Find the tangent line to the curve $y = 1-x^2$, $0\le x \le 1$, such that the triangle it makes with the coordinate axis has a minimum area.
So, I sketched the graph of $1-x^2$ and I'm supposing a point $a$ from $0$ to $1$. It's derivative will be $-2a$. Then, I mounted the equation of a line that passes in the point $(a,1-a^2)$ and has derivative $-2a$. The equation is $$\frac{y-1+a^2}{x-a} = -2a$$
By setting $y=0$ and $x=0$ I can get the vertex of the triangle. Then, by identifying the basis and height, I've found an equation which doesn't has a minimum. Am I doing it rigth?
Best Answer
By similar triangles, it can be proven that for $(x_0,y_0=f(x_0))$ with corresponding vertices $(x_c,0)$, and $(0,y_c)$ all lying on the tangent line, for any $y=f(x_0)$,
$$\frac{x_0}{x_c}+\frac{y_0}{y_c}=1$$
$x_c$, $y_c$, and $y_0$ can all be expressed as functions of $x_0$, so can their ratios.
The sum of the ratios in the above equation add up to one. As one ratio increases, the other decreases.
Principles of Linear Programming suggest finding optimal values where component terms take on extreme values or component terms are equal.
$x_c$ or $y_c$ take on extreme values when the opposite coordinate nears a vertex. Under these conditions, the area under the associated triangle gets arbitrarily large.
So the extremes will not give a minimum.
Now suppose $\frac{x_0}{x_c}=\frac{y_0}{y_c}$ from the other principle of Linear Programming.
Then $\frac{x_0}{x_c}=\frac{y_0}{y_c}=\frac{1}{2}$.
So $x_c=2x_0$ and $y_c=2y_0$, making $(x_0,y_0)$ the midpoint of the hypotenuse. By Thales' Theorem, the midpoint of a hypotenuse is equidistant from all vertices of the right triangle.
The area of the corresponding triangle is $\frac{1}{2}x_cy_c=\frac{1}{2}\cdot4x_0y_0=2x_0y_0$
So the triangle's area is minimized when the area enclosed by the rectangle with opposite corners the origin and the point on $f(x_0)$ is a minimum.
$$A(x)=x(1-x^2)=x-x^3$$
Which has a minimum at $1-3x^2=0$
We also know the length of the hypotenuse is twice the length of the distance of $(x_0,y_0)$ to the origin at the point of optimal area.
This means the intercepts must lie on a circle centered at $(x_0,y_0)$ and passing through the origin. On intercept will always be outside of the circle, and one always in. This allows for a searching algorithm with straight edge and compass.