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104/184 Quiz 1 Thursday September 24 First Name: Last Name: Student-No: Section: Grade: Very short answer questions 1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes. Marking scheme: 1 for each correct, 0 otherwise (a) Solve 2x = 1 5 for x. Answer: log2 log 5 5 − log = − ln . 2 ln 2 1 5 = − log2 5 = Solution: We have to allow several forms of the solution. 3 q − q2 (b) Compute lim . q→2 q+1 Answer: 4 3 Solution: Direct substitution. Short answer questions — you must show your work 2. 4 marks Each part is worth 2 marks. (a) Given that ln a = 3, ln b = 4, and ln c = −2, find the value of ln √abc . Answer: 2 Solution: √ a 1 ln √ = ln(a) − ln( bc) = ln a − ln(bc) 2 bc 1 = ln a − (ln b + ln c) 2 1 = 3 − (4 − 2) = 2 . 2 Marking scheme: If correct answer and some work, then 2 marks. If some work in the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If no solution, then 0. x2 + 3x − 4 x→1 4x2 − 7x + 3 (b) Compute the limit lim Answer: 5 Solution: Direct substitution yields 0/0, so we simplify first: x2 + 3x − 4 (x − 1)(x + 4) x+4 = = 2 4x − 7x + 3 (x − 1)(4x − 3) 4x − 3 x+4 = 5. x→1 4x − 3 Marking scheme: 1 for any correct factoring, 1 for answer. Hence the limit is lim Long answer question — you must show your work 3. 4 marks Pea Sea Inc. sells 200 laptops a month at the price of $640 each. For each $5 increase in price, the sales would drop by 2 laptops. Their factory costs $250,000 per month to operate and each laptop costs an additional $225 to make. Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for break even values or any optimal production values. Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get those 2 points. (a) Find the linear demand equation for the laptops. Use the notation p for the unit price and q for the monthly demand. Answer: q = − 25 p + 456, or p = 1140 − 52 q Solution: A data point is (p, q) = (price, demand). So two points on the linear ∆q demand curve are (640, 200) and (645, 198). Its slope is m = ∆p = 198−200 = − 52 . 645−640 Thus, q = − 25 p + b, b being a constant. Substitute in (640, 200) to get b = 456. Therefore, q = − 25 p + 456. or equivalently: p = 1140 − 25 q. (b) Find the cost function, C = C(q), for producing q laptops per month. Answer: C(q) = 250000 + 225q Solution: No work is needed (c) Find the monthly revenue function, R = R(q). Answer: R(q) = p · q = 1140q − 52 q 2 Solution: Either expression is acceptable. (d) Find the monthly profit (or net income) function P (q). Answer: P (q) = − 52 q 2 + 915q − 250000 Solution: P (q) = R(q)−C(q) = 1140q − 52 q 2 −(250000+225q) = 915q − 52 q 2 −250000. 104/184 Quiz 1 Thursday September 24 First Name: Last Name: Student-No: Section: Grade: Very short answer questions 1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes. Marking scheme: 1 for each correct, 0 otherwise (a) Solve 3y+1 = 2 for y. Answer: y = log3 2 − 1 Solution: 3y+1 = 2 ⇒ y + 1 = log3 2 ⇒ y = log3 2 − 1 (b) Compute lim p→(−3) p2 − p − 12 . p−3 Answer: 0 Solution: lim p→(−3) p2 − p − 12 p−3 lim (p2 − p − 12) = p→(−3) lim (p − 3) = 0 = 0. −6 p→(−3) Short answer questions — you must show your work 2. 4 marks Each part is worth 2 marks. (a) Given that log2 x = 5, log2 y = −3, find the value of log2 (4x3 y −2 ). Answer: 23 Solution: log2 4(x3 y −2 ) = log2 4 + log2 x3 − log2 y 2 = 2 + 3 log2 x − 2 log2 y = 2 + 3(5) − 3(−2) = 23 . Marking scheme: If correct answer and some work, then 2 marks. If some work in the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If no solution, then 0. 3x2 − x − 10 x→2 x2 − 4 (b) Compute the limit lim Answer: 11/4 Solution: Direct substitution yields 0/0, so we simplify first: 3x2 − x − 10 (x − 2)(3x + 5) 3x + 5 = = 2 x −4 (x − 2)(x + 2) x+2 3x + 5 = 11/4 . x→2 x + 2 Marking scheme: 1 for any correct factoring, 1 for answer. Hence the limit is lim Long answer question — you must show your work 3. 