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An investor has $826,000 to invest in bonds. Bond A yields an average of 5% and the bond B yields 6%. The investor requires that at least 3 times as much money be invested in bond A as in bond B. You must invest in these bonds to maximize his return. This can be set up as a linear programming problem. Introduce the decision variables: x = dollars invested in bond A y = dollars invested in bond B Compute x + y. 2$ Round to the nearest cent.

An investor has $826,000 to invest in bonds. Bond A yields an average of 5% and the bond B yields 6%. The investor requires that at least 3 times as much money be invested in bond A as in bond B. You must invest in these bonds to maximize his return. This can be set up as a linear programming problem. Introduce the decision variables: x = dollars invested in bond A y = dollars invested in bond B Compute x + y. 2$ Round to the nearest cent.

College Algebra (MindTap Course List)
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter6: Linear Systems
Section6.8: Linear Programming
Problem 32E
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An investor has $826,000 to invest in bonds. Bond A yields an average of 5% and the bond B yields 6%. The investor requires that at least 3 times as much money be invested in bond A as in bond B. You must invest in these bonds to maximize his return. This can be set up as a linear programming problem. Introduce the decision variables: x = dollars invested in bond A y = dollars invested in bond B Compute x + y. 2$ Round to the nearest cent.
Transcribed Image Text:An investor has $826,000 to invest in bonds. Bond A yields an average of 5% and the bond B yields 6%. The investor requires that at least 3 times as much money be invested in bond A as in bond B. You must invest in these bonds to maximize his return. This can be set up as a linear programming problem. Introduce the decision variables: x = dollars invested in bond A y = dollars invested in bond B Compute x + y. 2$ Round to the nearest cent.
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