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If a triangle is inscribed in an ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, Find the Maximum area of Triangle

My try:

Let $A(5\cos p, 4\sin p)$, $B(5\cos q, 4\sin q)$ and $C(5\cos r, 4\sin r)$ be vertices of the triangle

Its area is:

$$\Delta=0.5 \times\begin{vmatrix} 5\cos p &4 \sin p &1 \\ 5 \cos q& 4 \sin q &1 \\ 5 \cos r &4 \sin r & 1 \end{vmatrix}=10\begin{vmatrix} \cos p & \sin p &1 \\ \cos q & \sin q &1 \\ \cos r&\sin r & 1 \end{vmatrix}$$

$$\Delta=40\sin\left(\frac{p-q}{2}\right)\sin\left(\frac{q-r}{2}\right)\sin\left(\frac{r-p}{2}\right)$$

EDIT:

According to the mind blowing hint given by Mohammad Zuhair khan:

$p \gt q \gt r$

$$p=q+m$$

$$q=r+n$$

where $m,n \gt 0$

So we have

$$p-q=m$$

$$q-r=n$$

$$r-p=-(m+n)$$

Then

$$\Delta=40 \sin\left(\frac{m}{2}\right)\sin\left(\frac{n}{2}\right) \sin\left(\frac{m+n}{2}\right)$$

Ignoring the negative sign, since its is Area

Now

$$\Delta(m,n)=10(\sin m+\sin n-\sin(m+n))$$

Using Partial differentiation for optimization we have:

$$\frac{\partial \Delta}{\partial m}=0$$

$$\cos m=\cos(m+n)$$

$$m=2\pi-m-n$$

$$2m+n=2\pi$$

Like wise by symmetry:

$$2n+m=2\pi$$

So

$$m=n=\frac{2\pi}{3}$$

Hence

$$\Delta_{max}=10(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} +\frac{\sqrt{3}}{2})=15\sqrt{3}$$

But how to know using partial differentiation Maximum occurs?

Can any one give me a link when does partial differentiation applicable for optimization problems?

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  • $\begingroup$ Does the triangle have one of its vertices at one end of the major axis of the ellipse? $\endgroup$ – Ak19 Apr 29 at 16:58
  • $\begingroup$ No that information is not given $\endgroup$ – Umesh shankar Apr 29 at 16:59
  • $\begingroup$ @MohammadZuhairKhan thanks for the nice hint $\endgroup$ – Umesh shankar Apr 29 at 17:10
  • $\begingroup$ Maximize the Area in Auxiliary Circle. $\endgroup$ – DXT Apr 29 at 17:17
  • $\begingroup$ @Umeshshankar khanacademy.org/math/multivariable-calculus/… may interest you. $\endgroup$ – Mohammad Zuhair Khan Apr 29 at 17:49
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If we shrink the ellipse together with an inscribed triangle by a factor of $1/5$ along the $x$-axis and by a factor of $1/4$ along the $y$-axis, the ratio of the area of the triangle to the area of the ellipse remains unchanged. The ellipse is now transformed into the unit circle. It suffices to find the largest triangle inscribed in the unit circle.

Take an arbitrary chord of the unit circle as a side of the triangle. The triangle has the largest possible area if it is the point on the circle which is farthest away from the chord, which is the farther (from the chord) point of intersection of the perpendicular bisector of the chord and the circle. This implies that the triangle must be isosceles.

Let $x$ be the distance of the chord from the centre, then the area $A$ of the triangle is $(1+x)\sqrt{1-x^2}=\sqrt{(1-x)(1+x)^3}$.

$\dfrac{dA^2}{dx}=3(1-x)(1+x)^2-(1+x)^3=(1+x)^2(2-4x)$.

It is easy to check that $A$ is the maximum when $x=1/2$. The maximum area is $\dfrac{3\sqrt{3}}{4}$.

The maximum area of the inscribed triangle of the ellipse is $\displaystyle \dfrac{3\sqrt{3}}{4}\times 5\times 4=15\sqrt{3}$.

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  • 1
    $\begingroup$ Interesting. Very interesting! (+1) $\endgroup$ – HAMIDINE SOUMARE Apr 29 at 17:37
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Introduce new coordinates $(x',y') = (\frac{4}{5}x,y)$ so that the ellipse becomes a circle in these new coordinates. Any area $A$ in the original coordinate system will have the area $A'=\frac{4}{5}A$ in terms of the new coordinates. So the problem is reduced to finding the largest triangle area inscribed in a circle, and then converting this area back to the corresponding area in the original coordinate system.

The largest triangle inscribed in a circle is the only equilateral triangle that fits, which has the area $A'=3\frac{\sqrt{3}}{4}r^2$, where $r$ is the radius of the circle, and $r=4$ in this case. See the proof in CY Aries answer for how to find this area.

Then convert back to the old coordinates: $A=\frac{5}{4}A'$. The result is $A=15\sqrt{3}$.

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