Question

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

• A. a = $\frac{b}{4}$
• B. a = $\frac{b}{2}$
• C. a = 2b
• D. a = 4b

Answer 1

The correct option is D a = 4b

$f(x,y)=\frac{a{x}^2+b{y}^2}{xy}$

$\frac{\partial f}{\partial x}=\frac{xy\left(2ax\right)-(a{x}^2+b{y}^2)\left(y\right)}{x^2{y}^2}$

so ${\left(\frac{\partial f}{\partial x}\right)}_{(1,2)}=\frac{2a-8b}{4}$

$\frac{\partial f}{\partial y}=\frac{xy\left(2by\right)-(a{x}^2+b{y}^2)\left(x\right)}{x^2{y}^2}$

so ${\left(\frac{\partial f}{\partial y}\right)}_{(1,2)}=\frac{4b-a}{4}$

Given that $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}$

$\Rightarrow$ 2a - 8b = 4b - a

3a = 12b

$\Rightarrow$ a = 4b