4 marks A carver sells 80 carvings a month at the price of $160 each. For each $12 decrease in price, he can sell 4 more carvings. The workshop costs $20,000 per month to operate and each carving costs $40 to make. Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for break even values or any optimal production values. Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get those 2 points. (a) Find the linear demand equation for the carvings. Use the notation p for the unit price and q for the monthly demand. Answer: p=-3q+400 Solution: A data point is (q, p) = (quantity, price). So two points are (80, 160) and (84, 148). Since the rate of change is constant, the demand curve must be a line with 12 = 160−148 = −4 = −3. slope m = ∆p ∆q 80−84 Thus, p = −3q + K when K is some constant. Substitute in (80, 160) to get K = 400. Therefore, p = −3q + 400. (b) Find the cost function, C = C(q), for producing q carvings per month. Answer: C(q) = 20, 000 + 40q Solution: No work is needed (c) Find the monthly revenue function, R = R(q). Answer: R(q) = q · (−3q + 400) Solution: R = p · q = q · (−3q + 400) = −3q 2 + 400q (d) Find the monthly profit (or Net Income) function P (q). Answer: P (q) = −3q 2 + 360q − 20, 000 Solution: P (q) = R(q)−C(q) = −3q 2 +400q −(20, 000+40q) = −3q 2 +360q −20, 000 104/184 Quiz 1 Thursday September 24 First Name: Last Name: Student-No: Section: Grade: Very short answer questions 1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes. Marking scheme: 1 for each correct, 0 otherwise (a) Solve 5x = 1 3 for x. Answer: log5 Solution: 5x = (b) Compute lim q→2 1 3 ⇒ log5 5x = log5 1 3 ⇒ x = log5 1 3 1 3 3 = − log log 5 = log5 (3−1 ) = − log5 3 q3 − 6 . q+4 Answer: Solution: limq→2 q 3 −6 q+4 1 3 lim(q 3 − 6) = q→2 lim(q + 4) = 2 6 = 13 . q→2 Short answer questions — you must show your work 2. 4 marks Each part is worth 2 marks. √ (a) Given that log3 x = 7, log3 y = 5, and log3 z = 4, find the value of log3 xy . z Answer: 2 Solution: √ xy (xy)1/2 1 log3 = log3 = log3 (xy) − log3 z z z 2 1 = [log3 x + log3 y] − log3 z 2 1 = [7 + 5] − 4 = 2 . 2 Marking scheme: If correct answer and some work, then 2 marks. If some work in the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If no solution, then 0. x2 − 2x − 3 x→3 2x2 − 3x − 9 (b) Compute the limit lim Answer: 4/9 Solution: Direct substitution yields 0/0, so we simplify first: x2 − 2x − 3 (x − 3)(x + 1) x+1 = = 2 2x − 3x − 9 (x − 3)(2x + 3) 2x + 3 x+1 = 4/9 . x→3 2x + 3 Marking scheme: 1 for any correct factoring, 1 for answer. Hence the limit is lim Long answer question — you must show your work 3. 4 marks A manufacturer sells 100 tables per month at the price of $450 each. For each $10 decrease in price, he can sell 5 more tables per month. Their factory costs $150,000 per month to operate and each table costs an additional $325 to make. Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for break even values or any optimal production values. Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get those 2 points. (a) Find the linear demand equation for the tables. Use the notation p for the unit price and q for the monthly demand. Answer: p=-2q+650 Solution: A data point is (q, p) = (quantity, price). So two points are (100, 450) and (105, 440). Since the rate of change is constant, the demand curve must be a line with 10 = 450−440 = −5 = −2. Thus, p = −2q + K when K is some constant. slope m = ∆p ∆q 100−105 Substitute in (100, 450) to get K = 650. Therefore, p = −2q + 650. (b) Find the cost function, C = C(q), for producing q tables per month. Answer: C(q) = 150, 000 + 325q Solution: No work is needed (c) Find the monthly revenue function, R = R(q). Answer: R(q) = q · (−2q + 650) Solution: R = p · q = q · (−2q + 650) = −2q 2 + 650q (d) Find the monthly profit (or Net Income) function P (q). Answer: P (q) = −2q 2 + 325q − 150, 000 Solution: N I(q) = R(q) − C(q) = −2q 2 + 650q − (150, 000 + 325q) = −2q 2 + 325q − 150000 104/184 Quiz 1 Thursday September 24 First Name: Last Name: Student-No: Section: Grade: Very short answer questions 1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes. Marking scheme: 1 for each correct, 0 otherwise (a) Solve 6x = 2 5 for x. Answer: log6 Solution: 6x = (b) Compute lim q→4 2 5 ⇒ x = log6 2 5 2 5 = log 2/5 log 6 q2 + 2 . q−5 Answer: −18 Solution: Direct substituiton. Short answer questions — you must show your work 2. 