Дано: ABC - треугольник,
![\angle BAC=\angle BCA=\alpha,\,\,\,\, AB=BC](https://tex.z-dn.net/?f=%5Cangle+BAC%3D%5Cangle+BCA%3D%5Calpha%2C%5C%2C%5C%2C%5C%2C%5C%2C+AB%3DBC)
r - радиус вписанной окружности
R - радиус описанной окружности
![\angle ABC=180а-2 \alpha](https://tex.z-dn.net/?f=%5Cangle+ABC%3D180%D0%B0-2+%5Calpha+)
по т. Синусов:
![\frac{AC}{\sin\angle ABC} = \frac{BC}{\sin \angle BAC}](https://tex.z-dn.net/?f=+%5Cfrac%7BAC%7D%7B%5Csin%5Cangle+ABC%7D+%3D+%5Cfrac%7BBC%7D%7B%5Csin+%5Cangle+BAC%7D+)
![\frac{AC}{\sin(180а-2 \alpha )} = \frac{BC}{\sin \alpha }](https://tex.z-dn.net/?f=+%5Cfrac%7BAC%7D%7B%5Csin%28180%D0%B0-2+%5Calpha+%29%7D+%3D+%5Cfrac%7BBC%7D%7B%5Csin+%5Calpha+%7D+)
Откуда
![AC= \frac{BC\sin(180а-2 \alpha )}{\sin \alpha } = \frac{ BC\sin2 \alpha }{\sin \alpha } =2BC\cos \alpha](https://tex.z-dn.net/?f=AC%3D+%5Cfrac%7BBC%5Csin%28180%D0%B0-2+%5Calpha+%29%7D%7B%5Csin+%5Calpha+%7D+%3D+%5Cfrac%7B+BC%5Csin2+%5Calpha++%7D%7B%5Csin+%5Calpha+%7D+%3D2BC%5Ccos+%5Calpha+)
Из площади треугольника АВС имеем, что
![S=0.5AC\cdot BC\sin \angle BCA=BC\cos \alpha \cdot BC\sin \alpha =BC^2\cos \alpha \sin\alpha](https://tex.z-dn.net/?f=S%3D0.5AC%5Ccdot+BC%5Csin+%5Cangle+BCA%3DBC%5Ccos+%5Calpha+%5Ccdot+BC%5Csin+%5Calpha+%3DBC%5E2%5Ccos+%5Calpha+%5Csin%5Calpha+)
Радиус вписанной окружности вычисляется по формуле
![r= \frac{S}{p}](https://tex.z-dn.net/?f=r%3D+%5Cfrac%7BS%7D%7Bp%7D+)
![p= \frac{AB+BC+AC}{2}= \frac{2BC+AC}{2} = \frac{2BC+2BC\cos \alpha }{2} =BC(1+\cos \alpha )](https://tex.z-dn.net/?f=p%3D+%5Cfrac%7BAB%2BBC%2BAC%7D%7B2%7D%3D+%5Cfrac%7B2BC%2BAC%7D%7B2%7D++%3D+%5Cfrac%7B2BC%2B2BC%5Ccos+%5Calpha+%7D%7B2%7D+%3DBC%281%2B%5Ccos+%5Calpha+%29)
![\frac{BC}{\sin \angle BAC} =2R\,\,\,\,\,\,\, \Rightarrow\,\,\,\,\,\,\,\,\, R= \frac{BC}{2\sin \alpha }](https://tex.z-dn.net/?f=+%5Cfrac%7BBC%7D%7B%5Csin+%5Cangle+BAC%7D+%3D2R%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C+%5CRightarrow%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C+R%3D+%5Cfrac%7BBC%7D%7B2%5Csin+%5Calpha+%7D+)
Тогда отношение
![\frac{r}{R} = \frac{BC^2\sin \alpha \cos \alpha }{BC(1+\cos \alpha )} \cdot \frac{2\sin \alpha }{BC} = \frac{2\sin^2 \alpha \cos \alpha }{1+\cos \alpha }](https://tex.z-dn.net/?f=+%5Cfrac%7Br%7D%7BR%7D+%3D+%5Cfrac%7BBC%5E2%5Csin+%5Calpha+%5Ccos+%5Calpha+%7D%7BBC%281%2B%5Ccos+%5Calpha+%29%7D+%5Ccdot+%5Cfrac%7B2%5Csin+%5Calpha+%7D%7BBC%7D+%3D+%5Cfrac%7B2%5Csin%5E2+%5Calpha+%5Ccos+%5Calpha+%7D%7B1%2B%5Ccos+%5Calpha+%7D+)
