It takes 4 hours to produce the parts of one unit of T2, 2.5 hour to assemble and 1.5 hours to polish.
Per month, 7000 hours are available for producing the parts, 4000 hours for assembling the parts and 5500 hours for polishing the tables.
Vertices: A at intersection of \( 10x 30y = 60 \) and \( y = 0 \) (x-axis) coordinates of A: (6 , 0) B at intersection of \( 20x 20y = 90 \) and \( 10x 30y = 60 \) coordinates of B: (15/4 , 3/4) C at intersection of \( 40x 30y = 150 \) and \( 20x 20y = 90 \) coordinates of C : (3/2 , 3) D at at intersection of \( 40x 30y = 150 \) and \( x = 0 \) (y-axis) coordinates of D: (0 , 5) Evaluate the cost c(x,y) = 10 x 12 y at each one of the vertices A(x,y), B(x,y), C(x,y) and D(x,y).
At A(6 , 0) : c(6 , 0) = 10 (6) 12 (0) = 60 At B(15/4 , 3/4) : c(15/4 , 3/4) = 10 (15/4) 12 (3/4) = 46.5 At C(3/2 , 3) : c(3/2 , 3) = 10 (3/2) 12 (3) = 51 At D(0 , 5) : c(0 , 5) = 10 (0) 12 (5) = 60 The cost c(x , y) is minimum at the vertex B(15/4 , 3/4) where x = 15/4 = 3.75 and y = 3/4 = 0.75.
Choose the scales so that the feasible region is shown fully within the grid.
(if necessary, draft it out on a graph paper first.) Shade out all the unwanted regions and label the required region It also possible to test the vertices of the feasible region to find the minimum or maximum values, instead of using the linear objective function.Vertices of the solution set: A at (0 , 0) B at (0 , 1429) C at (1333 , 667) D at (2000 , 0) Calculate the total profit P at each vertex P(A) = 2 (0) 3 ()) = 0 P(B) = 2 (0) 3 (1429) = 4287 P(C) = 2 (1333) 3 (667) = 4667 P(D) = 2(2000) 3(0) = 4000 The maximum profit is at vertex C with x = 1333 and y = 667.Hence the store owner has to have 1333 toys of type A and 667 toys of type B in order to maximize his profit. It takes 2 hours to produce the parts of one unit of T1, 1 hour to assemble and 2 hours to polish.Let x be the total number of toys A and y the number of toys B; x and y cannot be negative, hence x ≥ 0 and y ≥ 0 The store owner estimates that no more than 2000 toys will be sold every month x y ≤ 2000 One unit of toys A yields a profit of while a unit of toys B yields a profit of , hence the total profit P is given by P = 2 x 3 y The store owner pays and for each one unit of toy A and B respectively and he does not plan to invest more than ,000 in inventory of these toys 8 x 14 y ≤ 20,000 What do we have to solve?Find x and y so that P = 2 x 3 y is maximum under the conditions \[ \begin \ x \ge 0 \ \ x \ge 0 \ \ x y \le 2000 \ \ 8 x 14 y \le 20,000 \ \end \] .Let x be the number of tables of type T1 and y the number of tables of type T2.Profit P(x , y) = 90 x 110 y \[ \begin \ x \ge 0 \ \ x \ge 0 \ \ 2x 4y \le 7000 \ \ x 2.5y \le 4000 \ \ 2x 1.5y \le 5500 \ \end \] .The store owner estimates that no more than 2000 toys will be sold every month and he does not plan to invest more than ,000 in inventory of these toys.How many units of each type of toys should be stocked in order to maximize his monthly total profit profit?The following videos gives examples of linear programming problems and how to test the vertices.Several word problems and applications related to linear programming are presented along with their solutions and detailed explanations.