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Mathematics 1090 2005 PRACTICE EXAM I-Solutions Spring 1. Find the slope-intercept equation for the following lines: (a) The line of equation 2y + 3x + 2 = 0 Solution: We solve for y in the general form of the equation 3 2y + 3x + 2 = 0 ⇐⇒ 2y = −3x − 2 ⇐⇒ y = − x − 1 2 This is then the slope intercept form of the equation. (b) The line passing through the points (1, 2) and (−1, 3) Solution: We find the slope: m= 1 3−2 =− −1 − 1 2 1 Point slope form y − 2 = − (x − 1) and we arrange it to find the slope-intercept from: 2 1 1 1 1 5 y − 2 = − (x − 1) ⇐⇒ y = − x + + 2 ⇐⇒ y = − x + 2 2 2 2 2 (c) The line of slope 2 passing through the point (1, 1) Solution: Point slope form: y − 1 = 2(x − 1) and we arrange it to find the slopeintercept from: y − 1 = 2(x − 1) ⇐⇒ y = 2x − 2 + 1 ⇐⇒ y = 2x − 1 (d) The line parallel to y = 2x + 3 and passing through (0, 1) Solution: As the line must be parallel to the line of equation y = 2x + 3 it must have the same slope m = 2. Moreover the line must pass through (0, 1) which on the y-axis, then the y-intercept is 1, so the slope-intercept equation is: y = 2x + 1 (e) The line perpendicular to y = x + 2 and passing through (1, 0) Solution: As the line must be perpendicular to the line of equation y = x + 2 its slope is m = − 11 = −1. Moreover the line must pass through (1, 0) so the point-slope equation is y − 0 = −(x − 1). Then the slope intercept form is y = −x + 1 2. Suppose that the cost of a business property is $720000. The owning company wants to use a straight-line depreciation schedule for a period of 120 months. If y is the value of the property after x months, write the linear equation of the depreciation schedule. Solution: Here x is the time in months, and y is the value of the building in dollars. The depreciation schedule is represented by a line passing through the point (0, 720000) which is the point corresponding to the purchase (time=0 and value=720000) and the point (120, 0) which is the end of the depreciation after 120 months, that is when the building is worth nothing. So the slope is m= 0 − 720000 = −6000 120 − 0 Moreover the y-intercept is 720000 so the equation of the depreciation schedule is y = −6000x + 720000 3. The total amount of a simple investment rate is given by A(t) = P + P rt where P is the principal investment, r is the yearly rate and t is the time (in years, i.e. t = 0 when you invest the money and for instance the amount of money after 10 years is A(10), i.e. when t = 10). You want to invest $10000, and your goal is to have $14000 after 8 years. For what rate r can you achieve this goal? Solution: We setup the equation representing our goal, that is having $14000 on the account after 8 years: A(8) = 14000 Now we solve it for r, using the fact that P = 10000: A(8) = 14000 ⇐⇒ 10000 + 10000r × 8 = 14000 ⇐⇒ 80000r = 4000 ⇐⇒ r = 5% So to achieve the goal we need to have a yearly rate of 5%. 4. You want to invest money so that if you decide to cash it out after 4 years you would have $8000 and if you finally decide to cash it out after 10 years you would have $15000. Find the equation of the line representing the amount of money A in function of the elapsed time t (since the investment) that would match your expectations. Then compare with the formula of the preceding exercise to determine the rate and the principal investment to achieve your goals. Solution: The amount of money is given by A(t) = P + P rt Then A is a linear function of t (the time). Our goal here is to have $8000 after 4 years, that