Equations and Graphs, Functions, and Systems of Equations Answers

Week 2 Quiz: Equations and Graphs, Functions, and Systems of Equations SGPE Summer School 2014 June 24, 2014 Lines: Slopes and Intercepts Question 1: Find the slope, y-intercept, and x-intercept of the following line: 15x − 3y = 60 (A) slope= 5, y-intercept= −20, and x-intercept= 4 (B) slope= 6, y-intercept= −22, and x-intercept= 5 (C) slope= 51 , y-intercept= 20, and x-intercept= −4 (D) slope= 5.5, y-intercept= −20, and x-intercept= 8 (E) None of the above Answer: (A) Given the form of the equation that we are given, it makes sense to go ahead and solve for the x and y intercepts. Solving for the y-intercept requires setting x = 0 in the above equation and then solving for y. Thus 15(0) − 3y = 60 which implies that y = −20. Similarly we can find the x−-intercept by setting y = 0 and solving for x, which yields 15x − 3(0) = 60, or x = 4. The easiest way to find the sole is just to convert the above equation to slope-intecept form (i.e, of the form y = mx + b). Follow the algebra: 15x − 3y = 60 −3y = −15x + 60 y = 5x − 20 Now we can just read off the slope from this expression. The slope is 5. Question 2: Find the equation of a straight line with slope of 12, and y-intercept of -33. (A) y = 1 12 x + 33 (B) y = −12x + 33 (C)y = 12x − 33 (D) y = 12x − 1 33 (E) None of the above Answer: (C) Simply substituting m = 12 and b = −33 into the slope intercept form of a line yields y = 12x − 33. Question 3: Find the equation of a straight line passing through the point (-2, 7) and perpendicular to a line with equation 24x + 6y = 30. (A) y = 4x + 13 2 (B) y = − 14 x + 15 2 (C)y = −4x + 15 1 (D) y = 1 12 x + 15 2 (E) None of the above Answer: (E) First we need to figure out the slope of our equation. We are told that it is perpendicular to a line whose equation is 24x + 6y = 30. What is the slope of this line? Solving for the slope intercept form yields y = −4x + 5. Thus the slope is −4, which means that the slope of our equation is 14 (i.e., the negative reciprocal). Now we need to find the y-intercept. To do this we first write down the point-slope form of the line: (y − y1 ) = m(x − x1 ). Now substitute for m, x1 and y1 using the slope (which we just calculated) and the point (which we were given). This yields: 1 y − 7 = (x − (−2)) 4 Now we simply re-arrange this equation to slope-intercept form to get the answer: 1 y − 7 = (x − (−2)) 4 1 1 y−7= x+ 4 2 1 15 y = x+ 4 2 Question 4: Find the x intercept of the following equation: −13x + 19y = −39 Answer: Substitute 0 for y: −13x + 19 ∗ 0 = −39 ⇔ x = 3 Therefore the x intercept is (3, 0). Question 5: Find the y intercept of the following equation: y = 79.5x − 250 Answer: Substitute 0 for x: y = 79.5 ∗ 0 − 250 ⇔ y = −250 Therefore the y intercept is (0, −250). Question 6: Find the x intercept of the following equation: 1 −5x − y = 55 2 Answer: Substitute 0 for y: −5x − 1 ∗ 0 = 55 ⇔ x = −11 2 Therefore the x intercept is (−11, 0). Question 7: Derive the equation for a straight line with: a) Slope: − 32 and b) y intercept: (0, 49). Answer: y = − 23 x + 49. 