feasible region calculator

Find the area of the region R enclosed between the x -axis and the graph of y = 3 x from x = 1 to x = 4 by using NINT on a graphing calculator. => 1.806179 square units or 2.57788 square units or 1.3862 square units or 2.34454 square units..

Shade the intersection of all the half planes which is the feasible region..

The graph of the linear system of constraints is called ______. => The constraint region or The feasible region or The linear region or The linear graph..

The optimal solution to a LPP, if it exists, occurs at the corners of the feasible region. The method includes the following st..

To maximise Z draw a line parallel to ax + by = k and farthest from the origin. This line should contain at least one point of the feasible region. Find the coordinates of this point by solving the equations of the lines on which it lies. To minimise Z draw a line parallel to ..

Corresponding to each constant, we obtain a shaded region. The intersection of all these shaded regions is the feasible region or feasible solution of the LPP. Let us find the feasible solution for the problem of a decorative item dealer..

Graphical Method of Solution of a Linear Programming Problem - So far we have learnt how to construct a mathematical model for a linear programming problem. If we can find the values of the decision variables x 1 , x 2 , x 3 , ..... x n , which can optimise (maximize or minimize) the objective func..

Find the optimal solution in the above problem of decorative item dealer whose objective function is Z = 50x + 18y. In the graph, the corners of the feasible region are O (0, 0), A (0, 80), B(20, 60), C(50, 0) At (0, 0) Z = 0 At (0, 80) Z = 50 (0) + 18(80) = Rs. 1440 At (20, 6..

The intersection of the half planes 2x + y 3 and x - y 0 is shown as shaded region in the figure. Feasible region is an unbounded convex region at A (0, 3), Z = 6 (0) + 3 = 3 At B (1, 1), Z = 6(1) + 1 = 7 Consider any point say (4, 5), the value of Z =..

Draw the graphs x + y = 1 - 0.5 -5y = - 10 Shade the half planes of the constraints x + y 1 (1) -0.5x - 5y -10 (2) Note that the origin (0, 0) does not satisfy the inequation (2) hence the required region is the upper half plane. From the graph, it is clear that the intersection of..