Explanation Given: Tabulate the given information. Models Cast iron Time (in min) Profit ( in $ ) A 3 lb 6 m $ 2.00 B 4 lb 3 m $ 1.50 Approach: Assume x is the no. of units to manufacture the model A on machine 1 and y is the no. of units to manufacture the model B on machine 2 . So, the total profit P (in dollars) is P = 2 x + 1.5 y and this is the objective function to be maximized. The total amount for Cast iron is given by 3 x + 4 y and must not exceed 1000 lb . So, the inequality is, 3 x + 4 y ≤ 1000 The total amount of time is given by 6 x + 3 y and must not exceed 1200 .So, the inequality is, 6 x + 3 y ≤ 1200 The total number of grates produced per day are x + y and must not less than 150 .So, the inequality is, x + y ≥ 150 As neither x nor y will be negative, so x ≥ 0 y ≥ 0 Calculation: Consider the given inequalities. 3 x + 4 y ≤ 1000 6 x + 3 y ≤ 1200 x + y ≥ 150 Replace the inequality sign with equality sign in the above linear inequalities and write them in intercept form. 3 x + 4 y = 1000 3 x 1000 + y 250 = 1 So, the coordinates are ( 33.33 , 0 ) and ( 0 , 250 ) . Also, 6 x + 3 y = 1200 x 200 + y 400 = 1 So, the coordinates are ( 200 , 0 ) and ( 0 , 400 ) . And, x + y = 150 x 150 + y 150 = 1 So, the coordinates are ( 150 , 0 ) and ( 0 , 150 ) . Plot the graph for the linear inequalities and shade the feasible set area. The vertices of the feasible set from the graph are A ( 0 , 250 ) , B ( 120 , 160 ) , C ( 200 , 0 ) D ( 150 , 0 ) and E ( 0 , 150 )

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Author: Tan

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Chapter 3.3, Problem 34E

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The number of grates of each model that should be produced in per day to maximize the Kane`s profit and optimal profit.

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Chapter 3.3, Problem 33E

Chapter 3.3, Problem 35E

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