Unformatted text preview: COMM 295 Math Review Solutions 1. Draw the following equation in a diagram with y on the vertical axis and x on the horizontal axis: 2x + 3y = 9. What is the vertical intercept and what is the slope? Rearrange: 3y = 9 – 2x or y = 9/3 – 2x/3 or y = 3 – 2x/3. This is a straight line with vertical intercept 3 and slope ‐2/3. y 3 4.5 x 2. Find the point of intersection of the following two lines, and then graph them: 2x + 3y = 9 x + 2y = 5 Solve each equation for y: y= 3 – 2x/3 and y = 5/2 – x/2. At the point of intersection, x and y must have the same values in both equations. Set: 3 – 2x/3 = 5/2 ‐ x/2. 3 – 5/2 = 2x/3 – x/2, so x/6 = ½ or x = 3. Substitute x = 3 into one of the original equations for y: y = 3 – 2x/3 = 3 – 6/3 = 1. So the two lines intersect where x = 3 and y = 1 Graphing is similar to that shown in Q1. 3. Find the point of intersection of the following two lines, and then graph them: x – 2y = 1 2x – 4y = ‐3 Solve each for x: x = 1 +2y and x = ‐3/2 + 2y. Set these expressions equal to solve for y: 1 + 2y = ‐3/2 + 2y. There is no solution to this equation. It is not possible for 2y + 1 to equal 2y – 3/2. If we try to isolate y, y simply drops from the equation. These two equations are parallel and therefore never cross. You can graph the lines in the usual way. 4. Graph the line: X = 5 This line is a vertical line at X=5. Notice that X is always equal to 5. 5. Graph the line: Y = 6 This line is horizontal at distance 6 on the Y axis. 6. Amy buys 10 videos each month regardless of the price. Write down the equation for Amy’s demand curve. If Q stands for quantity, the equation is Q = 10. 7. Brian cuts lawns. He faces a demand curve given by Q = 50 – 2P where Q is quantity and P is price. Write down the equation for Brian’s revenue, R, where R is a function of Q. Graph this revenue function for Q between 0 and 50. Invert the above function to get P = 25 ‐ 0.5Q. Revenue R= PQ so R = (25 ‐ 0.5Q)Q = 25Q – 0.5Q2. This is a standard quadratic function that takes on a value of 0 when Q = 0 and when Q = 50. The function reaches a maximum when Q = 25. 8. A demand curve is given by Q = a – P and a supply curve is given by Q = b + 2P. Suppose that a = 100 and b = 40. What is the point where the supply and demand curves cross? Illustrate the solution in a diagram. The equation for demand is Q = 100 – P and for supply is Q = 40 + 2P. At the point where these two curves cross, Q must be the same for both. Therefore, 100 – P = 40 + 2P at this intersection. Solving for P yields 100 ‐ 40 = 3P so P = 60/3 = 20. Substituting this value for P into the demand equation yields Q = 100 – P = 100 – 20 = 80. Thus the solution is Q = 80 and P = 20. 9. In general, the derivative of the functional form y = axb is given by dy/dx = baxb‐1. Most of the questions are based on this general form. a) 1 b) 2x c) 25x4 + 3 d) ‐1x‐2 = ‐1/x2 e) ‐2x‐3 ‐1/2 ‐2/3 ‐1/2 f) (1/2)x g) x h) (1/2)x i) 1/x j) 0 For j) note that x0 is equal to 1 (a constant). 10. If demand is given by Q = 100 – P, it follows that inverse demand is P = 100 – Q, Revenue, R is given by R = PQ = (100 – Q)Q = 100Q – Q2. The derivative, dR/dQ = 100 – 2Q. R reaches its highest value when Q = 50. The slope of the Revenue function at this point is 0. ...
View Full Document
- Winter '09
- vertical intercept, demand equation yields, Math Review Solutions, Revenue R= PQ