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Question

Cho x; y không âm với x + y = 1
Tìm GTLN, GTNN của:
$S = (4x^{2} + 3y)(4y^{2} +3x) +25xy$

score 3 · Answer 1

*) Tìm Max
Ta có:
$0\le xy\le \frac{(x+y)^2}{4}=\frac{1}{4}\Rightarrow -\frac{1}{16}\le xy-\frac{1}{16}\le\frac{3}{16}\Rightarrow |xy-\frac{1}{16}|\le\frac{3}{16}$
Suy ra:
$S=16(xy-\frac{1}{16})^2+\frac{191}{16}\le16.(\frac{3}{16})^2+\frac{191}{16}=\frac{25}{2}$
Max$S=\frac{25}{2}\Leftrightarrow x=y=\frac{1}{2}$

score 3 · Answer 2

Bình chọn tăng 3 Bình chọn giảm

*) Tìm Min
Ta có:
$S=16x^2y^2+12(x^3+y^3)+34xy$
    $=16x^2y^2+12(x+y)^3-36xy+34xy$
    $=16x^2y^2-2xy+12$
    $=16(xy-\frac{1}{16})^2+\frac{191}{16}\ge\frac{191}{16}$
Min$S=\frac{191}{16}\Leftrightarrow\left\{ \begin{array}{l} x+y=1\\ xy=\frac{1}{16} \end{array} \right. \Leftrightarrow (x,y)\in\{(\frac{2+\sqrt3}{4};\frac{2-\sqrt3}{4});(\frac{2-\sqrt3}{4};\frac{2+\sqrt3}{4})\}$