4 marks Each part is worth 2 marks. (a) Given that log5 x = 4, log5 y = 3, and log5 z = −2, find the value of log5 yz √ . 5 x Answer: -2 Solution: √ √ yz log5 √ = log5 (yz) − log5 (5 x) = log5 y + log5 z − (log5 5 + log5 x) 5 x = 3 + (−2) − (1 + 1/2 log5 x) = 1 − (1 + 1/2 × 4) = −2 . Marking scheme: If correct answer and some work, then 2 marks. If some work in the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If no solution, then 0. x2 + x − 12 x→3 2x2 − 3x − 9 (b) Compute the limit lim Answer: 7/9 Solution: Direct substitution yields 0/0, so we simplify first: x2 + x − 12 (x − 3)(x + 4) x+4 = = 2 2x − 3x − 9 (x − 3)(2x + 3) 2x + 3 x+4 = 7/9 . x→3 2x + 3 Marking scheme: 1 for any correct factoring, 1 for answer. Hence the limit is lim Long answer question — you must show your work 3. 4 marks A manufacturer sells 120 tables per month at the price of $400 each. For each $12 decrease in price, he can sell 6 more tables per month. Their factory costs $125,000 per month to operate and each table costs an additional $250 to make. Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for break even values or any optimal production values. Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get those 2 points. (a) Find the linear demand equation for the tables. Use the notation p for the unit price and q for the monthly demand. Answer: p=-2q+640 Solution: A data point is (q, p) = (quantity, price). So two points are (120, 400) and (126, 388). Since the rate of change is constant, the demand curve must be a line with 12 = −6 == −2. slope m = ∆p ∆q Thus, p = −2q +K when K is some constant. Substitute in (120, 400) to get K = 640. Therefore, p = −2q + 640. (b) Find the cost function, C = C(q), for producing q tables per month. Answer: C(q) = 125, 000 + 250q Solution: No work is needed (c) Find the monthly revenue function, R = R(q). Answer: R(q) = p · q = q · (−2q + 640) Solution: R = p · q = q · (−2q + 640) = −2q 2 + 640q (d) Find the monthly profit (or Net Income) function P (q). Answer: P (q) = −2q 2 + 390q − 125, 000 Solution: N I(q) = R(q) − C(q) = −2q 2 + 650q − (125, 000 + 250q) = −2q 2 + 400q − 125, 000 104/184 Quiz 1 Friday September 24 First Name: Last Name: Student-No: Section: Grade: Very short answer questions 1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes. Marking scheme: 1 for each correct, 0 otherwise (a) Solve 7x = 1 5 for x. Answer: log7 Solution: 7x = (b) Compute lim q→3 1 5 ⇒ x = log7 1 5 1 5 5 = − log log 7 = log7 (5−x ) = − log7 5 q3 − q + 1 . q+2 Answer: 5 Solution: lim q→3 q3 − q + 1 q+2 lim(q 3 − q + 1) = q→3 lim(q + 2) q→3 = 25 = 5. 5 Short answer questions — you must show your work 2. 4 marks Each part is worth 2 marks. √ (a) Given that log3 x = 3, log3 y = 5, and log3 z = 1, find the value of log3 xy . z Answer: 3 Solution: √ xy (xy)1/2 log3 = log3 = log3 (xy)1/2 − log3 z z z 1 = [log3 x + log3 y] − log3 z 2 1 = [3 + 5] − 1 = 3 . 2 Marking scheme: If correct answer and some work, then 2 marks. If some work in the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If no solution, then 0. 3x2 − 7x + 2 x→2 x2 − 5x + 6 (b) Compute the limit lim Answer: -5 Solution: Direct substitution yields 0/0, so we simplify first: 3x2 − 7x + 2 (x − 2)(3x − 1) 3x − 1 = = 2 x − 5x + 6 (x − 2)(x − 3) x−3 3x − 1 = −5 . x→2 x − 3 Marking scheme: 1 for any correct factoring, 1 for answer. Hence the limit is lim Long answer question — you must show your work 3. 