Question

A straight line L passes through the point of intersection of the lines \(x-y+1 = 0\) and \(2x+y-7 = 0\). If L intersects the positive x-axis at \(A(a,0)\) and the positive y-axis at \(B(0,b)\), then the minimum area of the triangle \(OAB\) (where \(O\) is the origin) is ....

• A. \(6\) square units.
• B. \(12\) square units.
• C. \(24\) square units.
• D. \(48\) square units.

Hint:

Find the point of intersection, write the intercept form, and minimise ab with AM-GM.

Answer 1

The Correct Option is B

Step 1: Calculus Route:
From $2/a+3/b=1$ we get $b=\dfrac{3a}{a-2}$ with $a>2$. Area $=\dfrac{ab}{2}=\dfrac{3a^2}{2(a-2)}$.

Step 2: Differentiate:
Let $A(a)=\dfrac{3a^2}{2(a-2)}$. Then $A'(a)=\dfrac{3}{2}\cdot\dfrac{2a(a-2)-a^2}{(a-2)^2}=\dfrac32\cdot\dfrac{a(a-4)}{(a-2)^2}$. It vanishes at $a=4$ and changes from negative to positive, so it is a minimum.

Step 3: Value:
$A(4)=\dfrac{3\cdot16}{2\cdot2}=12$. Option (B).

Final Answer:
Option (B). \[ \boxed{\text{(B) } 12\ \text{square units}} \]