# SOLUTION: 1)Two varieties of animal feed contain essential nutrients A and B. Feed I contains 2 units of A and 3 units of B per pound. Feed II contains 2 units of A and 5 units of B per po

Algebra ->  -> SOLUTION: 1)Two varieties of animal feed contain essential nutrients A and B. Feed I contains 2 units of A and 3 units of B per pound. Feed II contains 2 units of A and 5 units of B per po      Log On

 Ad: Mathway solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Click here to see ALL problems on Linear-systems Question 209247: 1)Two varieties of animal feed contain essential nutrients A and B. Feed I contains 2 units of A and 3 units of B per pound. Feed II contains 2 units of A and 5 units of B per pound. A farmer needs a feed mix that will give his animals a minimum of 16 units of A and 30 units of B. If Feed I costs \$3 per pound and Feed II costs \$4 per pound, how much of each should be bought to supply the proper nutrition while minimizing cost? Write the objective function and system of linear inequalities and solve it graphically using the method of corners. Answer by stanbon(60774)   (Show Source): You can put this solution on YOUR website!Two varieties of animal feed contain essential nutrients A and B. Feed I contains 2 units of A and 3 units of B per pound. Feed II contains 2 units of A and 5 units of B per pound. --------------------------- A farmer needs a feed mix that will give his animals a minimum of 16 units of A and 30 units of B. "A" Eq.: 2(I)+2(II) >= 16 "B" Eq.: 3(I)+5(II) >= 30 If Feed I costs \$3 per pound and Feed II costs \$4 per pound, how much of each should be bought to supply the proper nutrition while minimizing cost? Cost Eq: C(x) = 3(I) + 4(II) ---------------------------------- Write the objective function and system of linear inequalities and solve it graphically using the method of corners. I >=0 II >=0 "A"Eq: I >= -II+8 "B"Eq: I >= (-5/3)(II) + 10 --------------------------- ------------ (II,I) Corners: (0,8), (6,0), (3,5) --- Objective: C(x) = 3(I) + 4(II) (0,8) C(x) = 3*8+4*0 = 24 (6,0) C(x) = 3*0+4*6 = 24 (3,5) C(x) = 3*5+4*3 = 27 =================================== Minimal Cost : 8 of I or 6 of II ============================================= Cheers, Stan H. Reply to stanbon@comcast.net