4 marks A manufacturer sells 90 tables a month at the price of $450 each. For each $9 decrease in price, he can sell 3 more tables. Their factory costs $115,000 per month to operate and each table costs an additional $300 to make. Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for break even values or any optimal production values. Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get those 2 points. (a) Find the linear demand equation for the tables. Use the notation p for the unit price and q for the monthly demand. Answer: p=-3(q-90)+450=-3q+720 Solution: A data point is (q, p) = (quantity, price). So two points are (90, 450) and (93, 441). Since the rate of change is constant, the demand curve must be a line with 9 = 450−441 = −3 = −3. slope m = ∆p ∆q 90−93 Thus, p = −3q + K when K is some constant. Substitute in (90, 450) to get K = 720. Therefore, p = −3q + 720. (b) Find the cost function, C = C(q), for producing q tables per month. Answer: C(q) = 115, 000 + 300q Solution: No work is needed (c) Find the monthly revenue function, R = R(q). Answer: R(q) = q · (−3q + 720) Solution: R = p · q = q · (−3q + 720) = −3q 2 + 720q (d) Find the monthly profit (or Net Income) function P (q). Answer: P (q) = −3q 2 + 420q − 115, 000 Solution: P = R − C = −3q 2 + 720q − 300q − 115, 000 = −3q 2 + 420q − 115, 000. 104/184 Quiz #1 Friday September 24 First Name: Last Name: Student-No: Section: Grade: Very short answer questions 1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes. scheme: 1 for each correct, 0 otherwise Marking (a) Solve 2x = 25 for x. Answer: 2 log2 (5) Solution: 2x = 25 =⇒ x = log2 (2x ) = log2 (25) = log2 (52 ) = 2 log2 (5) (b) Compute lim r→−1 r2 −5r+1 r3 +4 . Answer: Solution: lim r→−1 r2 −5r+1 r3 +4 lim = r→−1 (r2 −5r+1) lim (r3 +4) r→−1 = 7 3 (−1)2 −5(−1)+1 (−1)3 +4 = 7 3 Short answer questions — you must show your work 2. 4 marks Each part is worth 2 marks. (a) Given that log5 x = 6, log5 y = 1, and log5 z = 3, find the value of log5 y √ 3 xz . Answer: −2 Solution: log5 y √ 3 xz = log5 y (xz)1/3 = log5 (y) − log5 (xz)1/3 1 = log5 (y) − (log5 (x) + log5 (z)) 3 1 = 1 − (6 + 3) = −2 3 Marking scheme: If correct answer and some work, then 2 marks. If some work in the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If no solution, then 0. 2 (b) Compute the limit lim 2xx2 −2x−3 . −2x−12 x→3 Answer: 2/5 Solution: Direct substitution yields 0/0, so we simplify first: x2 − 2x − 3 (x − 3)(x + 1) x+1 = = . 2 2x − 2x − 12 (x − 3)(2x + 4 2x + 4 2 x+1 Hence the limit is lim 2xx2 −2x−3 = lim 2x+4 = 4/10 = 2/5 −2x−12 x→3 x→3 Marking scheme: 1 for any correct factoring, 1 for answer. Long answer question — you must show your work 3. 4 marks A manufacturer sells 200 tables a month at the price of $350 each. For each $15 decrease in price, he can sell 10 more tables. There factory costs $50,000 per month to operate and each table costs an additional $125 to make. Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for break even values or any optimal production values. Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get those 2 points. (a) Find the linear demand equation for the tables. Use the notation p for the unit price and q for the monthly demand. Answer: p = − 32 q + 650 Solution: A data point is (q, p) = (quantity, price). So two points are (200, 350) and (210, 335). Since the rate of change is constant, the demand curve must be a line with 15 = 350−335 = −10 = − 23 . slope m = ∆p ∆q 200−210 Thus, p = − 32 q +K when K is some constant. Substitute in (200, 350) to get K = 650. Therefore, p = − 32 q + 650. (b) Find the cost function, C = C(q), for producing q tables per month. Answer: C(q) = 125q + 50, 000 Solution: No work is needed. (c) Find the monthly revenue function, R = R(q). Answer: R(q) = − 32 q 2 + 650q Solution: R = p · q = (− 23 q + 650) · q = − 23 q 2 + 650q. (d) Find the monthly profit (or Net Income) function P (q) (or N I(q)). Answer: P (q) = − 23 q 2 + 525q − 50, 000 Solution: P (q) = R(q) − C(q) = (− 32 q 2 + 650q) − (125q + 50, 000) = − 32 q 2 + 525q − 50, 000. 