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Page 1 Math 141, Summer 2014-copyright Joe Kahlig 1. (a) vertex: (50, 70) 9. (a) not possible (b) 70 (b) not possible (c) none (c) not possible 3x2 2. (a) R(x) = x ∗ p = 150x − −150 vertex at x = = 25 2 ∗ −3 plug into the price-demand function to get the price. Answer: $75 (b) break even when R(x) = C(x) or P (x) = 0. −3x2 + 150x = 42x + 780 −3x2 + 150x − 780 = 0 either factor or use quadratic formula Answer: 10 pairs and 26 pairs 3. (a) points (7, 1100) and (11, 900) Answer: y − 1100 = −50(x − 7) or y = −50x + 1450 (b) find y when x=0. Answer: 1450 4. (a) Profit = Rev - Cost P = A∗x−(5x+280) where A is the selling price of the sandwich. 320 = A ∗ 150 − (5 ∗ 150 + 280) A=9 Answer: $9 (b) solve 9x = 5x + 280 Answer: 70 sandwiches 5. use rref. Answer: x=2, y=5, and z=0 6. use rref. DVD Players: 40 " b 1 " 16 0 (d) (e) 2 0 5 11 # 10 3 # 10. Combine the matrices on the left side and you get this: " 2x-28 21-4w 2y-4z z-8 # " = 5 4 7 6 # since the matrices are equal, the corresponding enteries are equal. i.e. 2x − 28 = 5 2y − 4z = 7 21 − 4w = 4 z−8=8 now solve for the variables. Answer: x = 16.5, y = 31.5, z = 14, and w = 4.25 11. x = the number to type I cakes made. y = the number to type II cakes made. z = the number to type III cakes made. Objective function: P = 2x + 4y + 7z Constraints: 2x + 4y + 2z ≤ 280 2x + y + 3z ≤ 230 x ≥ 3(y + z) x, y, z ≥ 0 12. y + 700 = x + 500 z + 500 = 400 + y 300 + x = z + 600 price: 130 7. (a) x = 11, y = 44, and z = 7 13. use rref to get this matrix. (b) no solution " (c) x = 2 − 5y + w z = 3 − 7w y, w =any number 8. points are in the form (x, p) 1 0 0 1 −1 2 −50 200 # From this we know the parametric solution is x = z − 50 y = 200 − 2z z =any number. restrictions on Z: Z = 50, 51, 52, ......, 100 (200, 40) and (275, 60) Check the back of the page for more problems.