2 Linear Equations in Business and Economics Question 8: Find the total cost of producing (a) 25 units, and (b) 40 units of output for a firm that has fixed costs of $3200 and a marginal cost of $500 per unit. (A) 15,700 and 23,200 (B) 16,000, and 23, 000 (C) 24,000 and 54,000 (D) 32,000 and 43,500 (E) None of the above Answer: (A) First need to find the total cost function. Total cost, T C, is the sum of total fixed costs, T F C, and total variable costs, T V C. Total variable costs, T V C, depend on the level of output, Q, that the firm chooses to produce. Mathematically, we say that T V C is a function of Q, and write T V C(Q). Since T C is a function of T V C(Q), this means that T C is also a function of Q. This is denoted by T C(Q). Now we can write a general total cost functions as follows: T C(Q) = T F C + T V C(Q) We are told that T F C = $3200 and that the only type of variable costs are marginal costs, and that these marginal costs amount to $500 per unit output, which implies that T V C(Q) = $500Q. Substituting into the above equation yields a total cost function of: T C(Q) = 3200 + 500(Q) To answer the question we simply need to evaluate the total cost function we have derived at Q = 25, and Q = 40. This yields answers of T C = 3200 + 25(500) = 15, 700 and T C = 3200 + 40(500) = 23, 500. Question 9: Find the profit level of a firm in perfect competition that has a fixed cost of $900, a variable cost of $80, and a selling price for their output of $90 per unit if they sell (a) 55 units, and (b) 100 units. (A) 350 and 100 (B) -350 and -100 (C) -350 and 100 (D) 4950 and 9000 (E) None of the above Answer: (C) First need to find the profit function. Profit (π) is always Revenue (R) less Total Costs (TC). π = R − TC From solution to the previous question, we know that T C(Q) = T F C + T V C(Q). Thus T C(Q) = 900 + 80Q. What about revenue? First note that a firm’s revenue is a function of the quantity of output produced (i.e., R(Q)). In this case we are told that the firm brings in $90 per unit output, thus their revenue function is R(Q) = 90Q. Now we have enough information to write the firm’s profits as a function of quantity produced: π = R(Q) − T C(Q) = 90Q − (900 + 80Q) = 10Q − 900 Now we need only to evaluate the profit function at Q = (a) 55 units, and (b) 100 units. Which yields profits of π = 10(55) − 900 = −350 (Negative profits indicates firm is operating at a loss!); and π = 10(100) − 900 = 100. Question 10: Assume that there are no salvage (or liquidation) costs. Find (a), the value of an asset after 6 years and (b), the salvage (sometimes also called the liquidation) value after 100 years of an asset whose current value A after t years is: A = 1, 500, 000 − 1, 500t 3 (A) 1,492,000 and 1,360,000 (B) 1,491,000 and 1,350,000 (C) 1,591,000 and 1,356,000 (D) 1,191,000 and 1,400,000 (E) None of the above Answer: (B) Note that given there are no liquidation costs, the salvage (or liquidation) value of an asset after t years is simply the current value of the asset after t years. Thus we need only evaluate the expression for current value after 6 and 100 years respectively. This yields A = 1, 500, 000 − 1, 500(6) = 1, 491, 000 and A = 1, 500, 000 − 1, 500(100) = 1, 350, 000. Question 11: With $100,000 to invest, how much should a broker invest in a Portuguese bond that pays 11% and how much in an Italian bond that pays 15% in order to earn an expected return of 14% on the