Question

With usual notations, if the perimeter of a triangle is \(2s\), then its maximum area is:

• A. \(\frac{s^{2}}{3}\)
• B. \(\frac{\sqrt{3}s^{2}}{4}\)
• C. \(\frac{s^{2}}{\sqrt{3}}\)
• D. \(\frac{s^{2}}{3\sqrt{3}}\)

Hint:

Use Heron's formula with AM-GM; the maximum occurs for an equilateral triangle.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Heron's formula gives area \(\Delta=\sqrt{s(s-a)(s-b)(s-c)}\).

Step 2: Apply AM-GM
\((s-a)(s-b)(s-c)\le\left(\dfrac{3s-2s}{3}\right)^3=\dfrac{s^3}{27}\).

Step 3: Maximum Area
\(\Delta\le\sqrt{s\cdot\dfrac{s^3}{27}}=\dfrac{s^2}{3\sqrt{3}}\), with equality for an equilateral triangle.

Step 4: Compare the options
This is option (D).

Final Answer:
The maximum area is \(\dfrac{s^2}{3\sqrt{3}}\). \[ \boxed{\text{(D) }\dfrac{s^{2}}{3\sqrt{3}}} \]