Find the extremum of function u = x + y + Z under the condition of 1 / x + 1 / y + 1 / z = 1, x > 0, Y > 0, z > 0

Find the extremum of function u = x + y + Z under the condition of 1 / x + 1 / y + 1 / z = 1, x > 0, Y > 0, z > 0

It belongs to conditional extremum
Using Lagrange least squares
Constructor:
F(x,y,z)=x+y+z+λ(1/x+1/y+1/z-1)
X, y, Z, respectively
Fx'(x,y,z)=1-λ/x^2
Fy'(x,y,z)=1-λ/y^2
Fz'(x,y,z)=1-λ/y^2
And make it zero
Then x ^ 2 = y ^ 2 = Z ^ 2 = λ
But x > 0, Y > 0, z > 0
1/x+1/y+1/z=1
Then x = y = z = 3
be
x+y+z=9

Find the extremum of the function z = XY (a-x-y). (a > 0)

Because (a + B + C) / 3 > = Open cubic (a * b * c);
So: XY (a-x-y)

Finding the extremum of function z = XY (a-x-y)
High number of review questions, solution method

Z = XY (a-x-y) (a > 0)
=axy-x²y-xy²
zx=ay-2xy-y²=0
zy=ax-x²-2xy=0
Solution
x=y=0,x=y=a/3
zxx=-2y,zxy=a-2x-2y,zyy=-2x
Obviously (0,0) is not an extreme point
（a/3,a/3）
A=zxx=-2a/3
B=-a/3
C=-2a/3
therefore
AB-C²=4a²/9-a²/9=a²/3>0
A

The extremum of the function z = XY under the additional condition x + y = 1 is

Because x + y = 1, y = 1-x, z = XY, z = x (1-x), that is Z = 1 / 4 - (x-1 / 2) ^ 2
So, when x = 1 / 2, y = 1 / 2, Z reaches the maximum value of 1 / 4

High number: find the extremum of function z = XY under appropriate additional conditions x + y = 1. I can find the extremum, but I don't know how to judge whether it is a maximum or a minimum

x+y=1=(x+y)^2=x^2+y^2+2xy≥4xy(xy>0)
Z = xy = XY / 1 = XY / (x + y) ^ 2 ≤ 1 / 4, this is the maximum,

Find the extremum of the following function z = x ^ 2-xy + y ^ 2-2x + y

This is the result of the following \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(α & # 178; Z / α x

Can I help you again? Find the extremum of the function z = XY under the condition x + y = 2
Find the extremum of the function z = XY under the condition x + y = 2.

From x + y = 2, y = 2-x,
z=xy =x（2-x)=-x^2+2x=-(x^2-2x+1)+1=-(x-1)^2+1
When x = 1, there is a maximum z = 1

Given the function y = ax ^ 3 + BX ^ 2, when x = 1, there is extremum 3; find the value of a, B, find the minimum extremum of function y
Urgent

Finding the extremum of function f (x, y) = - 3x2 + 3y2-9x
I can't understand what it says!
\X09 = 3x2-6x-9 = 6y, the stationary point is (3,0) (- 1,0)
\x09 =6x-6 =0 =6
\X09 is at point (3,0), a = 12 > 0, B = 0, C = 6, △ = - 72

Seeking partial derivative
Partial f / partial x = - 6x - 9
Partial f / partial y = 6y
Let these two partial derivatives be 0, respectively
x = - 3/2
y = 0
Then f (- 3 / 2, 0) = 27 / 4 is the extremum

Find the extremum and monotone interval of function f (x) = (x-1) 2 (x + 1) 3

f(x)=(x-1)^2 * (x+1)^3
f'(x) = (x-1)^2 * 3(x+1)^2 + (x+1)^3 * 2(x-1) = (x+1)^2 * (x-1) * { 3(x-1)+2(x+1) }
= (x+1)^2 * (x-1) * (5x-1)
= 5(x+1)^2 * (x-1/5) * (x-1)
When x ＜ 1 / 5 or X ＞ 1, f '(x) ＞ 0, f (x) monotonically increases; when 1 / 5 ＜ x ＜ 1, f' (x) ＜ 0, f (x) monotonically decreases
Monotone increasing interval: (- infinity, 1 / 5) U (1, + infinity); monotone decreasing interval (1 / 5,1)
When x = 1 / 5, there is a maximum f (1 / 5) = (1 / 5-1) ^ 2 * (1 / 5 + 1) ^ 3 = 3456 / 3125
When x = 1, there is a minimum f (1) = 0