Question
A variable line $\mathrm{L}$ 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 :
✓
Answer
a
$ \frac{x}{a}+\frac{y}{b}=1 $
$ \frac{x}{a}+\frac{y}{b}=1 $
$ \frac{3}{a}+\frac{5}{b}=1 \Rightarrow b=\frac{5 a}{a-3}, a>3$
$Image$
$\mathrm{A}=\frac{1}{2} \mathrm{ab}=\frac{1}{2} \mathrm{a} \frac{5 \mathrm{a}}{(\mathrm{a}-3)}=\frac{5}{2} \cdot \frac{\mathrm{a}^2}{\mathrm{a}-3}$
$ =\frac{5}{2}\left(\frac{a^2-9+9}{a-3}\right) $
$ =\frac{5}{2}\left(a+3+\frac{9}{a-3}\right) $
$ =\frac{5}{2}\left(a-3+\frac{9}{a-3}+6\right) \geq 30$
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