Examveda

Let $${\text{f}}\left( {{\text{x, y}}} \right) = \frac{{{\text{a}}{{\text{x}}^2} + {\text{b}}{{\text{y}}^2}}}{{{\text{xy}}}},$$     where a and b are constants. If $$\frac{{\partial {\text{f}}}}{{\partial {\text{x}}}} = \frac{{\partial {\text{f}}}}{{\partial {\text{y}}}}$$   at x = 1 and y = 2, then the relation between a and b is

A. $${\text{a}} = \frac{{\text{b}}}{4}$$

B. $${\text{a}} = \frac{{\text{b}}}{2}$$

C. a = 2b

D. a = 4b

Answer: Option D


This Question Belongs to Engineering Maths >> Calculus

Join The Discussion

Related Questions on Calculus