Question

The points (x, y) lies on the line $2x+3y=6$. The smallest value of the quantity $\sqrt{x^2+y^2}$, is?

• A. $\frac{6\sqrt{13}}{13}$
• B. $6$
• C. $\frac{1}{2}\sqrt{3}$
• D. $\sqrt{13}$

Answer 1

Answer: (A)

$\frac{6\sqrt{13}}{13}$

It is given that point $(x,y)$ lies on $2x+3y=6$.

$\therefore x=\frac{6-3y}{2}$ ...(1) (rearranging)

Now,

$\sqrt{x^2+y^2}=\sqrt{{\left(\frac{6-3y}{2}\right)}^2+y^2}$ (substituting from (1))

$\therefore \sqrt{x^2+y^2}=\sqrt{\left(\frac{36-36y+9y^2}{4}\right)+y^2}$

$\therefore \sqrt{x^2+y^2}=\sqrt{\frac{36-36y+13y^2}{4}}$

Differentiating expression wrt $y$ and equating to 0, we get

$\frac{1}{2}\frac{1}{\sqrt{\frac{36-36y+13y^2}{4}}}\frac{26y-36}{4}=0$

$\therefore 26y-36=0$ (ensuring $36-36y+13y^2\ne 0$ for obtained solution)

$\therefore y=\frac{18}{13}$

Hence, minima of function $\sqrt{x^2+y^2}$ occurs at $y=\frac{18}{13}$.

$\therefore \sqrt{x^2+y^2}=\sqrt{\frac{36-36\times \frac{18}{13}+13{\left(\frac{18}{13}\right)}^2}{4}}$

$\therefore \sqrt{x^2+y^2}=\sqrt{\frac{9\times 13-9\times 18+9\times 9}{13}}$

$\therefore \sqrt{x^2+y^2}=\sqrt{\frac{36}{13}}$

$\therefore \sqrt{x^2+y^2}=6\frac{\sqrt{13}}{13}$

This is the required answer.