The problem: two straight fences intersect at an arbitrary angle. A post lies somewhere between them. What third straight fence, constructed using the post, encloses the minimal area? Call the fences F1 and F2 and the post P, the diagram is what's given.
This is an old problem, and I've seen it before. My question is not what the solution is, but whether a particular kind of calculus-based solution exists.
One solution I already know from John Casey's Sequel to Euclid: construct a triangle with P the midpoint of the base and show that any deviation from point P as the midpoint yields a greater triangle. Construct BP parallel to one fence, mark BC=AB and from C through P draw CD. Since AB=BC and AD is parallel to BP, CP=PD and P is the midpoint of CD. For any other line EG through P, with the help of a line CF parallel to AD, we can show the area gained exceeds the area lost, so the original construction has the minimal area.
I was asked to solve the problem via calculus. The simplest I came up with is this, based on a parallelogram construction.
The post and and fences form a parallelogram $A_0$. Let one side have length $r$. Then for any added length $s$, a triangle through P can be formed with $r+s$ as its length, and the parallelogram completed as shown. Minimizing the triangle is equivalent to minimizing the parallelogram, in which $A_0$ is fixed. I express areas in terms of $A_0$.
$\frac{A_1}{A_0}=\frac{s}{r}$ and $\frac{A_2}{A_1}=\frac{r^2}{s^2}$
Total area is $A=A_1+A_2$, and in terms of $A_0$ this is:
$A = \frac{A_1}{A_0}+\frac{A_2}{A_0} = \frac{A_1}{A_0}+\frac{A_2}{A_1}\frac{A_1}{A_0} = \frac{s}{r} + \frac{r^2}{s^2} \frac{s}{r} = \frac{s}{r} + \frac{r}{s}$
So $A(s) = \frac{s}{r} + \frac{r}{s}$
$A'(s) = \frac{1}{r} + \frac{-r}{s^2}$
$A''(s) = \frac{2r}{s^3} > 0$
Hence when $A'(s)=0$ we have a minimum, which occurs when $r=s$.
My question: What I like about the optimization on $s$ is that it is constructive, as opposed to the Euclidean approach that starts with the answer and proves that no other answer is satisfactory. Also it is fast and simple. I found other optimizations that would work but that were more complicated. One approach I thought was probably simple and nice, but that I could not ultimately see the way to implement, was to view the base as a chord in the circumcircle. I can "see" in my minds eye that the circumradius is minimized when P is the midpoint of the chord, and I suspect this can be the basis of a slick minimization of the triangle area, but I am having trouble coming up with the actual construct. The diagram below shows the idea. Any input would be appreciated.
The analytic solution below is simple, we use (possibly non-rectangular) axes along the lines / fences $F_1$, $F_2$. Let their intersection be the origin. We draw rays / axes $OX$ along $F_1$, and $OY$ along $F_2$, so that $P$ has positive coordinates $p,q>0$, obtained by drawing parallels to $OY$, $OX$ through $P$, and measuring the distances obtained at intersection on the axes $OX$ (the value $p$), and $OY$ (the value $q$).
Then consider a segment $AB$ through $P$ with $A,B$ on the two positive semiaxes, $A\in [OX$, $B\in [OY$. Such a configuration is possible when $A,B$ have coordinates $A=(p+x,0)$ and $B=(0,q+y)$ with positive values $x,y>0$. (A negative value of $x$ makes $B$ move on the negative part of the axis $OY$. A negative value of $y$ makes $A$ move on the negative part of the axis $OX$.)
In the figure, the triangles with sides $q,x,PA$ and $y,p,BP$ are similar. So we obtain a simple condition joining the variables $x,y$, which is $xy=pq$ because of: $$ \frac xp=\frac{AP}{PB}=\frac qy\ . $$ And we want to maximize the area of $\Delta OAB$ which is $$ [OAB] = OA\cdot OB\cdot\sin\widehat{AOB} =(p+x)(q+y)\cdot\sin\widehat{AOB}\ . $$ The above involved angle between the fences is constant. So we want to maximize $(p+x)(q+y)=(pq+xy)+(py+qx)=2pq + (py+qx)$. So we maximize $py+qx$ We apply the $AM-GM$ inequality, $$ py+qx\ge2\sqrt{py\cdot qx}=2pq\ ,\\ $$ and the minimal value is obtained in the case of equal terms, $py=qx$, which leads to $x=p$ and $y=q$. So in the above similarity, we have $AP:PB=1$, so $P$ is the mid point of $AB$.
$\square$