tìm gtln của p=x2 y2 z2-4x-4y-z

12x + 10y + 15z ≤ 60

⇒ \(\frac{x}{5} + \frac{y}{6} + \frac{z}{4} \le 1\)

Đặt \(\frac{x}{5} = a;\frac{y}{6} = b;\frac{z}{4} = c\left( {0 \le a,b,c \le 1} \right),a + b + c \le 1\)

Khi đó T = 25a² + 36b² + 16c² – 20a – 24b – 4c

\(25a\left( {a - \frac{{32}}{{25}}} \right) \le 0\)⇒ 25a² ≤ 32a

36b(b – 1) ≤ 0 ⇒ 36b² ≤ 36b

16c(c – 1) ≤ 0 ⇒ 16c² ≤ 16c

Suy ra: T ≤ 32a + 36b + 16c – 20a – 24b – 4c = 12(a + b + c) ≤ 12

Vậy T_max = 12 khi \(\left\{ \begin{array}{l}a = 0\\b = 0\\c = 1\end{array} \right.\) hoặc \(\left\{ \begin{array}{l}a = 0\\b = 1\\c = 0\end{array} \right.\)