Найдите наибольшее значение функции y=8x-4tgx-2pi+2 на отрезке [-pi/3; pi/3]​

y=8x-4tgx-2π+2

y' = 8 - 4/cos²x;

y' = 0; 8 - 4/cos²x = 0;

4/cos²x = 8;

cos²x = 4/8;

cos²x = 1/4;

2cos²x = 1/2;

1 + cos2x = 1/2;

cos2x = 1/2 - 1;

cos2x = -1/2;

2x = ±arccos( -1/2) + 2πn;

2x = ±( 2π/3) + 2πn;

x = ±( π/3) + πn;

критические точки, принадлежащие данному отрезку: ± π/3.

y(-π/3) = 8(-π/3)-4tg(-π/3)-2π+2 = -8π/3 + 4√3 - 2π + 2 = 4√3 - 14π/3 + 2

y(π/3) = 8π/3-4tg(π/3)-2π+2 = 8π/3 - 4√3 - 2π + 2 = -4√3 + 2π/3 + 2

miny(x) = y(-π/3) = 4√3 - 14π/3 + 2

[-π/3; π/3]

miny(x) = y(π/3) = -4√3 + 2π/3 + 2

[-π/3; π/3]

200-100=100

100+100=200 100гр кофе и 100 гр молока

200-100=100 50гр кофе и 50 гр молока

и того 1/4 часть кофе