Minimising the area between a line and an exponential curve We need to find a suitable k , for which the area between the two curves $y=e^x, y=k(x-1) + \frac{1}{2}(e^2+1)$ is minimum.
The line passes through a fixed point $(\frac{1}{2}(e^2+1))$, and has variable slope. But, I don"t think its possible to determine the intersection points of the two curves , so I had assumed them to be $\alpha$ and $\beta$.
I performed the integral $\int (k(x-1) + 1/2(e^2+1) -e^x)dx$ from $\alpha$ to $\beta$, (the line lies above the curve for this interval) and tried to make use of the fact that : $e^\alpha= k(\alpha-1)+\frac{1}{2}(e^2+1)$ and likewise for $\beta$, but was still unable to get the area explicitly in terms of $k$. 
There might be a geometric argument that minimizes the area but the ways these curves are, I fail to realize it.
Can we have generalized method to minimize/maximize the are enclosed by them? Without explicitly knowing their intersection points?
 A: Let $f(k,x)=k(x-1)+\frac12(e^2+1)-e^x$, so $\alpha,\beta$ satisfy $f(k,\alpha)=f(k,\beta)=0$ for some fixed $k$. Minimising the area is equivalent to setting the derivative of the integral to be $0$:
$$\frac d{dk}\int_\alpha^\beta f(k,x)\,dx=\int_\alpha^\beta\frac\partial{\partial k}f(k,x)\,dx=\int_\alpha^\beta(x-1)\,dx=(\beta^2-\alpha^2)/2+\alpha-\beta$$
$$=\frac12(\beta-\alpha)(\alpha+\beta-2)=0$$
From this we get $\alpha+\beta=2$ and
$$f(k,\alpha)+f(k,\beta)=k(\alpha-1+\beta-1)+e^2+1-e^\alpha-e^{2-\alpha}$$
$$=k(\alpha+\beta-2)+e^2+1-e^\alpha-e^{2-\alpha}$$
$$=-e^\alpha-\frac{e^2}{e^\alpha}+e^2+1=0$$
This last equation rearranges into a quadratic in $e^\alpha$:
$$e^{2\alpha}-(e^2+1)e^\alpha+e^2=(e^\alpha-1)(e^\alpha-e^2)=0$$
So we get $\alpha=0$, $\beta=2$ and $k=\frac12(e^2-1)$. The minimum area is $2$.
A: Perhaps the geometrical trick or insight is as follows.
Suppose the fixed point is P and the line intersects the curve at A and B. 
When the enclosed area is minimised, then a small change in the slope $k$ makes no difference to that area. As the line pivots about P through a small angle $\delta \theta$ the triangular area $AP \delta\theta$ removed on one side equals the triangular area $PB\delta\theta$ added on the other side. So $AP=PB$ - ie P is then the midpoint of AB.
(Note : Since this argument does not depend on the form of the curve, it applies for all curves. Also, the minimum area may be a local rather than global minimum. It may even be a local or global maximum since the method only finds turning points.)

The fixed point has co-ordinates $x=1, y=\frac12 (e^2+1)$. 
Suppose the co-ordinates of A and B are $(\alpha, e^{\alpha})$ and $(\beta, e^{\beta})$. Then 
$$\frac12 (\alpha + \beta) = 1$$
$$\frac12 (e^{\alpha}+e^{\beta})=\frac12 (e^2+1)$$
$$k=\frac{e^{\beta}-e^{\alpha}}{\beta-\alpha}$$
from which 
$$e^{(\alpha+\beta)}=e^{\alpha}e^{\beta}=e^2$$
$$e^{\alpha}(e^{\alpha}+e^{\beta})=(e^{\alpha})^2+e^2=e^{\alpha}(e^2+1)$$
$$(e^{\alpha})^2-(e^2+1)e^{\alpha}+e^2=(e^{\alpha}-e^2)(e^{\alpha}-1)=0$$
If $e^{\alpha}=1$ then $\alpha=0, \beta=2, e^{\beta}=e^2$.
Conversely if $e^{\alpha}=e^2$ then $\alpha=2, \beta=0, e^{\beta}=1$.
Therefore $$k=\frac12 (e^2-1)$$
Although this method finds the slope $k$ of the chord through P which minimises (or maximises) area, it does not find the area.
A: After a brilliant idea by Parcly Taxel to invoke feynman's trick, I think it can be now be solved without the lambert W function.
After setting the derivative=0,we get $\alpha + \beta =2$.(since $\alpha$ and $\beta$ are distinct).
Originaly, we had the equations $e^\alpha= k(\alpha-1)+\frac{1}{2}(e^2+1)$ and $e^\beta= k(\beta-1)+\frac{1}{2}(e^2+1)$.
Adding these 2, and using $\alpha + \beta=2$, we get a simple quadratic equation which gives $\alpha=0, \beta=2$ and thus $k=\frac{1}{2}(e^2-1)$.
