Kane Manufacturing has a division that produces two models of grates, model A and model B. To produce each model A grate requires 3 pounds of cast iron and 6 minutes of labor. To produce each model B grate requires 4 pounds of cast iron and 3 minutes of labor. The profit for each model A grate is $2.00, and the profit for each model B grate is $1.50. Available for grate production each day are 1,560 pounds of cast iron and 22 labor-hours. Because of an excess inventory of model A grates, management has decided to limit the production of model A grates to no more than 180 grates per day.
Let x denote the number of model A grates and y the number of model B grates produced. Then, the problem can be reduced to a linear programming problem with the objective function P = 2x + 1.5y and constraints

Find the range of values that resource 2 (the constant on the right-hand side of constraint 2) can assume.

Question

Kane Manufacturing has a division that produces two models of grates, model A and model B. To produce each model A grate requires 3 pounds of cast iron and 6 minutes of labor. To produce each model B grate requires 4 pounds of cast iron and 3 minutes of labor. The profit for each model A grate is $2.00, and the profit for each model B grate is $1.50. Available for grate production each day are 1,560 pounds of cast iron and 22 labor-hours. Because of an excess inventory of model A grates, management has decided to limit the production of model A grates to no more than 180 grates per day. ​
Let x denote the number of model A grates and y the number of model B grates produced. Then, the problem can be reduced to a linear programming problem with the objective function P = 2x + 1.5y and constraints
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Find the range of values that resource 2 (the constant on the right-hand side of constraint 2) can assume.
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Quizplus · Accepted Answer

Constraint 2 is the labor-hour constraint, which is 22 hours. We need to express this constraint in terms of x and y. For model A, 6 minutes of labor are needed to produce each grate. Therefore, for x model A grates, 6x minutes of labor are needed. We know 1 hour = 60 minutes, so 6x minutes = (6/60)x = 0.1x hours. For model B, 3 minutes of labor are needed to produce each grate. Therefore, for y model B grates, 3y minutes of labor are needed. Again, 1 hour = 60 minutes, so 3y minutes = (3/60)y = 0.05y hours. So the labor-hour constraint can be expressed as: 0.1x + 0.05y ≤ 22 Multiplying both sides by 20 to eliminate decimals, we get: 2x + y ≤ 440 This is equivalent to: y ≤ -2x + 440 Now we can see that the answer must be a value for y that satisfies this constraint. Since y represents labor-hours, it must be a whole number. So we need to find the largest whole number that satisfies this inequality. We can graph the constraint and shade in the region below the line y = -2x + 440: (https://i.imgur.com/OQDWzpv.png) The largest whole number of labor-hours that satisfies this inequality is 37. Therefore, the range of values for resource 2 is: 3x + 2y ≤ 1560 2x + y ≤ 440 y ≤ 37 Solving the third constraint for y, we get: y ≤ 37 Substituting this into the second constraint, we get: 2x + 37 ≤ 440 Solving for x, we get: x ≤ 201.5 Since x must also be a whole number, the largest integer x that satisfies this inequality is 201. Therefore, the range of values for resource 2 is: 1,170 ≤ C_resourse2 ≤ 1,845. So the answer is B.

Kane Manufacturing Has a Division That Produces Two Models of Grates