Given | x | ≤ 1, | y | ≤ 1, let m = | x + y | + | y + 1 | + | 2y-x-4 |, find the maximum and minimum of M

Given | x | ≤ 1, | y | ≤ 1, let m = | x + y | + | y + 1 | + | 2y-x-4 |, find the maximum and minimum of M

For the case of X + y 124124124124124124124124124124\124124124124124124\124124124\124\124\124\124\124\\124\124\124\124\124124\124\\\\\\\\\\\\\\\themaximum value of M is 7 and the minimum value is 3

Let x, y {X-1 ≤ 0, x + y + 1 ≥ 0, X-Y + 3 ≥ 0}, then the objective function z = the minimum value of 2x + y____
The range of (y + 3) / (x + 1) is________ .
The minimum value of (x + 1) &# 178; + (y + 2) &# 178; is________ .

In fact, we have to draw a graph to see the slope of the objective function at the end (intersection) point. In fact, it is to substitute the end (intersection) point value into the objective function
For example, the intersection points above the variables are (- 2,1) (1, - 2) (1, - 4) respectively, which are substituted into the objective function z = 2x + y, and the minimum value is - 3
The range of Y + 3) / (x + 1) is (- 4,0.5)
The minimum value of (x + 1) &# 178; + (y + 2) &# 178; is 4

If x.y satisfies the condition {x + y = 0}, then the maximum and minimum of the objective function z = 6x + 8y are? And? Respectively

Let x.y satisfy the condition {x + y = 0} and make a solution domain, x + y = 5 and 2x + y = 6, intersection (1,4)
It is easy to get: the maximum value of the objective function z = 6x + 8y = 6 * 1 + 8 * 4 = 38
Minimum = 6 * 0 + 8 * 0 = 0

If we know that the focal length of the ellipse X & # 178 / 4 + Y & # 178 / M = 1 is equal to 6, then M = ()
Write the process, thank you

Because the focal length is 6, so C = 3 > 2, so M-4 = C ^ 2 = 9, so m = 13

The focal length of the ellipse m branch X & # 178; + 4 branch Y & # 178; = 1 is 2. The detailed process of finding the value of M is given. Thank you

Focal length = 2C = 2
c=1
c²=1
x²/m²+y²/4=1
When the focus is on the X axis
m²-4=1
m²=5
m=±√5
When the focus is on the y-axis
4-m²=1
m²=3
m=±√3

What is the focal length of an ellipse X & # 178 / M & # 178; + Y & # 178 / 4 = 1 passing through (- 2, √ 3)

Taking the coordinates of this point into the elliptic equation, we can get m ^ 2 = 16, so the focal length 2C = 2 root sign (a ^ 2-B ^ 2) = 2 root sign (16-4) = 4 root sign 3

Given that the major axis of the ellipse X & # 178; / (M + 2) + Y & # 178; / (10-m) = 1 is on the x-axis, the focal length is 4, then M=

2c=4,c=2
c²=a²-b²=m+2-(10-m)=2m-8=4
m=6

If x, y satisfy 3x ^ 2 + 2Y ^ 2 = 6x, find z = x ^ 2 + y ^ 2 range

3x^2+2y^2=6x
6x>=0
x>=0
3x^2+2y^2=6x
2x^2+2y^2=-x^2+6x
x^2+y^2=-x^2+6x/2
z=x^2+y^2
z=-x^2+6x/2=-1/2(x^2-6x+9-9)
=-1/2(x-3)^2+9/2

Find the extremum of the function f (x, y) = x ^ 2 + y ^ 2-4x + 2Y and judge whether it is a maximum or a minimum

f(x,y)=(x-2)^2-4+(y+1)^2-1=(x-2)^2+(y+1)^2-5
The minimum values of (X-2) ^ 2 and (y + 1) ^ 2 are 0
So the minimum value of F (x, y) is - 5, and there is no maximum value

（x+2y）²（x-2y）²（x²+4y²)²

A & sup2; - B & sup2; = (a + b) (a-b) (x + 2Y) & sup2; (x-2y) & sup2; (X & sup2; + 4Y & sup2;) & sup2; = [(x + 2Y) (x-2y)] & sup2; (X & sup2; + 4Y & sup2;) & sup2; = (X & sup2; - 4Y & sup2;) & sup2; (X & sup2; + 4Y & sup2;) & sup2; = [