Найдите наименьшее значение функции y=18x-4x√x на отрезке [4; 16]

Y`=(18x-4x√x)`=(18x-4√x³)`=(18x-4x³/²)`=18-4*(3/2)*x¹/²=18-6√x=0 6√x=18  |÷6 √x=3 (√x)²=3² x=9 y(9)=18*9-4*9*√9=162-36*3=162-108=54. y(4)=18*4-4*4√4=72-4*4*2=72-32=40. y(16)=18*16-4*16√16=288-4*16*4=288-256=32=ymin ответ: ymin=32.    (16; 32).

pusti budet x i y storoni preamougolinovo

{2x+2y=26,x*y=36

{x+y=13,x*y=36

{x=13-y,(13-y)y=36

{x=13-y,13y-y^2=36

{x=13-y,y^2+13y-36=0

y^2+13y-36=0

d=169-4*1*(-36)=25

y1=-13-5/2=-9

y2=-13+5/2=4

{x=13-4=9,y=4

otvet: storoni preamougolinika rovni 4 cm i 9 cm

(1/27)*x^3=0.001

1*x^3=27*0.001

x^3=0.027

x^3=(0.3)^3

x=0.3