Electronis Discussion

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

the relation between $\text{a}$ and $\text{b}$ is

- $a=\frac{b}{4}$
- $a=\frac{b}{2}$
- $a=2b$
- $a=4b$

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