both conditions must be satisfied simultaneously:
- $ rt - s^2 > 0 $
- $ r > 0 $
- This leads us to the answer:

Correct Option
- The correct answer is option **'D'**: $ rt - s^2 > 0 $ and $ r > 0 $.

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