is we want A(4) = 8000 and also $15000 after 10 years, that is A(10) = 15000. So that means that the graph of the linear function A(t) is a line passing through the points (4, 8000) and (10, 15000). We the slope of this line: m= 7000 15000 − 8000 = 10 − 4 6 Then we find the point-slope form of the equation (here A is y-coordinate t is xcoordinate): A − 8000 = 7000 14000 7000 7000 (t − 4) ⇐⇒ A = 8000 + t− = t + 10000/3 6 6 3 6 Then as we know that A(t) = P + P rt then we must have P = 10000/3 = $3333.33 and P r = 3 7000 7 7000 ⇐⇒ r = × = = 0.35 = 35% 6 10000 6 20 5. You want to repaint your house. You calculated that you need 15 gallons of paint. At the paint store they only sell a thick paint that you must dilute with a special solvent. The salesman tells you that you must prepare a solution containing 10% of solvent to have a good result. How many gallons of pure paint and solvent must you buy to prepare the 15 gallons of diluted paint? Solution: We let x be the quantity of paint and y the quantoty of slovent, both in gallons. Then first we must prepare a total of 15 gallons with the paint and the solvent so we have the equation x + y = 15 Moreover the quantity of solvent must be 10% of the paint, so we have y = 0.1x ⇐⇒ x = 10y So we must solve the following system ( x = 10y x + y = 15 by substitution the second equation becomes: 10y + y = 15 ⇐⇒ 11y = 15 ⇐⇒ y = 15/11 = 1.4 and the x = 10y = 13.6. So you must buy 13.6 gallons of paint and 1.4 gallons of solvent. 6. Find ( the market equilibrium for the following laws of supply and demand. p = − 51 q + 200 , calculate also the M.E. in the case of a tax. of $6 per sold unit. a) p = q + 80 Solution: To find the market equilibrium before tax by solving: 6 1 − q + 200 = q + 80 ⇐⇒ 120 = q ⇐⇒ q = 100 5 5 then the corresponding price is p = 100 + 80 = 180. So the Market Equilibrium is reached for the price of $180 and a quantity if 100. With taxation, the new law of supply is p = q + 80 + 6 = q + 86. We find the market equilibrium after tax by solving: 1 6 − q + 200 = q + 86 ⇐⇒ 114 = q ⇐⇒ q = 95 5 5 then the corresponding price is p = 95 + 86 = 181. So the Market Equilibrium is reached for the price of $181 and a quantity if 95. ( 2p − q = 50 b) pq = 100 + 2q Solution: To find the market equilibrium we solve the system by substitution: from the first equation we get 2p − q = 50 ⇐⇒ q = 2p − 50 and we put this in the second equation: p(2p − 50) = 100 + 2(2p − 50) ⇐⇒ 2p2 − 50p = 100 + 4p − 100 ⇐⇒ 2p2 − 54p = 0 ⇐⇒ 2p(p − 27) = 0 ⇐⇒ p = 0 or p = 27 Then we compute the corresponding quantities: for p = 0 we find that q = −50 < 0, so we discard this solution. For p = 27 we find q = 54 − 50 ⇐⇒ q = 4. So the market equilibrium is for p = 27 and q = 4. 7. A company selling dryers has fixed costs of $5250, and variable costs of 100 + 0.1x per unit. a) Write the formula for the cost function C(x). Solution: We have C(x) = 5250 + x(100 + 0.1x) = 5250 + 100x + 0.1x2 . b) If the selling price for a dryer is $210, calculate the profit function P (x) and find the break-even point(s). Solution: As the revenue function is R(x) = 210x, we have P (x) = R(x) − C(x) = 210x − 5250 − 100x − 0.1x2 = −0.1x2 + 110x − 5250 The break-even points are given by solving the equation P (x) = 0 that is by solving −0.1x2 + 110x − 5250 = 0 We compute the discriminant ∆ = 1102 −4×(−0.1)×(−5250) = 12100−2100 = 10000. This is positive so we have 2 solutions √ √ −110 + 10000 −10 −110 − 10000 x1 = = = 50 and x2 = = 1050 2 × (−0.1) −0.2 −0.2 So the break-even points occur when selling 50 or 1050 dryers.