104/184 Quiz 1 Friday September 25 First Name: Last Name: Student-No: Section: Grade: Very short answer questions 1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes. Marking scheme: 1 for each correct, 0 otherwise (a) Solve 7x = 1 2 for x. Answer: log7 Solution: 7x = (b) Compute lim t→1 1 2 ⇒ log7 7x = log7 1 2 ⇒ x = log7 1 2 1 2 2 = − log log 7 = log7 (2−1 ) = − log7 2 3 − t2 . 1 + t3 Answer: 1 Solution: lim t→1 3 − t2 1 + t3 lim(3 − t2 ) = t→1 3 lim(1 + t ) t→1 = 2 = 1. 2 Short answer questions — you must show your work 2. 4 marks Each part is worth 2 marks. (a) Given that log2 x = 5, log2 y = −1, and log2 z = 3, find the value of log2 x . y2z Answer: 4 Solution: log2 x y2z = log2 x − log2 y 2 − log2 z = log2 x − 2 log2 y − log2 z = 5 − 2(−1) − 3 = 4 Marking scheme: If correct answer and some work, then 2 marks. If some work in the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If no solution, then 0. 2x2 − 7x − 15 x→5 x2 − 6x + 5 (b) Compute the limit lim Answer: 13/4 Solution: Direct substitution yields 0/0, so we simplify first: 2x2 − 7x − 15 (x − 5)(2x + 3) 2x + 3 = = 2 x − 6x + 5 (x − 1)(x − 5) x−1 2x + 3 = 13/4 . x→5 x − 1 Marking scheme: 1 for any correct factoring, 1 for answer. Hence the limit is lim Long answer question — you must show your work 3. 4 marks A manufacturer sells 100 chairs per month at the price of $520 each. For each $15 decrease in price, they can sell 5 more chairs per month. Their factory costs $125,000 per month to operate and each chair costs an additional $370 to make. Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for break even values or any optimal production values. Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get those 2 points. (a) Find the linear demand equation for the chairs. Use the notation p for the unit price and q for the monthly demand. Answer: p = −3q + 820 Solution: A data point is (q, p) = (quantity, price). Two points are (100, 520) and 15 = 520−505 = −5 = −3. (105, 505). The demand curve is a line with slope m = ∆p ∆q 100−105 Thus, p = −3q + K where K is some constant. Substitute (100, 520) to get K = 820. Therefore, p = −3q + 820. (b) Find the cost function, C = C(q), for producing q chairs per month. Answer: C(q) = 125, 000 + 370q Solution: No work is needed (c) Find the monthly revenue function, R = R(q). Answer: R(q) = q · (−3q + 820) Solution: R = p · q = q · (−3q + 820) = −3q 2 + 820q (d) Find the monthly profit (or Net Income) function P (q). Answer: P (q) = −3q 2 + 450q − 125, 000 Solution: P (q) = R(q) − C(q) = −3q 2 + 820q − (125, 000 + 370q) = −3q 2 + 450q − 125, 000 104/184 Quiz 1 Friday September 25 First Name: Last Name: Student-No: Section: Grade: Very short answer questions 1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes. Marking scheme: 1 for each correct, 0 otherwise (a) Solve 5x = 3 2 for x. Answer: log5 (3/2) = Solution: 5x = 3/2 ⇒ x = log5 (3/2) ⇒ x = (b) Compute lim z→3 log(3/2) log 5 log(3/2) . log 5 z−5 . z2 + 4 Answer: Solution: lim z→3 z−5 z2 + 4 lim (z − 5) = z→3 2 lim (z + 4) z→3 = −2 13 −2 −2 = . 13 13 Short answer questions — you must show your work 2. 4 marks Each part is worth 2 marks. (a) Given that log3 x = 7, log3 y = 5, and log3 z = 4, find the value of log3 xy √ . z Answer: 10 Solution: √ xy 1 log3 √ = log3 (xy) − log3 ( z) = log3 x + log3 y − log3 z 2 z 1 = 7 + 5 − ∗ 4 = 10 . 2 Marking scheme: If correct answer and some work, then 2 marks. If some work in the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If no solution, then 0. 2x2 − 9x − 5 x→5 x2 − 4x − 5 (b) Compute the limit lim Answer: 11/6 Solution: Direct substitution yields 0/0, so we simplify first: 2x2 − 9x − 5 (x − 5)(2x + 1) 2x + 1 = = 2 x − 4x − 5 (x − 5)(x + 1) x+1 2x + 1 = 11/6 . x→5 x + 1 Marking scheme: 1 for any correct factoring, 1 for answer. Hence the limit is lim Long answer question — you must show your work 3. 