Use the technique developed in this section to solve the minimization problem. Minimize C = −3x − 2y − z subject to −x + 2y − z ≤ 20 x − 2y + 2z ≤ 25 2x + 4y − 3z ≤ 30 x ≥ 0, y ≥ 0, z ≥ 0 The minimum is C = at (x, y, z) = .

C= -145, (35/4, 295/8, 45)

Step-by-step explanation:

Use Gaussian elimination to find the values of x, y and z

Eq 1: -x+2y-z=20

Eq 2: x-2y+2z=25

Eq 3: 2x+4y-3z=30

Multily Eq1 by 1 and add to Eq 2

Eq 1: (-x+2y-z=20 ) × 1

Eq 2:  x-2y+2z=25

Eq 3:  2x+4y-3z=30

⇒ Eq1: -x+2y-z=20

    Eq2:         z= 45

   Eq 3: 2x+4y-3z=30

Multiply Eq 1 by 2 and then add to Eq 3

Eq1:  (-x+2y-z=20 ) × 2

Eq2:            z= 45

Eq3:   2x+4y-3z=30

⇒ Eq1:  -x+2y-z=20

   Eq2:            z= 45

  Eq3:      8y-5z= 70

swap Eq 2 and Eq 3

Eq 1: -x+2y-z=20

Eq 3:     8y-5z= 70

Eq 2:       z= 45

Solve Eq 2 for z

Z=45

solve Eq Eq 3 for y.

y= 295/8

Using the value z=45 and y= 295/8, substitue in Eq 1 to get value of x

x= 35/4

Substitue values of x,y and z in C= -3x-2y-z to get minimum value of C

C= -145

Ok here we go ..there are plenty of ways to solve such math problems but we will go for the smartest and easiest ..5x+2y=7  -2x+6y=9we rearrange them into the following ..by multiplying the 1st equation by 2 it becomes 10x + 4y = 14we rearrange it and so it becomes 10x = 14 - 4yas for the second equation we multiply it by 5 and rearrange it so it becomes as to the following ..-10x + 30y = 45rearranging so 10x = 30y -45now 10x equals both 14 - 4y and 30y -45so   14 - 4y = 30y -4531 = 26yy= 31/26 = 1.19 or around 1.2so by now we can calculate x easily as  -2x+6y=9so 2x = 6y -9 = 6 * 31/26 - 9 = -1.84  then x = - 0.92 or around -0.9best luck!