Question

The minimum value of $ax+by$,where $xy=r^2$ is $(r,ab>0)$

• A. $2r\sqrt{ab}$
• B. $2ab\sqrt{r}$
• C. $-2r\sqrt{ab}$
• D. None of these

Answer 1

The correct option is A $2r\sqrt{ab}$

Given, $p=ax+by⟶\left(1\right)$

and $xy=r^2\Rightarrow y=\frac{r^2}{x}$

Putting the value of $y$ in equation $\left(1\right)$ we get,

$p=ax+\frac{{br}^2}{x}\Rightarrow \frac{dp}{dx}=a-\frac{{br}^2}{x^2}⟶\left(2\right)$

For minimum value of $p$,

$\frac{dp}{dx}=0$

So, from equation $\left(2\right)$ we have,

$a-\frac{{br}^2}{x^2}=0\Rightarrow a=\frac{{br}^2}{x^2}\Rightarrow a{x}^2={br}^2\Rightarrow x^2=\left(\frac{b}{a}\right)r^2\Rightarrow x=r{\left(\frac{b}{a}\right)}^{\frac{1}{2}},-r{\left(\frac{b}{a}\right)}^{\frac{1}{2}}$

But, $x=-r{\left(\frac{b}{a}\right)}^{\frac{1}{2}}$ is imaginary.

Again,

$\frac{d^{2}p}{d{x}^2}=\frac{2(b{r}^2)}{x^3}$

Putting $x=r{\left(\frac{b}{a}\right)}^{\frac{1}{2}}$ we get

$\frac{d^{2}p}{d{x}^2}=positive$

$\therefore$ Minimum value of $p$ at $x=r{\left(\frac{b}{a}\right)}^{\frac{1}{2}}$,

$ax+\frac{{br}^2}{x}=ar{\left(\frac{b}{a}\right)}^{\frac{1}{2}}+\frac{{br}^2}{r{\left(\frac{b}{a}\right)}^{\frac{1}{2}}}=r{\left(ab\right)}^{\frac{1}{2}}+\left(br\right){\left(\frac{a}{b}\right)}^{\frac{1}{2}}=r{\left(ab\right)}^{\frac{1}{2}}+r{\left(ab\right)}^{\frac{1}{2}}=2r{\left(ab\right)}^{\frac{1}{2}}=2r\sqrt{ab}.$