total investment? (A) 23,000 in Portuguese bond and 77,000 in Italian bond (B) 50,000 in Portuguese bond and 50,000 in Italian bond (C) 75,000 in Portuguese bond and 25,000 in Italian bond (D) 35,000 in Portuguese bond and 65,000 in Italian bond (E) None of the above Answer: (E) If x denotes the amount of wealth invested in the Portuguese bond, then (100, 000 − x) denotes the amount of wealth invested in the Italian bond. We can express the the total annual interest on the portfolio, r as: r(100, 000) = .11x + .15(100, 000 − x) = 15, 000 − 0.04x Given that our broker wants to achieve an expected return of r = 14%, we simply need to solve for x as follows: .14(100, 000) =15, 000 − 0.04x 14, 000 =15, 000 − .04x .04x = 1, 000 x = 25, 000 Thus our broker should invest $25,000 in the Portuguese bond paying 11% and $75,000 in the Italian bond paying 15%. Solving Quadratic Equations Question 12: Solve the following quadratic equation: x2 + 6x + 9 = 0 (A) x = 3 and x = −3 (B) x = 3 (C) x = −3 (D) x = 2 and x = 3 (E) None of the above Answer: (C) Easiest way to solve this equation is NOT to use the quadratic formula, but to simply factor the polynomial on the left-hand-side of the equation. This yields (x + 3)2 = 0, which has a double root of x = −3. 4 Question 13: Solve the following quadratic equation: 5x2 + 47x + 18 = 0 (A) x = − 25 and x = −9 (B) x = − 52 and x = 9 (C) x = 2 5 and x = 9 (D) x = 2 5 and x = −9 (E) None of the above Answer: (A) Solve by applying the quadratic equation. Follow the algebra: p −47 ± 472 − 4(5)(18) x= 2(5) √ −47 ± 2209 − 360 = √10 −47 ± 1849 = 10 −47 ± 43 = 10 4 2 −90 = − = − and = −9 10 5 10 Question 14: Solve the following quadratic equation: 1 3 x2 + x + = 0 4 8 Answer: Solve by applying the quadratic equation formula: q
3 2 − 34 ± − 4(1) 4 x= 2(1) q 9 − 34 ± 16 − 12 = 2 q 3 1 − 4 ± 16 = 2 − 34 ± 14 = 2 1 1 1 =−2 =− ∨=− 2 4 2 Question 15: Solve the following quadratic equation: 4x2 − 31x − 45 = 0 5 1 8
Answer: Solve by applying the quadratic equation formula: p 31 ± (−31)2 − 4(4)(−45) x= 2(4) √ 31 ± 961 + 720 = √8 31 ± 1681 = 8 31 ± 41 = 2 72 10 = = 9∨ = − = 1.25 8 8 Question 16: Solve the following quadratic equation: x2 − 67.5x + 126 = 0 Answer: Solve by applying the quadratic equation formula. Note that the square root is not an exact square in this case, and has been rounded to 2 decimal places: p 67.5 ± (67.5)2 − 4(1)(126) x= 2(1) √ 67.5 ± 4556.25 − 504 = √ 2 67.5 ± 4052.25 = 2 67.5 ± 63.66 = 2 131.16 3.84 = = 65.58∨ = = 1.92 2 2 Non-linear Functions Question 17: Items such as automobiles are subject to accelerated depreciation whereby they lose value faster than they do under linear depreciation. Suppose that a car with initial value of $100, 000 value depreciates at 10% per year (continuously compounded) (i.e., the cars value after t years is V (t) = 100000e−0.10t ). Further, suppose that there always exists the option to sell the car for scrap to a “chop-shop” for $2000. After how many years will in be optimal to sell the car for scrap? (A) 39 (B) 41 (C) Never! (D) 40 (E) None of the above Answer: (D) We want to find after how many years t will the scrap value of the car exceed its value in year t, 6 V (t). To solve this question we need to solve the following inequality: 2000 ≥ 100000e−0.10t 1 ≥ 50e−0.10t 1 ≥ e−0.10t 50  
