A variable line $L$ passes through the point $(3,5)$ and intersects the positive coordinate axes at the points $\mathrm{A}$ and $\mathrm{B}$. The minimum area of the triangle $\mathrm{OAB}$, where $\mathrm{O}$ is the origin, is :

• Correct answer
  A

  30
• B

  25
• C

  40
• D

  35

Solution

$\begin{aligned} & \frac{x}{a}+\frac{y}{b}=1 \\ & \frac{3}{a}+\frac{5}{b}=1 \Rightarrow b=\frac{5 a}{a-3}, a>3\end{aligned}$
$A=\frac{1}{2} a b=\frac{1}{2} a \frac{5 a}{(a-3)}=\frac{5}{2} \cdot \frac{a^2}{a-3}$ $\begin{aligned} & =\frac{5}{2}\left(\frac{\mathrm{a}^2-9+9}{\mathrm{a}-3}\right) \\ & =\frac{5}{2}\left(\mathrm{a}+3+\frac{9}{\mathrm{a}-3}\right) \\ & =\frac{5}{2}\left(\mathrm{a}-3+\frac{9}{\mathrm{a}-3}+6\right) \geq 30\end{aligned}$

Hence, the correct answer is option: 'A'.