2 friends discussing

Calculus Level 5

Ashutosh and Kevin are best friends. Ashutosh is humble and kind-hearted guy and Kevin is a very intelligent guy who could solve any problem. They both met at a ground. Ashutosh asked Kevin a doubt, about which he was thinking the whole week. He drew an ellipse on the mud and asked Kevin to find the area of the greatest isosceles triangle that can be inscribed in the given ellipse having its vertex coincident with one end of the major axis.

Kevin thought for a while and gave the correct answer. What answer did Kevin give?

Note: The equation of ellipse is $\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } =1$ .

Also try this set .

The answer is 7.79.

1 solution

Vishwak Srinivasan
Jun 12, 2015

From the diagram is the very evident that the height (the length of the perpendicular from $C$ to $AB$ is $3 + 3\sin(\theta)$ and the base $AB$ is $4\cos(\theta)$ .

Let the area of the variable isosceles triangle be a function of $\theta$ , say $f(\theta$ .

$f(\theta) = \frac{1}{2} (3 + 3\sin(\theta)) (4\cos(\theta)$ )

We need to maximize this function, and hence find $f'(\theta)$ and equate to $0$ .

$\Rightarrow f'(\theta) = -6\sin(\theta) + 6\cos(2\theta) = 0$

Solving $\sin(\theta) = \cos(2\theta)$ we get $\theta = \frac{\pi}{6}$

Hence, $f(\frac{\pi}{6}) = \boxed{\frac{18\sqrt{3}}{4} } \approx \huge\boxed{7.794} sq.units$

2 friends discussing

Also try this set .

The answer is 7.79.

1 solution

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