PRECALCULUS:GRAPHICAL,...-NASTA ED.
10th Edition
ISBN: 9780134672090
Author: Demana
Publisher: PEARSON
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Expert Solution & Answer
Chapter 7.4, Problem 39E
Solution
To calculate: The minimize cost to place the order by customer.
40 operations at Refinery 1, 60 at Refinery 2, $48000, or 120 operations at Refinery 1, 20 at Refinery 2, $48000.
Given Information:
Two oil refineries produce three grades of gasoline: A, B, and C. At each refinery, the three grades of gasoline are produced in a single operation in the following proportions: Refinery 1 produces 1 unit of A, 2 units of B, and 1 unit of C; Refinery 2 produces 1 unit of A, 4 units of B, and 4 units of C. For the production of one operation, Refinery 1 charges $300 and Refinery 2 charges $600. A customer needs 100 units of A, 320 units of B. and 200 units of C.
Calculation:
Consider the given information,
Suppose that x the number of operations produced at Refinery 1 and y the number of operations produced at Refinery 2.
Write the constraints as a system of inequalities using the given information.
Write the objective function by using the given information.
Use the graphing calculator to draw the inequality and find the corner points.
The corner points are
Now, find the value of the objective function on corner points.
Substitute 0 for x and 100 for y in the objective function.
Substitute 40 for x and 60 for y in the objective function.
Substitute 200 for x and 0 for y in the objective function.
Substitute 120 for x and 20 for y in the objective function.
The cost is minimized by producing 40 operations at Refinery 1 and 60 at Refinery 2, and by producing 120 operations at Refinery 1 and 20 at Refinery 2 and the both cost is $48000.
As the feasible region is unbounded and has
Therefore, 40 operations at Refinery 1, 60 at Refinery 2, $48000, or 120 operations at Refinery 1, 20 at Refinery 2, $48000, If conservation of operations is favoured, the first option would be the best choice..
/EXPLANATION>-->
Chapter 7 Solutions
PRECALCULUS:GRAPHICAL,...-NASTA ED.
Ch. 7.1 - Prob. 1QRCh. 7.1 - Prob. 2QRCh. 7.1 - Prob. 3QRCh. 7.1 - Prob. 4QRCh. 7.1 - Prob. 5QRCh. 7.1 - Prob. 6QRCh. 7.1 - Prob. 7QRCh. 7.1 - Prob. 8QRCh. 7.1 - Prob. 9QRCh. 7.1 - Prob. 10QR
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