Find the greatest area of triangle with base as major axis of ellipse $3x^2+2y^2=6$ and one of the vertex lies on cicumference $$\frac{x^2}{2}+\frac{y^2}{3}=1$$
The length of major axis of ellipse is $2\sqrt 3$
For area to be maximum, the vertex will probably lie on the minor axis, so the length be $\sqrt 2$
$$\Delta =\sqrt 6$$
The correct answer is 6. What’s going wrong?
Note- I tried using calculus but was unable to figure it out, so if you want to use it, it’s fine.
 A: Using calculus:
Our base is obviously maximized when it’s the whole major axis. Our hieght function is given by $y^2=\frac{(6-3x^2)}{2}$ and since the eclipse is symmetric, we will work with positive $y$. so 
$y=\sqrt{\frac{(6-3x^2)}{2}}$ our area of the triangle is given by base times height and the base is fixed. So all we have to do it maximize height.
Take it derivative to maximize:
$y’=\frac{3x}{N}$ where $N$ represents junk (doesn’t matter). Clearly the max height occurs at $x=0$ and $y=\sqrt{3}$ and base is $2 \sqrt{2}$
So area is:
$\frac{\sqrt{6}2}{2}$ which gives $\sqrt{6}$
A: Hint:
$$\dfrac{x^2}2+\dfrac{y^2}3=1$$
So, the other two vertices will be $$(-\sqrt2,0);(+\sqrt2,0)$$
WLOG the third vertice be $$P(\sqrt2\cos t,\sqrt3\sin t)$$
As the length of the base is fixed $2\cdot\sqrt2$
we effectively need to maximize the perpendicular distance of $P$ from the base $y=0$
A: For the area we have $$\frac{2\sqrt3\sqrt{2-\frac{2}{3}y^2}}{2}\leq\sqrt6.$$
The equality occurs for $y=0$, which says that we got a maximal value.
A: The area will be maximum f the triangle is an isosceles and its base is fixed
as the major axis $2\sqrt{3}$ its height is semi-minor axis $\sqrt{2}$.
So the maximum area of the triangle is $\frac{1}{2} 2 \sqrt{3} \sqrt{2}=\sqrt{6}$.
A: The lengths of the axis are $2\sqrt 3$ and $2\sqrt2$. The largest area is achieved with the third vertex at the largest distance of the major axis, which is obviously an endpoint of the minor axis.
Hence the area
$$\frac{2\sqrt 3\cdot\sqrt2}2.$$
