A matter of perspective

You probably already know that the triangle of maximum area inscribed in a circle is an equilateral triangle (This is much easier to prove than the current problem and I leave it up to you).

Let's calculate the ratio $\frac{\triangle}{\bigcirc}$ in a circle of radius $r$ .

$\triangle=3 \times \frac{1}{2}r^2 \sin(\frac{2\pi}{3})=\frac{3\sqrt{3}}{4}r^2$ ; $\qquad \bigcirc=\pi r^2 \qquad \implies \dfrac{\triangle}{\bigcirc}=\dfrac{3\sqrt{3}}{4 \pi}\approx0.413497$

Now, here is the thing. What does this circle with its inscribed triangle of maximum area look like when you view it from an angle?

It looks exactly like what we are looking for: an ellipse with an inscribed triangle, which must have maximum area since the ratio hasn't been altered.

The given ellipse has an area of $\pi ab$ where $a=5$ and $b=4$ . The maximum area of the triangle is then $\triangle=0.413497\times 20\pi\approx\boxed{25.981}$

There are infinitely many of these triangles we can generate by simply rotating the circle.

The triangle with maximum area must be symmetric about the minor axis of the ellipse. So let the coordinates of the vertices of the triangle be (0,4), (h, -k) and (-h, -k). Then we have to maximize the expression h(4+k) subject to the condition 16h^2+25k^2=400. Differentiating the first, we get dk/dh=-(4+k)/h and the second we get dk/dh=-(16h/25k). Solving we get k=2, h=(5√3)/2. Therefore the maximum area of the triangle is (5√3/2)(4+2)=15√3=25.98076

$Circle~radius ~5~will~have~equivalent~\Delta~sdes~5\sqrt3~that~ has~ the~maximum ~area.\\ This~maximum~area= \dfrac{\sqrt3} 4*(5\sqrt3)^2.\\ When ~this~circle~is~compressed~to~an~ellipse~minor~axis~4,\\ the~area~of~the~\Delta~becomes, \dfrac{\sqrt3} 4*(5\sqrt3)^2*\frac 4 5=15\sqrt3=\Large \color{#D61F06}{25.9808}.$