Given that X and y satisfy the constraint condition x + y − 2 ≥ 0 x ≤ 2 y ≤ 2, find the maximum and minimum of Z = 2x + y

Given that X and y satisfy the constraint condition x + y − 2 ≥ 0 x ≤ 2 y ≤ 2, find the maximum and minimum of Z = 2x + y

Make the plane region represented by the inequality system x + y − 2 ≥ 0x ≤ 2Y ≤ 2 (as shown in the figure), that is, the row region ------ (4 points) change the objective function z = 2x + y into y = - 2x + Z ------ (5 points) make z = 0, make the line l0: y = - 2x, translate the line l0, pass through the point a in the feasible region, the value of Z is the minimum, pass through the point C in the feasible region, the value of Z is the maximum. - - (7 points) get a from x + y − 2 = 0y = 2 (0, 2), from y = 2x = 2, C (2, 2), --- - (8 points) at this time, Zmin = 2 × 0 + 2 = 2, Zmax = 2 × 2 + 2 = 6 ------ (10 points)

What is the minimum value of X + Y - 4x + 6y + 15?

= (x-2)^2 + (y+3)^2 + 2
So min = 2

1-(x-1)/3=(x+3)/2 （2y+1）/4-1=y-（10y+1）/12 3x+（x+1）/2=3-（2x-1）/3

1 - (x-1) / 3 = (x + 3) / 2 multiply both sides by 6 to get 6-2 (x-1) = 3 (x + 3) 6-2x + 2 = 3x + 93x + 2x = 6 + 2-95x = - 1x = - 1 / 5 (2Y + 1) / 4-1 = y - (10Y + 1) / 12 multiply both sides by 12 to get 3 (2Y + 1) - 12 = 12Y - (10Y + 1) 6y + 3-12 = 12y-10y-16y-12y + 10Y = - 1-3 + 124y = 8y = 23x + (x + 1) /

What does the positive integer solution of X 1,2,3 represent

0

Ask a question: it is known that △ ABC is all equal to △ a'b'c ', ad and a'd' are the middle lines of △ ABC and △ a'b'c ', respectively

Proof: ? △ ABC ≌ △ a'b'c '(known)
{B =} B ′, BC = B ′ C ′, ab = a ′ B ′ (the corresponding sides and angles of congruent triangles are equal)
And ∵ AD and a'd 'are the midlines of △ ABC and △ a'b'c' respectively
∴BD=B′D′
In △ abd and △ a ′ B ′ D ′,
∵ AB=A′B′
｛∠B=∠B′
BD=B′D′
∴△ABD≌△A′B′D′（SAS）
Ψ ad = a ′ D ′ (the corresponding sides of congruent triangles are equal)

Let a, B, C ∈ R, a + B + C equal o, ABC > 0, prove 1 / A + 1 / B + 1 / C
Another question: compare the quartic power of 1 + 2x with the square of 2x + the square of X?

1、
a+b+c>0
So there are positives and negatives
Let a > = b > = C
abc>0
So one positive and two negative
a>0>b>=c
1/a+1/b+1/c=(bc+ac+ab)/abc
abc>0
Looking at molecules
a+b+c=0
a=-(b+c)
So molecule = BC + a (B + C)
=bc-(b+c)²
=bc-b²-2bc-c²
=-(b²+bc+c²)
=-[(b+c/2)²+c²/4]=2x³+x²
Where x = 1 is equal

Let a, B, C ∈ R and a + B + C equal to o, ABC < 0, and prove 1 / A + 1 / B + 1 / C > o
I'm in a hurry

The denominator ABC is negative and the numerator AB + BC + AC. the square of a + B + C is calculated and observed. The square term subtracted from the result must be less than zero

If ABC > 0, B + C / A + C + A / B + A + B / C is greater than or equal to 6
Proved by arithmetic mean or geometric mean.

(b+c)/a+(c+a)/b+(a+b)/c
=b/a + c/a + c/b + a/b + a/c + b/c
=( b/a + a/b ) + ( c/a + c/a) + (c/b+b/c)
>= 2 + 2 + 2
>=6

Given a >, b > 0, C > O, and a + B + C = 1, it is proved that (1-A) (1-B) (1-C) is greater than or equal to ABC
It's known that it's a > o

(1-a )(1-b)(1-c)
=1-c-b+bc-a+ac+ab-abc
=[1-(a+b+c)]+bc+ac+ab-abc
=ab+bc+ac-abc
Because a > 0, b > 0, C > 0
So (1-A) (1-B) (1-C) > = ABC

A times b times 6 is equal to a times c times 3

A can be as much as B 3C 6