Find the maximum of - X & # 178; - 6x + 7

Find the maximum of - X & # 178; - 6x + 7

-x²-6x+7=
-(x²+6x+9)+9+7
=-(x-3)²+16
（x-3）²》=0
-(x-3)²《=0
So the maximum is 16

If 0

00
y=3x(3-x)
≤3 *[(x+3-x)/2]²
=3*(9/4)
=27/4
If and only if x = 3-x, that is, x = 3 / 2, the equal sign holds
So the maximum value of 3x (3-x) is 27 / 4

Known quadrilateral ABCD is an inscribed rectangle of circle X & # 178; + Y & # 178; = 9, find the maximum perimeter of rectangle ABCD? Do it with mean value theorem!

Let a be in the first quadrant, and the four sides of the rectangle a (x, y) (x > 0, Y > 0) are parallel to the axis of symmetry. Then the perimeter is 4x + 4Y. By using the mean inequality X & # 178; + Y & # 178; ≥ 2XY ｜ 2 (X & # 178; + Y & # 178;) ≥ (x + y) &# 178; ｜ 2 * 9 ≥ (x + y) &# 178; ｜ x + y ≤ 3 √ 2 ｜ 4x + 4Y ≤ 12 √ 2, that is, the perimeter is