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?

  • 1
    $\begingroup$ @saran please clarify if it is $\frac{1}{2(e^2+1)}$.?! $\endgroup$
    – Z Ahmed
    Apr 7, 2020 at 5:31
  • $\begingroup$ edited. its supposed to be $ \frac{1}{2}e^2-1$ $\endgroup$
    – satan 29
    Apr 7, 2020 at 5:45
  • $\begingroup$ ~satan Oh! no Your edit shows $\frac{1}{2}(e^2+1).$ $\endgroup$
    – Z Ahmed
    Apr 7, 2020 at 5:48
  • $\begingroup$ oops my bad, its supposed to be + only. $\endgroup$
    – satan 29
    Apr 7, 2020 at 5:49
  • $\begingroup$ Do you have the solution of this problem (since you gave the value for $k$) ? If you do, make an answer to your own question. $\endgroup$ Apr 7, 2020 at 7:00

3 Answers 3


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 $2$.

  • $\begingroup$ I was working the same way. Just out of curiosity, did you compute the area ? I did and it is very surprising. Cheers :-) $\endgroup$ Apr 7, 2020 at 6:32
  • $\begingroup$ @ClaudeLeibovici I added that into the answer. $\endgroup$ Apr 7, 2020 at 6:36
  • $\begingroup$ Unfortunately, this was on a test without calculators so there has to be another way....and just for context, the correct value of k (which matches with yours) happens to be $\frac{1}{2}(e^2-1)$. $\endgroup$
    – satan 29
    Apr 7, 2020 at 6:40
  • $\begingroup$ What properties of the roots can make this result ? Isn't it strange ? $\endgroup$ Apr 7, 2020 at 6:41
  • $\begingroup$ @ClaudeLeibovici Answer that yourself. $\endgroup$ Apr 7, 2020 at 6:41

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.

  • $\begingroup$ +1,Very nice! Although I Am slightly struggling regarding why the pivot thing works. Seems intuitive, but is there a rigorous proof? $\endgroup$
    – satan 29
    Apr 7, 2020 at 8:34
  • $\begingroup$ @satan29 No, it is just intuitive! $\endgroup$ Apr 7, 2020 at 8:35
  • $\begingroup$ Lets take it the other way then. Say the area isnt a minima. What can we say about the area of the small segments obtained after rotating by $d\theta$ about the pivot?, $\endgroup$
    – satan 29
    Apr 7, 2020 at 8:39
  • $\begingroup$ @satan29 They are not equal? ... Minimum area means that a small change makes no difference. It is similar to the Principle of Virtual Work : at a position of equilibrium a small change in any free parameter does no net work. $\endgroup$ Apr 7, 2020 at 8:44

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 $k=\frac{1}{2}(e^2-1)$.


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