A coffee merchant sells three blends of coffee. a bag of house blend contains 30

Question

A coffee merchant sells three blends of coffee. a bag of house blend contains 300 grams of Colombian beans, 50 grams of Kenyan beans, and 150 grams of French roast beans and a bag of gourmet blend contains 100 grams Colombian beans, 350 of Kenyan beans, and 50 grams of French roast beans, and a bag of special blend contains 200 grams of Colombian beans, 200 grams of Kenyan beans, and 100 grams of French rost beans. The merchant has on hand 30 kilograms of Colombian beans, 15 kilograms of Kenyan beans, and 15 kilograms of French roast beans. Suppose one bag of the house blend produces a profit of $0.50, one bag of the special .blend produces a profit of $1.50, and one bag of the gourmet blend produces a profit of $2.00. How many bags of each type should the merchant prepare if he wants to use up all of the beans and maximize his profit? what is the maximum profit?

Explanation / Answer

Let's say he'll make x bags of house, y bags of gourmet, and z bagsof special. Because he wants to use all his beans, we can write the followingequations: Colombian: 30,000=300x+100y+200z Kenyan:      15,000=50x+350y+200z French:       15,000=150x+50y+100z The first and third equations say exactly the same thing, so weonly need to pay attention to one of them Let's use the Colombian and Kenyan equations. After reducingthem: 3x+y+2z=300 x+7y+4z=300 Lets write this system of equations as: 3x+y=300-2z x+7y=300-4z Solve for y: make it so that the x's cancel when we subtractthe 2 equations 3x+y=300-2z 3x+21y=900-12z 20y=600-10z, y=(60-z)/2, since y must be nonnegative, z

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