1 ln ≥ ln e−0.10t 50 ln(1) − ln(50) ≥ −0.10t t ≥ 10ln(50) t ≥ 39.12 ≈ 40 years! So it would be optimal to sell the car for scrap after about 40 years. Question 18: A factory’s cost function C(Q) is a function of the number (quantity) of units produced; suppose that the quantity of units produced is itself a function of time, Q(t). Specifically, assume that C(Q) = 1900 + 50Q and that Q(t) = 16t − 14 t2 . Find the function that expresses the factory’s cost as a function of time, and then find out the factory’s costs are after t = 1 and t = 10 periods. (A) 1900 + 300t − 12t2 ; C(1) = 2188; C(10) = 1000 (B) 1900 + 800t − 25 2 2 t ; C(1) = 2687.5; C(10) = 40650 (C) 1900 + 800t − 10t2 ; C(1) = 2690; C(10) = 31900 (D) 1900 + 300t − 25 2 2 t ; C(1) = 2187.5; C(10) = 3650 (E) None of the above Answer: (E) To find the factory’s costs as a function of time, which is a composition of the factory’s cost function and the quantity function. Simply substitute the quantity function Q(t) in for Q in the cost function as follows: 1 25 C(t) = C(Q(t)) = 1900 + 50(16t − t2 ) = 1900 + 800t − t2 4 2 Now we need only evaluate this composite function C(t) = C(Q(t)) at t = 1 and t = 10. C(1) = 1900 + 800 − 25 2 = 25 2 2700 − 12.5 = 2687.5; C(10) = 1900 + 800(10) − 2 (10) = 1900 + 8000 − 1250 = 8650. Question 19: The average level of carbon monoxide in the air is given by the function L(n) = (1 + 0.6n) parts per million (ppm), where n is the number of people (in thousands). Assuming the population in thousands at time t is N (t) = 400 + 30t + 0.15t2 , a) express the level of carbon monoxide in their air as a function of time, and b) estimate the level of carbon monoxide at time t = 5. Answer: First substitute the function N (t) for n in the function L(n), to get the composite function L(N (t)): L(N (t)) = 1 + 0.6(400 + 30t + 0.15t2 ) = 241 + 18t + 0.09t2 . Then to solve part b) find L(N (5)) by substituting 5 got t: L(N (5)) = 241 + 18 ∗ 5 + 0.09 ∗ (5)2 = 333.25ppm. Systems of Equations Question 20: Suppose that supply and demand are described by the following set of equations: Supply: Qs = 2P − 2 8 Demand: Qd = − P + 16 5 7 Find the market clearing price and quantity. (A) Pe = 6; Qe = 9 (B) Pe = 4; Qe = 7 (C) Pe = 5; Qe = 8 (D) Pe = 8; Qe = 5 (E) None of the above Answer: (C) To find the market clearing price, Pe , simply equate quantity demanded with quantity supplied and solve for P . Thus... 8 (Supply) Qs = 2Pe − 2 = − Pe + 16 = Qd (Demand) 5 8 2P + Pe = 16 + 2 5 18 Pe = 18 5 Pe = 5 To find Qe , simply plug Pe = 5 into both the equation for supply and demand. Why both equation? Shouldn’t you get the same answer by plugging the market clearing price into either equation? YES! But plugging the market clearing price into both equations is an easy way to check and make sure that you haven’t made a silly maths mistake somewhere. Thus Qd = 2(5) − 2 = 8, and Qs = − 58 (5) + 16 = 8. And we have found Qe = Qs (Pe ) = Qd (Pe ) = 8. Question 21: Assume a revenue function of R(Q) = 48Q − 3Q2 and a cost function of C(Q) = 6Q + 120. Find the break even level (or perhaps levels) of Q. (A) 3 and 11 (B) 4 and 10 (C) 2 and 9 (D) 6 and 13 (E) None of the above Answer: (B) There are multiple ways to solve this problem. I