Linear Programming Solved Problems

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?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).

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Joanne can carry not more than 3.6 kg of fruits home.

a) Write 3 inequalities to represent the information given above.

A farmer plans to mix two types of food to make a mix of low cost feed for the animals in his farm.

A bag of food A costs $10 and contains 40 units of proteins, 20 units of minerals and 10 units of vitamins.

A bag of food B costs $12 and contains 30 units of proteins, 20 units of minerals and 30 units of vitamins.

How many bags of food A and B should the consumed by the animals each day in order to meet the minimum daily requirements of 150 units of proteins, 90 units of minerals and 60 units of vitamins at a minimum cost?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.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.In these lessons, we will learn about linear programming and how to use linear programming to solve word problems.Many problems in real life are concerned with obtaining the best result within given constraints.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 \[ \begin \ x \ge 0 \ \ x \ge 0 \ \ x y \le 2000 \ \ 8 x 14 y \le 20,000 \ \end \] .Linear programming deals with this type of problems using inequalities and graphical solution method. She must buy at least 5 oranges and the number of oranges must be less than twice the number of peaches.An orange weighs 150 grams and a peach weighs 100 grams.

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