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.
The equation of the tangent line is given by $y-a=f'(b)(x-b)=e^b(x-b),$ where $x\le 0$ and $f(x)=e^x,$ and also the point of tangency is given by $(b,a).$ Thus the intercepts on both axes, taken absolutely, give the lengths of the legs of the right triangle, namely $|a-be^b|$ and $|-ae^{-b}+b|$ respectively.
Hence the area of this triangle is given by $$\frac12|(a-be^b)(b-ae^{-b})|,$$ where $a$ is given in terms of the $x$-coordinate $b$ by $a=e^b.$ Thus our objective function in terms of $b$ alone is given by $$\frac{e^b|(1-b)(b-1)|}{2}=\frac{e^b(b-1)^2}{2},$$ defined for all $b\le 0.$
Can you now see that this function never vanishes on its domain? Also, can you now find the minimum of this function on its domain?
Best Answer
First, we compute the derivative of the function $ f(x) = \frac{1}{x} $, which is actually $ -\frac{1}{x^2} $. (If you want to know more about this, try looking up "Quotient Rule" in Google!)
So the tangent line at $ (a,b) = (a, \frac{1}{a}) $ should satisfy the following:
With those conditions in mind, we can set up the equation of the line: $ y=-\frac{1}{a^2}(x-a)+\frac{1}{a} $. Now compute the x-intercept and y-intercept. Those are $ 2a \text{ and } \frac{2}{a} $, respectively. Therefore the area of the triangle equals to $ 2a \times \frac{2}{a} \times \frac{1}{2} $, which is 2 units.
Hope this helped!