4 marks A manufacturer sells 100 tables a month at the price of $400 each. For each $10 decrease in price, they can sell 5 more tables. Their factory costs $200,000 per month to operate and each table costs an additional $275 to make. Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for break even values or any optimal production values. Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get those 2 points. (a) Find the linear demand equation for the tables. Use the notation p for the unit price and q for the monthly demand. Answer: p=-2q+600 Solution: A data point is (q, p) = (quantity, price). So two points are (100, 400) and (105, 390). Since the rate of change is constant, the demand curve must be a line with 10 = 400−390 = −5 = −2. slope m = ∆p ∆q 100−105 Thus, p = −2q + K where K is some constant. Substitute in (100, 400) to get 400 = −2(100) + K, implying K = 600. Therefore, p = −2q + 600. (b) Find the cost function, C = C(q), for producing q tables per month. Answer: C(q) = 200, 000 + 275q Solution: No work is needed (c) Find the monthly revenue function, R = R(q). Answer: R(q) = −2q 2 + 600q Solution: R = p · q = q · (−2q + 600) = −2q 2 + 600q (d) Find the monthly profit (or Net Income) function P (q). Answer: P (q) = −2q 2 + 325q − 200, 000 104/184 Quiz 1-AD Friday September 25 First Name: Last Name: Student-No: Section: Grade: Very short answer questions 1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes. Marking scheme: 1 for each correct, 0 otherwise (a) Solve 3x = 5 2 for x. Answer: log3 Solution: 3x = (b) Compute lim t→2 5 2 5 2 ⇒ log3 3x = log3 ⇒ x = log3 5 2 5 2 3 − t2 . 1 + t3 Answer: −1/9 Solution: lim t→2 3 − t2 1 + t3 lim(3 − t2 ) = t→2 3 lim(1 + t ) t→2 = −1 = 1. 9 Short answer questions — you must show your work 2. 4 marks Each part is worth 2 marks. (a) Given that log2 x = 5, log2 y = −3, and log2 z = 3, find the value of log2 x . y2z Answer: 8 Solution: log2 x y2z = log2 x − log2 y 2 − log2 z = log2 x − 2 log2 y − log2 z = 5 − 2(−3) − 3 = 8 Marking scheme: If correct answer and some work, then 2 marks. If some work in the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If no solution, then 0. 2x2 − 13x + 15 x→5 x2 − 6x + 5 (b) Compute the limit lim Answer: 7/4 Solution: Direct substitution yields 0/0, so we simplify first: 2x2 − 13x + 15 (x − 5)(2x − 3) 2x − 3 = = 2 x − 6x + 5 (x − 1)(x − 5) x−1 2x − 3 = 7/4 . x→5 x − 1 Marking scheme: 1 for any correct factoring, 1 for answer. Hence the limit is lim Long answer question — you must show your work 3. 4 marks A manufacturer sells 100 chairs per month at the price of $520 each. For each $15 decrease in price, they can sell 5 more tables per month. Their factory costs $125,000 per month to operate and each chair costs an additional $370 to make. Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for break even values or any optimal production values. Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get those 2 points. (a) Find the linear demand equation for the chairs. Use the notation p for the unit price and q for the monthly demand. Answer: p = −3q + 820 Solution: A data point is (q, p) = (quantity, price). Two points are (100, 520) and 15 = 520−505 = −5 = −3. (105, 505). The demand curve is a line with slope m = ∆p ∆q 100−105 Thus, p = −3q + K where K is some constant. Substitute (100, 520) to get K = 820. Therefore, p = −3q + 820. (b) Find the cost function, C = C(q), for producing q chairs per month. Answer: C(q) = 125, 000 + 370q Solution: No work is needed (c) Find the monthly revenue function, R = R(q). Answer: R(q) = q · (−3q + 820) Solution: R = p · q = q · (−3q + 820) = −3q 2 + 820q (d) Find the monthly profit (or Net Income) function P (q). Answer: P (q) = −3q 2 + 450q − 125, 000 Solution: P (q) = R(q) − C(q) = −3q 2 + 820q − (125, 000 + 370q) = −3q 2 + 450q − 125, 000