prefer to use the revenue and cost functions to derive the profit function, and then fund those values of Q for which profits are zero. If profits are zero, then revenue exactly covers costs and the firm is breaking even. Profits, π(Q) = R(Q)−C(Q) = 48Q−3Q2 −6Q−120 = 42Q − 3Q2 − 120. Setting profits equal to zero, and solving the quadratic equation will gives us the break-even levels of Q. Thus set π(Q) = −3Q2 + 42Q − 120 = 0 and solve. p −42 ± 422 − 4(−3)(−120) Q= 2(−3) √ −42 ± 324 Q= −6 −42 ± 18 Q= −6 Q = 4 and Q = 10 Question 22: Solve the following system of equations for the equilibrium levels of income, Y and interest rate, i (round your answers to the nearest hundredth): 0.1Y + 80i − 75 = 0 0.4Y − 124i − 260 = 0 8 (A) Ye = 677.93; ie = 0.09 (B) Ye = 505.84; ie = 0.22 (C) Ye = 604.83; ie = 0.11 (D) Ye = 605.99; ie = 0.24 (E) None of the above Answer: (A) Easiest way to solve the above equation is to use the substitution method. Solving the first equation for Y yields: 0.1Y + 80i − 75 = 0 Y + 800i − 750 = 0 Y = −800i + 750 Substituting this expression for Y into the second equation allows us to solve for the equilibrium level of the interest rate. 0.4(−800ie + 750) − 124ie − 260 = 0 −320ie + 300 − 124ie − 260 = 0 −444ie + 40 = 0 40 ie = ≈ 0.09or9% 444 (1) 40 Now using ie = 444 , we can substitute it into either (or both!) of
the above equation to find the equilibrium level of
40 40 income Y . Thus 0.4Ye − 124 444 − 260 = 0 and 0.1Ye + 80 444 − 75 = 0, both equations imply that Ye ≈ 677.93. Question 23: Solve the following system of equations for the the equilibrium level of income, Ye , in terms of given levels of government spending, G = G0 , and investment, I = I0 . Y =C +I +G C = C0 + bY where 0 < b < 1 (A) Ye = 1 b−1 (C0 + I0 + G0 ) (B) Ye = 1 1−b (C0 + I0 + G0 ) (C) Ye = 1−b (C0 +I0 +G0 ) (D) Ye = (1 − b)(C0 + I0 + G0 ) (E) None of the above Answer: (B) Substitute for C, G, and I in the first equation above, and then solve for Y. Y =C +I +G Ye = (C0 + bYe ) + I0 + G0 Ye − bYe = C0 + I0 + G0 Ye (1 − b) = C0 + I0 + G0 1 (C0 + I0 + G0 ) Ye = 1−b Question 24: Solve the following system: ( 24x − 7y = 37 −6x + 9y = 27 9 Answer: Use substitution: ( ( 24x − 7y = 37 24x − 7y = 37 ⇔ −6x + 9y = 27 x = 23 y − 92 (
24 23 y − 92 − 7y = 37 ⇔ x = 32 y − 92 ( 29y − 145 = 0 ⇔ x = 23 y − 92 ( y=5 x = 23 ∗ 5 − ⇔ ( y=5 ⇔ x=3 9 2 Question 25: Solve the following system: ( y = 0.75x − 4 y = 2.5x − 11 Answer: In this case, as both equations as given with y already isolated, simply set the two left hand sides equal to each other to find x. Then substitute x in either equation to find y. ( ( y = 0.75x − 4 x=4 ⇔ 0.75x − 4 = 2.5x − 11 ⇔ 1.75x = 7 ⇔ x = 4;y = 2.5 ∗ 4 − 11 = −1 ⇔ y = 2.5x − 11 y = −1 Question 26: Solve the following system: ( 18x − 2y = 32 12x + 5y = −23 Answer: Use substitution: ( ( 18x − 2y = 32 y = 9x − 16 ⇔ 12x + 5y = −23 12x + 5y = −23 ( y = 9x − 16 ⇔ 12x + 5(9x − 16) = −23 ( y = 9x − 16 ⇔ ⇔ 57x − 57 = 0 ( ( y = 9 ∗ 1 − 16 y = −7 ⇔ x=1 x=1 Question 27: Solve the following system: ( 11x + 3y = 53 4x − 18y = 172 Answer: Use substitution: ( ( 11x + 3y = 53 11x + 3y = 53 ⇔ 4x − 18y = 172 x = 43 + 4.5y ( 11(43 + 4.5y) + 3y = 53 ⇔ x = 43 + 4.5y ( 52.5y + 420 = 0 ⇔ x = 43 + 4.5y ( y = −8 x = 43 + 4.5 ∗ −8 10 ⇔ ( y = −8 ⇔ x=7