Question

If x<0 , then what is the maximum value of 9/x+x/9 ?

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$\displaystyle \: y = \frac{9}{x} + \frac{x}{9}$

Differentiating both sides with respect to x

$\displaystyle \: \frac{dy}{dx} = - \frac{9}{ {x}^{2} } + \frac{1}{9}$

Again Differentiating both sides with respect to x

$\displaystyle \: \frac{ {d}^{2} y}{d {x}^{2} } = \frac{18}{ {x}^{3} }$

For maximum

$\displaystyle \: \frac{dy}{dx} = 0$

$\implies \: \displaystyle \: - \frac{9}{ {x}^{2} } + \frac{1}{9} = 0$

$\implies \: {x}^{2} = 81$

$\implies \: x \: = \pm \: 9$

Since $x < 0$

So $x = - 9$

Now for

$x = - 9 \: \: \displaystyle \: we \: \: have \: \: \: \frac{ {d}^{2} y}{d {x}^{2} } = \frac{18}{ {( - 9)}^{3} } < 0$

So Y has a maximum at x = - 9

And the maximum value is

$= \displaystyle \: \frac{9}{ - 9} + \frac{ - 9}{9} = - 1 - 1 = - 2$

Answer 2

differentiate and get x = -9

ans: -2