Minimization problem of $e^x$ and straight line.

Question

Given the curve $C$: $y=e^{x}$ and a point $P$: $(x_0,y_0)$ with $y_0 > e^{x_{0}}$, a straight line $L$ of slope $k$ goes through the point $P$ and intersects $C$ at two points $A$ and $B$, try to find out: $(1).$ $k$ that minimizes the distance of $A$ and $B$, $(2).$ k that minimizes the area of triangle OAB ($O$ is the origin $(0,0)$).

Edit:

I've tried the Lagrange multiplier method and only solved part of the problem(the minimal triangle when P on yAxis, in this case, $k=y_0$). However, the steps are quite complicated (for high school students). I'd like to know if there is some elementary methods to approach this problem. Thanks!

Edit 2:

(add the image)

I took the following Steps:

1. as shown in the chart, points P,Q,A,B are colinear. then we got the constraits of variables a,b and q, which are: $g(a, b) = 0, g(a,q) = 0, g(b,q) = 0$.
2. the distance between A and B is given by $f(a,b) = sqrt((b-a)^2+(e^b-e^a)^2)$
3. the area of the triangle OAB is given by $f(a,b,q)=1/2 * q * (b-a)$
4. ${\mathcal {L}}(a,b,\lambda) = f(a,b)+\lambda*g(a,b)$
5. solve equations: $\nabla _{a,b,\lambda }{\mathcal {L}}(a,b,\lambda )=0$

The last step was the part that is very complicated.

Answer 1

From the constraint $$\dfrac{e^b-y_0}{b-x_0} = \dfrac{y_0-e^a}{x_0-a}$$ it is not difficult to solve it for $b$. This gives $$b(a)=x_0+\frac{y_0 (x_0-a)}{e^a-y_0}-W\Big[\frac{x_0-a}{e^a-y_0}\exp\left(\frac{e^a x_0-a y_0}{e^a-y_0} \right)\Big]$$ The minimum of $$f(a) = \big[a-b(a)\big]^2+\big[e^a-e^{b(a)}\big]^2$$ will happen at a value of $a$ a bit larger than $\log(y)$.

We do not need optimization since the problem is now to find the root of $f'(a)=0$.

$$f'(a)=(a-b(a)) \left(1-b'(a)\right)+\left(e^a-e^{b(a)}\right) \left(e^a-e^{b(a)} b'(a)\right)$$ The edit gives the formal expression of $b'(a)$.

Now, using a very poor starting point $a_0=3$, Newton iterates are $$\left( \begin{array}{cc} n & a_n \\ 0 & 3.000000000 \\ 1 & 2.518474855 \\ 2 & 2.048112714 \\ 3 & 1.597475428 \\ 4 & 1.194726874 \\ 5 & 1.004001060 \\ 6 & 1.029256086 \\ 7 & 1.031630200 \\ 8 & 1.031646684 \end{array} \right)$$

If we want a better starting point, compute $f'(a)$ for $a=k \log(y)$ with $k >1$ until $f'(a)> 0$. For the worked example, this would give $$\left( \begin{array}{cc} k & f'(k \log(y)) \\ 1.1 & -324.676 \\ 1.2 & -55.3709 \\ 1.3 & -18.4489 \\ 1.4 & -5.89996 \\ 1.5 & +0.61317 \end{array} \right)$$

So, our starting guess $a_0=1.5\log(2) \sim 1.03972$ which is more than decent. Now, Newton iterates will converge without any overshoot since $f'(a_0)$ is positive and $f'''(a)>0$ (by Darboux theorem). $$\left( \begin{array}{cc} n & a_n \\ 0 & 1.039720771 \\ 1 & 1.031465818 \\ 2 & 1.031646591 \\ 3 & 1.031646684 \end{array} \right)$$

In short, the problem seems to be simple from a numerical point of view. Beside the root finder algorithm (using numerical derivatives), the only requirement is a routine for the computation of Lambert function (there are many which are available).

Edit

For the formal calculation of $b'(a)$ defining three intermediate functions of $a$ $$u=\frac{x-a}{e^a-y} \qquad \qquad v=\frac{e^a x-a y}{e^a-y}\qquad \qquad w=W(u\, e^v)$$ to make $$b(a)=x+y\,u-w\implies b'(a)=y u'-w'$$ $$u'=\frac{e^a (a-x-1)+y}{\left(e^a-y\right)^2}\qquad \qquad v'=y\, u'\qquad \qquad w'=\frac w {1+w}\left(\frac {u'}u +v'\right)$$

Answer 2

This is not an answer, just an illustration. The picture below represents the contour lines for $f(a,b) = (a-b)^2+(e^a-e^b)^2$ and, in red, the constraint $\dfrac{e^b-y_0}{b-x_0} = \dfrac{y_0-e^a}{x_0-a}$. This was obtained for $(x_0,y_0) = (\frac 12, 2)$

As you can see, the minimum will be attained at a point where the contour line of $f$ is tangent to the constraint. This is in fact the thinking behind Lagrange multipliers. I believe that Lagrange multipliers is probably the best way to go but, even in this case, we need handle the numerical solution of one or more equations.