### Production Optimization ProblemTwo factories produce three different types of kitchen cutting appliances. The following table summarizes the production capacities, the numbers of each type of appliance ordered, and the daily operating costs for the factories. How many days should each factory operate to fill the orders at minimum cost? Find the minimum cost.*(Use $ x $ for Houston Texas factory and $ y $ for Salem Oregon factory.)*| | Houston Texas Factory | Salem Oregon Factory | Number Ordered ||---------------------|-----------------------|----------------------|----------------|| **Personal Blender**| 80/day | 20/day | 5390 || **Family Blender** | 15/day | 15/day | 1575 || **Food Processor** | 60/day | 25/day | 4550 || **Daily Costs** | $14,000 | $9,000 | |**Objective:**Minimize $ C = $ (cost function to be determined)**Constraints:**- Personal Blenders: $ \geq 5390 $- Family Blenders: $ \geq 1575 $- Food Processors: $ \geq 4550 $**Instructions:**Enter the solution to the simplex matrix below. If there is no solution, enter 'DNE' in the boxes below. If more than one solution exists, enter only one of the multiple solutions below. If needed, round days to 1 decimal place and cost to 2 decimal places.- Number of days to run the Houston Texas factory is _______- Number of days to run the Salem Oregon factory is _______- Minimum cost is $ _______

There are many ways to solve a linear programming problem. If there are only two variables in an LPP, then the most efficient way to solve the problem is the graphical method. In this method, the constraints are treated as equalities and are graphed. Then the common points are found, which are the vertices. The value of the objective function at each vertex is found and eh minimum or maximum value is taken as solution.First formulate the LPP. MinC=14000x+9000ySub to 80x+20y≥539015x+15y≥157560x+25y≥4550x,y≥0 Using graphical method, solve the LPP. First consider the inequalities and equality and graph them. 80x+20y=539015x+15y=157560x+25y=4550Now, using a graphing calculator graph these equalities and find the vertices. The shaded region is the feasible region. The vertices are the points 0,269.5, 54.833, 50.167, 55,50 and 105,0.Now evaluate the value of the cost function at each vertices and find the minimum. x,y C=14000x+9000y 0,269.5 140000+9000269.5=2425500 54.69, 50.75 14000(54.69)+9000(50.75)=1222375 (55,50) 14000(55)+ 9000(50) = 1220000 105,0 14000(105)+9000(0)=1470000 The minimum value is obtained at the vertex 55,50 and the minimum value is 1220000. Therefore, The number of days to run the factory HT is 55. The number of days to run the factory SO is 50. The minimum cost is $1220000.