Нужно найти наибольшее значение функции
![\frac{r}{R} = \frac{2\sin^2 \alpha \cos \alpha }{1+\cos \alpha }](https://tex.z-dn.net/?f=+%5Cfrac%7Br%7D%7BR%7D+%3D+%5Cfrac%7B2%5Csin%5E2+%5Calpha+%5Ccos+%5Calpha+%7D%7B1%2B%5Ccos+%5Calpha+%7D+)
на промежутке
![\alpha \in (0; \frac{\pi}{2})](https://tex.z-dn.net/?f=%5Calpha+%5Cin+%280%3B+%5Cfrac%7B%5Cpi%7D%7B2%7D%29+)
![( \frac{r}{R} )'=2\cdot \frac{(\sin^2 \alpha \cos \alpha )'(1+\cos \alpha )-(1+\cos \alpha )'\sin^2 \alpha \cos \alpha }{(1+\cos \alpha )^2} =\\ \\ = 2\cdot\frac{\sin \alpha (2\cos^2 \alpha -\sin^2 \alpha )(1+\cos \alpha )+\sin^2 \alpha \cos \alpha }{(1+\cos \alpha )^2}=\\ \\ =2\cdot \frac{\sin \alpha (2\cos^2 \alpha -1+\cos^2 \alpha )(1+\cos \alpha )+(1-\cos \alpha )(1+\cos \alpha )\cos \alpha }{(1+\cos \alpha )^2} =\\ \\ =2\cdot \frac{\sin \alpha (3\cos^2 \alpha -1+\cos \alpha -\cos^2 \alpha )}{1+\cos\alpha}=](https://tex.z-dn.net/?f=%28+%5Cfrac%7Br%7D%7BR%7D+%29%27%3D2%5Ccdot+%5Cfrac%7B%28%5Csin%5E2+%5Calpha+%5Ccos+%5Calpha+%29%27%281%2B%5Ccos+%5Calpha+%29-%281%2B%5Ccos+%5Calpha+%29%27%5Csin%5E2+%5Calpha+%5Ccos+%5Calpha+%7D%7B%281%2B%5Ccos+%5Calpha+%29%5E2%7D+%3D%5C%5C+%5C%5C++%3D+2%5Ccdot%5Cfrac%7B%5Csin+%5Calpha+%282%5Ccos%5E2+%5Calpha+-%5Csin%5E2+%5Calpha+%29%281%2B%5Ccos+%5Calpha+%29%2B%5Csin%5E2+%5Calpha+%5Ccos+%5Calpha+%7D%7B%281%2B%5Ccos+%5Calpha+%29%5E2%7D%3D%5C%5C+%5C%5C+%3D2%5Ccdot+%5Cfrac%7B%5Csin+%5Calpha+%282%5Ccos%5E2+%5Calpha+-1%2B%5Ccos%5E2+%5Calpha+%29%281%2B%5Ccos+%5Calpha+%29%2B%281-%5Ccos+%5Calpha+%29%281%2B%5Ccos+%5Calpha+%29%5Ccos+%5Calpha+%7D%7B%281%2B%5Ccos+%5Calpha+%29%5E2%7D++%3D%5C%5C+%5C%5C+%3D2%5Ccdot+%5Cfrac%7B%5Csin+%5Calpha+%283%5Ccos%5E2+%5Calpha+-1%2B%5Ccos+%5Calpha+-%5Ccos%5E2+%5Calpha+%29%7D%7B1%2B%5Ccos%5Calpha%7D%3D)
![= \frac{2\sin \alpha (2\cos^2 \alpha +\cos \alpha -1)}{1+\cos \alpha }](https://tex.z-dn.net/?f=%3D+%5Cfrac%7B2%5Csin+%5Calpha+%282%5Ccos%5E2+%5Calpha+%2B%5Ccos+%5Calpha+-1%29%7D%7B1%2B%5Ccos+%5Calpha+%7D+)
Приравниваем к нулю
