Application of basic inequalities. Finding the maximum value of function y = (x-1) / (x ^ 2-2x + 10), (x > 1)

Application of basic inequalities. Finding the maximum value of function y = (x-1) / (x ^ 2-2x + 10), (x > 1)

y=(x-1)/(x^2-2x+10)＝(x-1)/（(x－1）^2+9)
If u = X-1 > 0
y＝u/（u^2+9）=1/(u+9/u)
(U + 9 / U) (U * 9 / U) ^ 1 / 2 = 3
So y "1 / 3"
That is, the maximum value of Y is 1 / 3

Known 0

∵0＜x＜1/3,∴1－3x＞0
[method 1]
y＝x(1－3x)＝1/3•3x•(1－3x)≤1/3[ ( 3x＋(1－3x) )/2 ] ²＝ 1/12
If and only if 3x = 1-3x, i.e. x = 1 / 6, take the equal sign
When x = 1 / 6, the function obtains the maximum value of 1 / 12
[method 2]
∵0＜x＜1/3,∴1/3－x＞0
∴y＝x(1－3x)＝3•x(1/3－x)≤3[ ( x＋(1/3－x) )/2 ] ²＝ 1/12
The equal sign holds if and only if x = 1 / 3-x, i.e. x = 1 / 6
When x = 1 / 6, the function obtains the maximum value of 1 / 12

The real numbers x and y are known to satisfy the system of inequalities x-y+2≥0 x+y-4≥0 2x-y-5≤0 ， If the only optimal solution when the objective function z = y-ax reaches the maximum is (1,3), the value range of real number a is () A. （-∞，-1） B. （0，1） C. [1，+∞） D. （1，+∞）

The plane area is made according to the meaning of the topic,
Change z = y-ax into y = ax + Z, Z is equivalent to the longitudinal intercept of the straight line y = ax + Z,
It can be seen from the figure that if the objective function z = y-ax is maximized, the only optimal solution is B (1,3),
Then a > 1,
Therefore, D

It is known that the function f (T) is an odd function and an increasing function on R. if x and y satisfy the inequality f (x2-2x) ≤ - f (y2-2y), the maximum value of x2 + Y2 is () A. three B. 2 two C. 8 D. 12

∵f（x2-2x）≤-f（y2-2y），
∴f（x2-2x）≤f（-y2+2y），
∵ f (x) is an increasing function
‡ x2-2x ≤ - Y2 + 2Y, finishing (x-1) 2 + (Y-1) 2 ≤ 2
If the coordinates of point P are (x, y), point P is centered on (1, 1),
2 is the point on and within the circle of radius, and this circle passes through the origin
be
x   2+y   2 is the distance from point P to the origin,
∵ the circle crosses the origin,
∴
x   2+y   The maximum value of 2 is the diameter of the circle 2
two
The maximum value of x2 + Y2 is 8
So choose C

Satisfaction inequality system 2x+y−3≤0 7x+y−8≤0 x，y＞0 ， Then the maximum value of the objective function K = 3x + y is __

First draw the group satisfying the inequality
2x+y−3≤0
7x+y−8≤0
x，y＞0 The flat area of the,
Make the objective function K = 3x + y and the translation line k = 3x + y
When passing point a (1, 1), the objective function K = 3x + y takes the maximum value of 4,
So kmax = 4
So the answer is: 4

Known 0

x(3-2x)=2x[(3/2)-x]≤2(x+1.5-x) ²/ 4=9/8
When x = 3 / 4, the equal sign holds