Explanation Given: The total hectare for growing both the crop is 500,000. The contract requires at least 100,000 hectares for coffee and at least 200,000 to at most 270,000 hectares for cocoa. The number of workers required for the production of coffee and cocoa is 2 per hectare and 5 per hectare respectively. The profit made from coffee is $220 per hectare and from cocoa is $550 per hectare. The total number of workers for growing both the crop is 1,750,000. Calculation: Let x and y represents the number of hectares of coffee and cocoa crops respectively. From the given data, the following observations are made. x + y ≤ 500 , 000 . x ≥ 100 , 000 200 , 000 ≤ y ≤ 270 , 000 Since 2 workers are required per hectare, the total number of workers needed for the production of coffee is 2 x . Similarly, the total number of workers needed for the production of cocoa is 5 y . Hence the number of workers needed for both the crop is 2 x + 5 y Since the total number of workers is 1,750,000, the resulting inequality is 2 x + 5 y ≤ 1 , 750 , 000 Given the profit made from coffee is $220 per hectare and from cocoa is $550 per hectare. That is, z = 220 x + 550 y From the above information, the linear programing problem can be stated as follows, Maximize z = 220 x + 550 y (1) Subject to the constraints, x + y ≤ 500 , 000 (2) x ≥ 100 , 000 (3) 200 , 000 ≤ y ≤ 270 , 000 (4) 2 x + 5 y ≤ 1 , 750 , 000 (5) x ≥ 0 y ≥ 0 Consider the equation (2). Since the given inequality in the equation is ≤ , the solid line is used to join the coordinates of this equation. Now, replace the symbol ≤ by the symbol = in the equation (2). Then the equation becomes, x + y = 500 , 000 . Obtain any 2 coordinates of the above equation to plot on the line. Substitute x = 0 in the linear equation x + y = 500 , 000 . 0 + y = 500 , 000 y = 500 , 000 Thus, the value of y is 5000000 at x = 0 . Hence, the first coordinate is ( 0 , 500000 ) . Substitute x = 500000 in the linear equation x + y = 500 , 000 . x + 500000 = 500 , 000 x = 500 , 000 − 500 , 000 x = 0 Hence, the second coordinate is ( 500000 , 0 ) . Therefore, the coordinates are ( 0 , 500000 ) and ( 500000 , 0 ) . Use the test point to determine the half plane which is to be shaded. Let the test point be ( 0 , 0 ) . Now substitute the test point ( 0 , 0 ) in x + y ≤ 500 , 000 . 0 + 0 ≤ 500 , 000 0 ≤ 500 , 000 It is noticed that the inequality x + y ≤ 500 , 000 is true. Thus, shade the half-plane that contain the test point ( 0 , 0 ) . That is, the half plane below the line. Consider the equation (3). The given inequality in the equation is ≤ . Therefore, the solid line is used to join the coordinates of this equation. Then the equation (3) becomes, x = 100 , 000 . The equation x ≥ 100 , 000 represents the set of all points whose x coordinates are greater than 100000. That is, all points lies on the right side of the solid vertical line x = 100000 . Determine whether the inequality is true or false using the test point. Let the test point be ( 0 , 0 ) . Substitute the test point ( 0 , 0 ) in x ≥ 100 , 000 , 0 ≥ 100 , 000 The given inequality is false for the test point ( 0 , 0 ) , thus shade the half-plane that does not contain the test point ( 0 , 0 ) . That is, the half-plane right to the line. Consider the equation (4). The given inequality in the equation is ≤ . Therefore, the solid line is used to join the coordinates of this equation. Then the equation (4) becomes, 200 , 000 = y = 270 , 000 . Thus, draw the vertical lines at ( 0 , 200000 ) and ( 0 , 270000 ) Determine whether the inequality is true or false using test point. Let the test point be ( 0 , 0 ) . Substitute the test point ( 0 , 0 ) in 200 , 000 ≤ y ≤ 270 , 000 , 200 , 000 ≤ 0 ≤ 270 , 000 The given inequality 200 , 000 ≤ y is false for the test point ( 0 , 0 ) , so shade the half-plane that does not contain the test point ( 0 , 0 ) . That is, the half-plane left to the line. Also, the given inequality y ≤ 270000 is true for the test point ( 0 , 0 ) , so shade the half-plane that contain the test point ( 0 , 0 ) . That is, the half-plane right to the line. Consider the equation (5)

Finite Mathematics and Calculus with Applications

1st Edition

ISBN: 9781323188361

Author: Margaret Lial

Publisher: Pearson Education

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Chapter 3.3, Problem 10E

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The number of hectares that is to be devoted to each crops in order to maximize the profit and obtain the maximum profit.

Solution Summary: The author explains that the maximum profit of 192,500,000 is obtained by growing 200,000 hectares of coffee, and 270,000 for cocoa. The number of workers required for both the crop is 1,750,000.

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To find: The number of hectares that is to be devoted to each crops in order to maximize the profit and obtain the maximum profit.

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