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

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