Given that 2×2
Answer 2×2 =4 ( 2 added 2 times) or simple multiplication.
For 2x2=we have to multiply 2 with 2. Or also table if 2 we can use
What are the zeros of f(x) = 2x2 - 2x -220?
Min Z = 6X1 + 4x2 Subject to Xi + 2x2 > 2 -X1 + 2x2 5 4 3x1 + 2x2 < 12 X1, X2 > 0
Solve this problem using the two-phase method. What special case do you observe? Max Zz4X1-2X2+X3 X1+2X2+X3 3 2X1-3X2+6X3 100 X1,X2,X3>0
What are the x-coordinate(s) of the x-intercept(s) of the function? g(x)4x3 2x2 -12x What are the x-coordinate(s) of the x-intercept(s) of the function? g(x)4x3 2x2 -12x
MAX X Z = 3X1 + 4x2 s.t 2x1 + 2x2 ≤ 8 1x1 + 2x2 ≤ 6 2x2 ≥ 1 please graph with the optimal solution. Then dual price for the first constraint by adding one. Then dual price for the third constraint adding one. Please also graph these with the same graph showing the new optimal solutions Also please show the iso-z lines for the initial problem.
3. Consider the following LP. Maximize u = 4x1 + 2x2 subject to X1 + 2x2 < 12, 2x1 + x2 = 12, X1, X2 > 0. (a) Use simplex tableaux to find all maximal solutions. (b) Draw the feasible region and describe the set of all maximal solutions geometrically.
(b) (6 marks] Given that f(z) = 2x2 + y - 4y2 +i(4xy – 2x2 + 32). (1) Is f differentiable? 1
Consider the following LP problem: Minimize Cost = 3x1 + 2x2 s.t. 1x1 + 2x2 ≤ 12 2x1 + 3 x2 = 12 2 x1 + x2 ≥ 8 x1≥ 0, x2 ≥ 0 A) What is the optimal solution of this LP? Give an explanation. (4,0) (2,3) (0,8) (0,4) (0,6) (3,2) (12,0) B)Which of the following statements are correct for a linear programming which is feasible and not unbounded? 1)All of the above. 2)Only extreme points may be optimal....
Determine the Dual of the following Linear Programming Problems Max 4x1 - 22 + 2.T3 Subiect to: 2x1 + x2 7 Min 4 + 2x2 - T3 Subject to: x1 + 2x2-6 Max 4x1 - 22 + 2.T3 Subiect to: 2x1 + x2 7 Min 4 + 2x2 - T3 Subject to: x1 + 2x2-6
Consider the following. Xi' = 3x1 - 2x2 x1(0) = 3 xz' = 2x1 – 2x2, *2(0) = (a) Transform the given system into a single equation of second order by solving the first equation for x2 and substitute into the second equation, thereby obtaining a second order equation for X1. (Use xp1 for xı' and xpP1 for x1".) xpP1 – xP1 – 2x1 = 0 (b) Find X1 and x2 that also satisfy the initial conditions. *2(t) =