Two straight fences meet at a point, but not necessarily at right angel between them. A post stand in the angle between them. If a triangular corral is constructed by building a new straight fence containing this post, show that the fenced off triangle has minimal area when the old post is in the center of the new fence. I draw the triangle, labeled first two fence as x and y and angle between them as $t$ then 3th fence divided two pieces and labeled $a$ and $b$. $\frac{xysin{t}}{2}=AREA_{min}$ All coming my mind is cos theorem but it get rids of a and b while i need at least one of them. (Sorry i dont know how to draw triangle.)
2026-05-16 08:05:29.1778918729
Calculus fence and corral
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Hint: Put the post at the origin and the intersection of the two fence segments on the circle of constant distance from the post. Then the $x$-axis contains the third fence section. Labelling the two parts of the third fence $a$ and $b$ as you did, we have a triangle with base $a+b$ and height $r \cdot \sin \theta$. Now maximize area with respect to varying $\theta$. Once you have the optimal $\theta$, show that this makes $a = b$.