Solve the LP problem. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. Maximize p=x+ysubject to

x+3y≤4

3x+y≤4

x≤0

y≤0

P = 4

Step-by-step explanation:

The LP is:

Maximize p = x+y

x+3y≤4

3x+y≤4

x ≥ 0

y ≥ 0

Solving graphically using the geogebra graphing calculator which is attached, the points are A(0, 4), B(0, 1.33), C(1.33, 0), D(4, 0) and E(1, 1)

The maximum objective is:

For point A(0, 4):  Maximize p = x + y = 0 + 4 = 4

For point B(0, 1.33):  Maximize p = x + y = 0 + 1.33 = 1.33

For point C(1.33, 0):  Maximize p = x + y = 1.33 + 0 = 1.33

For point D(4, 0):  Maximize p = x + y = 4 + 0 = 4

For point E(1, 1):  Maximize p = x + y = 1 + 1 = 2

Hence, the maximum point is at A(0, 4) which gives P = 4

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation:

the answer is b.

step-by-step explanation:

identity property is where 0+whatever number=whatever number. basically 0+ a number.