Предмет: Математика, автор: andarang3

найдите точку максимума функции y = (10 - x)^2 * (6-x) + 7

Ответы

Автор ответа: mukus13

y= $y=(10-x)^2*(6-x)+7=(100+x^2-20x)*(6-x)+7$
$y'=(100+x^2-20x)*(6-x)+7)'=$ $(2x-20)*(6-x)-1*(100+x^2-20x)+0=12x-2x^2-120+20x$ $-100-x^2+20x=-3x^2+52x-220$
$-3x^2+52x-220=0$
D=2704-2640=64
x1=7 1/3
x2=9 2/3
x max=9 2/3

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