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Page 1 Math 141, Summer 2014-copyright Joe Kahlig 1. (a) vertex: (26, −100) " 9. (a) (b) none 2x+4y 2+7y # (b) not possible (c) −100 2. (a) R(x) = x ∗ p = 88x − 2x2 −88 vertex at x = = 22 2 ∗ −2 plug into the price-demand function to get the price. Answer: $44 (b) break even when R(x) = C(x) or P (x) = 0. −2x2 + 88x = 24x + 350 −2x2 + 64x − 350 = 0 either factor or use quadratic formula Answer: 7 pairs and 25 pairs 3. (a) points (4, 2300) and (6, 2000) Answer: y − 2300 = −150(x − 4) or y = −150x + 2900 (b) find y when x=0. Answer: 2900 4. (a) Profit = Rev - Cost P = A∗x−(4x+520) where A is the selling price of the sandwich. 680 = A ∗ 300 − (4 ∗ 300 + 520) A=8 Answer: $7 (b) solve 8x = 4x + 520 Answer: 130 sandwiches 5. use rref. Answer: x=0.5, y=0, and z = −2.5 6. use rref. DVD Players: 60 price: 150 7. (a) x = 20, y = 22, and z = 10 m (c) k 3 1 4 3 (d) not possible " (e) 4 6 12 1 # 10. Combine the matrices on the left side and you get this: " 2x - 28 21-4w 3y-4z z-8 # " = 10 21 2 4 # since the matrices are equal, the corresponding enteries are equal. i.e. 2x − 28 = 10 3y − 4z = 2 21 − 4w = 21 z−8=4 now solve for the variables. Answer: x = 19, y = 50/3, z = 12, and w = 0 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 = 5x + 3y + 2z Constraints: 2x + 4y + 2z ≤ 120 2x + y + 3z ≤ 52 y ≥ 2(x + z) x, y, z ≥ 0 12. w + 700 = v + 300 m + 800 = 880 + w 450 + v = m + 950 13. use rref to get this matrix. (b) no solution (c) x = 5 − 3y + 2w z = 8 − 4w 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 (300, 25) and (350, 40) Check the back of the page for more problems.