Finding the two points on a heart curve which have maximal distance between them

Question

How do I find two points on this curve that have maximal distance between them?

I tried to use Lagrange multiplier to solve this, but solve equations is diffcult.

Answer 1

Not an answer

$(x^2+y^2-1)^3-x^2y^3=0$

$x^2+y^2-1=x^{2/3}y$

Using quadratic formula on $y$

$$y=\frac{x^{\frac 23}\pm\sqrt{x^{\frac 43}-4x^2+4)}}{2}$$

For any $x$ value, we have $2$ different $y$ values (except at the leftmost or rightmost points), so this equation represents all points.

Let $(x_1,y_1),(x_2,y_2)$ be the points, from the above equation, we reduced it to two unknowns: $x_1,x_2$

$f(x_1,x_2)=(x_2-x_1)^2+(\frac{x_2^{\frac 23}\pm\sqrt{x_2^{\frac 43}-4x_2^2+4}}{2} - \frac{x_1^{\frac 23}\pm\sqrt{x_1^{\frac 43}-4x_1^2+4}}{2} )^2$

Note that we have two $\pm$ signs, so in our case, we need to consider all $4$ of them. But let's take derivative first

$$f_{x_2}=2(x_2-x_1)+2(y_2-y_1)(\frac 13 x_2^{-\frac 13}\pm 2 (x_2^{\frac 43}-4x_2^2+4)^{-\frac 12} (\frac 13 x_2^{\frac 13} -2x_2))$$

$$f_{x_1}=-2(x_2-x_1)-2(y_2-y_1)(\frac 13 x_1^{-\frac 13}\pm 2 (x_1^{\frac 43}-4x_1^2+4)^{-\frac 12} (\frac 13 x_1^{\frac 13} -2x_1))$$

$x_1=x_2$, $y_1=y_2$ is the obvious minimum

The above equations are pretty hard to solve. I don't have access to any mathematical software for now, maybe someone could help plugging it in?