![\frac{2\sin \alpha (2\cos^2 \alpha +\cos \alpha -1)}{1+\cos \alpha }=0\\ \\ \sin \alpha =0;\,\,\,\,\,\, \Rightarrow\,\,\,\,\, \alpha = \pi n,n \in Z\\ \\ 2\cos^2 \alpha +\cos \alpha -1=0](https://tex.z-dn.net/?f=+%5Cfrac%7B2%5Csin+%5Calpha+%282%5Ccos%5E2+%5Calpha+%2B%5Ccos+%5Calpha+-1%29%7D%7B1%2B%5Ccos+%5Calpha+%7D%3D0%5C%5C+%5C%5C+%5Csin+%5Calpha+%3D0%3B%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C+%5CRightarrow%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C++%5Calpha+%3D+%5Cpi+n%2Cn+%5Cin+Z%5C%5C+%5C%5C+2%5Ccos%5E2+%5Calpha+%2B%5Ccos+%5Calpha+-1%3D0+)
пусть
![\cos \alpha =t(|t| \leq 1)](https://tex.z-dn.net/?f=%5Ccos+%5Calpha+%3Dt%28%7Ct%7C+%5Cleq+1%29)
, тогда имеем
![2t^2+t-1=0\\ D=b^2-4ac=1^2-4\cdot2\cdot(-1)=9\\ t_1=-1\\ t_2=0.5](https://tex.z-dn.net/?f=2t%5E2%2Bt-1%3D0%5C%5C+D%3Db%5E2-4ac%3D1%5E2-4%5Ccdot2%5Ccdot%28-1%29%3D9%5C%5C+t_1%3D-1%5C%5C+t_2%3D0.5)
Обратная замена
![\cos \alpha =-1;\,\,\,\,\,\,\, \Rightarrow\,\,\,\,\,\, \alpha = \pi +2 \pi n,n \in Z\\ \cos \alpha =0.5;\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\, \alpha=\pm \frac{\pi}{3}+2 \pi n,n \in Z](https://tex.z-dn.net/?f=%5Ccos+%5Calpha+%3D-1%3B%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C+%5CRightarrow%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C++%5Calpha+%3D+%5Cpi+%2B2+%5Cpi+n%2Cn+%5Cin+Z%5C%5C+%5Ccos+%5Calpha+%3D0.5%3B%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C+%5CRightarrow+%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C+%5Calpha%3D%5Cpm+%5Cfrac%7B%5Cpi%7D%7B3%7D%2B2+%5Cpi+n%2Cn+%5Cin+Z+)
На промежутке при n=0 корень
![x= \frac{\pi}{3}](https://tex.z-dn.net/?f=x%3D+%5Cfrac%7B%5Cpi%7D%7B3%7D+)
удовлетворяет.
(0)___+__(π/3)__-___(π/2)
В т. х=π/2 производная функции меняет знак с (+) на (-), следовательно х=π/3 - точка максимума.
Найдем наибольшее значение этого отношения
![\frac{2\sin^2 \frac{\pi}{3} \cos\frac{\pi}{3} }{1+\cos\frac{\pi}{3} }=0.5](https://tex.z-dn.net/?f=+%5Cfrac%7B2%5Csin%5E2+%5Cfrac%7B%5Cpi%7D%7B3%7D+%5Ccos%5Cfrac%7B%5Cpi%7D%7B3%7D+%7D%7B1%2B%5Ccos%5Cfrac%7B%5Cpi%7D%7B3%7D+%7D%3D0.5+)
Ответ: наибольшее значение равно 0,5 при
![\alpha =\frac{\pi}{3}](https://tex.z-dn.net/?f=+%5Calpha+%3D%5Cfrac%7B%5Cpi%7D%7B3%7D+)
8:4=2-Свай за 1 годину
звідси 2*6=12-Свай за 6 годин
-7р+3n
если р=2,1 и n=3, то
-7*2,1+3*3=-14,7+9=-5,7
Составим пропорцию
35%-13,5
100%-х
х=(13,5*100)/35=1350/35=38 и 4/7