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© Scarborough MATH 141 Week In Review 1 KEY Spring 2013 1 1. The demand for a VCR is 8000 units when the price is $230. If the price drops down to $210 then 9000 are sold. The manufacturer will not market VCRs if the price drops to $100. For each $60 increase in price, the manufacturer will supply an additional 1300 VCRs. How many VCR's are made at the equilibrium point? Demand (x, p) (8000, 230) Supply (x, p) (0, 100) (9000, 210) (0+1300, 100+60) = (1300, 160) or m 210 230 1 9000 8000 50 1 x 8000 p 230 50 1 p 230 x 160 50 1 p x 390 price in $ for x VCRs demanded 50 m 160 100 3 1300 0 65 3 x 0 p 100 65 3 p 100 x 65 3 p x 100 price in $ for x VCRs supplied 65 m Demand = Supply at the equilibrium point 1 3 x 390 x 100 50 65 43 x 290 650 188500 x 4384 43 Therefore 4384 VCRs are made at the equilibrium point. 2. A tractor is originally purchased for $48,000. After 8 years, the tractor is worth $19360. a. Find a linear equation for the value, V, of the tractor as a function of time, t, in years. ( ): ( ), ( ) ( ( ) ) dollars in value of the tractor t years after purchase b. What is the rate of depreciation of the tractor? Since m = –3580, the depreciation rate is $3580 per year. c. What is the tractor worth after 10 years? V 10 35800 48000 $12, 200 60 3 1300 65 © Scarborough MATH 141 Week In Review 1 KEY Spring 2013 2 3. It cost a company $22,500 to make 50 gadgets and $26,000 to make 120 gadgets. This company sells the gadgets for $80 each. What is the profit function? Compute the profit or loss when 2500 gadgets are produced and sold. C ( x) cx F is the linear cost function. R( x) sx is the linear revenue function. P( x) R( x) C( x) is the linear profit function. Cost x, C (50, 22500) (120, 26000) 26000 22500 50 120 50 C 22500 50( x 50) or C 26000 50 x 120 m C x 50 x 20000 dollars in cost for x gadgets Revenue R( x) 80 x dollars in revenue for x gadgets Profit P( x) R( x) C ( x) 80 x 50 x 20000 80 x 50 x 20000 Therefore P( x) 30 x 20000 dollars in profit/loss for x gadgets P 2500 30 2500 20000 55000 Therefore there is a profit of $55,000 when 2500 gadgets are produced and sold. © Scarborough MATH 141 Week In Review 1 KEY Spring 2013 3 4. A GPS manufacturer has a fixed monthly production cost of $59,985. If 360 GPS’s are produced and sold, there is a loss of $43,785. A GPS is sold for $210. What is the break-even quantity? Cost function: Revenue function: Profit function: C ( x) cx F R( x) sx P( x) R( x) C( x) C ( x) cx 59985 R( x) 210 x P( x) 210 x cx 59985 P(360) 210 360 c 360 59985 43785 360c 59400 c 165 Note: At the break-even point R( x) C ( x) and P( x) 45x 59985 0 . R( x ) C ( x ) OR 210 x 165x 59985 45x 59985 x 1333 Therefore the break-even quantity is 1333 GPS’s. P( x) 45x 59985 0 45x 59985 x 1333 © Scarborough MATH 141 Week In Review 1 KEY Spring 2013 4 5. Due to recycling by Aggies, trash has been decreasing in Maroon City. The table gives the amount of trash in thousands of pound over the time period of 2000 to 2007. Let x = 0 represent the year 2000 and y represent thousands of pounds of trash. Year x Trash, in thousands of pounds 2000 0 400 2002 2 360 2004 4 320 2005 5 310 2006 6 290 L1 L2 a. Find the equation of the least-squares (best fit, regression) line, in slope-intercept form, for this data. Give your coefficients to five decimal places. State any special features or programs you use on your calculator. LinReg L1, L2, Y1 y = –18.18966 x + 397.84483 thousands of pounds of trash for x years after 2000 b. Use the unrounded-coefficients of the linear regression line to estimate the pounds of trash, to the nearest whole number, that will be produced in the year 2009. Y1(9) = 234.137931 or Y1(2009 – 2000) = 234.137931 and (234.137931)(1000) = 234137.931 OR 1000 * Y1(9) = 234137.931 or 1000 * Y1(2009 – 2000) = 234137.931 234137.931 234138 pounds of trash c. What is the value, to 5 decimal places, and interpretation of the correlation coefficient? r –0.99715, which indicates a strong negative relation between the data points and the regression line. NOTE: Clear/Reset your calculator: MEM (2nd +), Reset, ALL, Reset To turn on the correlation coefficient: Catalog (2nd 0), DiagnosticOn, Enter, Enter © Scarborough MATH 141 Week In Review 1 KEY Spring 2013 5 6. The table shows the thousands of bacteria in a culture based upon the temperature. x, degrees C y, thousands of bacteria 35 80 40 65 45 52 50 40 55 20 a. What is the best-fitting (linear regression) line that represents the data? Do not round your coefficients. LinReg L1, L2, Y1 y = –2.9x + 181.9 thousands of bacteria for x degrees C b. Using the line of best fit, how many bacteria, to the nearest whole number, would you predict if the temperature was 38 degrees C? Y1(38) = 71.7 this is in thousands Predict 71,700 bacteria at 38 degrees C c. Using the line of best fit, at what temperature, to the nearest degree, would you expect to have 30,000 bacteria? Y2 = 30 Intersect (52.37931…., 30) OR 30 = –2.9 x + 181.9 1519 x= 52.37931034 29 At 52 degrees C, would predict to have 30,000 bacteria. d. What is the value, to 3 decimal places, and interpretation of the correlation coefficient? r –0.996, which indicates a strong negative relation between the data points and the regression line. © Scarborough MATH 141 Week In Review 1 KEY Spring 2013 6 7. If a water recirculation system is priced at $690, the quantity demanded is 310. For each $210 drop in price, an additional 310 are demanded. Suppliers of the water recirculation system will supply 930 systems if the price is $540 and will supply 1550 if the price is $660. a. Find the linear demand equation. (x, p): (310, 690) 210 21 21 or use the point (310 + 310, 690 – 210) = (620, 480) p 690 m x 310 310 31 31 21 Therefore p x 900 price in dollars for x systems demanded 31 METHOD II (Calculator): Put points (310, 690) and (620, 480) in L1 and L2. Then do LinReg L1, L2, Y1 b. Find the linear supply equation. (x, p): (930, 540), (1550, 660) 660 540 120 6 m 1550 930 620 31 6 p 540 x 930 31 6 Therefore p x 360 price in dollars for x systems supplied 31 METHOD II (Calculator): Put points (930, 540) and (1550,660) in L3 and L4. Then do LinReg L3, L4, Y2 c. Above what price will there be no demand? 21 p x x 900 31 21 p 0 0 900 900 31 When priced at $900 or more, there will be no demand for the system. METHOD II (Calculator): Y1(0) = 900 © Scarborough MATH 141 Week In Review 1 KEY Spring 2013 7 d. What quantity would be demanded if the system was free? 21 p x x 900 31 21 0 x 900 31 21 x 900 31 9300 x 1328.57 7 Therefore 1329 systems would be demanded if the system was free. METHOD II (Calculator): Find the x-value of the zero/root [calc, zero] of Y1; a good viewing window is [–200, 1600] by [–200, 1000] e. Above what price will the system be marketed? 6 p x x 360 31 6 p 0 0 360 360 31 Therefore, when priced at $360 or more, the system will be marketed. METHOD II (Calculator): Y2(0) = 360 f. If the system price is $840, how many systems will be marketed? 6 p x x 360 31 6 840 x 360 31 6 x 480 31 x 2480 Therefore, when the system price is $840, 2480 systems will be marketed. METHOD II (Calculator): Let Y3 = 840. Find the x-value of the intersection point of Y2 and Y3 [calc, intersection]; a good viewing window is [–200, 3000] by [–200, 1000] © Scarborough MATH 141 Week In Review 1 KEY Spring 2013 8 g. Find and interpret the equilibrium point. Solve the system of equations. 21 p x 900 31 6 p x 360 multiply equation by –1 31 21 x 900 31 6 p x 360 add the two equations 31 ______________________ p 0 27 x 540 31 27 x 540 31 6 21 620 360 480 or p 620 620 900 480 31 31 Therefore, the equilibrium point is (620, 480). x 620 p 620 When 620 water recirculation systems are produced and sold at a price of $480, both consumers and producers are satisfied. METHOD II (Calculator): Find the intersection point of Y1 and Y2 [calc, intersection]; a good viewing window is [–200, 1600] by [–200, 1000]