JEE Main 2024 (Online) 9th April Morning ShiftQuestion 1
A variable line $L$ passes through the point $(3,5)$ and intersects the positive coordinate axes at the points $A$ and $B$. The minimum area of the triangle $OAB$, where $O$ is the origin, is :
Differentiation and its Application
Solution
Concept. Write the line in intercept form; the condition through $(3,5)$ plus AM-GM bounds the product of the intercepts. Formulas used. (1) Line with intercepts $p,q$: $\frac xp+\frac yq=1$; area of $\triangle OAB=\frac{pq}{2}$(2) AM-GM: $u+v\ge2\sqrt{uv}$ for $u,v>0$Step 1 - Condition. $(3,5)$ is on the line: $\frac3p+\frac5q=1$. Step 2 - AM-GM. By (2), $$1=\frac3p+\frac5q\ge2\sqrt{\frac{15}{pq}}\ \Longrightarrow\ pq\ge60.$$ Equality when $\frac3p=\frac5q=\frac12$, i.e. $p=6$, $q=10$. Step 3 - Minimum area. $\frac{pq}{2}\ge30$. Answer: (D)