Exam A. Exam 3. (e) Two critical points; one is a local maximum, the other a local minimum.

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1 1.(6 pts) The function f(x) = x 3 2x 2 has: Exam A Exam 3 (a) Two critical points; one is a local minimum, the other is neither a local maximum nor a local minimum. (b) Two critical points; one is a local maximum, the other is neither a local maximum nor a local minimum. (c) Two critical points, both local maxima. (d) Two critical points, both local minima. (e) Two critical points; one is a local maximum, the other a local minimum. 2.(6 pts) For a continuous function f(x), consider the following conditions: f (x) > 0 on ( 2, 1) (0, 1). f (x) < 0 on (, 2) ( 1, 0) (1, 2) (2, + ). Which of the following statements must be true, for any function satisfying the above conditions? (a) The function has at least 4 critical points, two of which must be local maxima, and two must be local minima. (b) The function has at most 4 critical points, which could be local maxima or minima. (c) The function has at most 3 critical points, one of which is a local minimum, the others local maxima. (d) The function has at most 3 critical points, and no local maximum. (e) The function has no critical points, since it is continuous everywhere.

2 x 3.(6 pts) Let m [0, 2] be the point where the function f(x) =, restricted to the x interval [0, 2], attains a global maximum. In other words, f(m) f(x) for all x [0, 2]. Which of the following statements about m is correct? (a) m = 2. (b) 1/2 < m < 3/2. (c) 3/2 < m < 2. (d) 0 < m < 1/4. (e) 1/4 < m < 1/2. 4.(6 pts) Suppose that a company determines that given a price p per unit of a commodity, the number of units sold q is q = 20e 0.5p. Determine the maximal revenue R = pq. (a) 20e 0.75 (b) 40e 1.0 (c) 30e 1.0 (d) 20e 0.25 (e) 10e 0.5

3 5.(6 pts) Consider the curve given by Determine dy dx at the point (1, 2). x 2 + y 3 + 2xy + xy 2 = 17. (a) 1 2 (b) 1 2 (c) 5 9 (d) 5 9 (e) 0 6.(6 pts) Determine all of the inflection points for the function f(x) = x 4 2x 3 12x 2 5. (a) x = 2 (b) x = 2, x = 1 (c) x = 1 (d) x = 2, x = 1 (e) f does not have any inflection points.

4 7.(6 pts) The maximum value of the function y = x 3 3x 1 on the interval [0, 3] is (a) 1 (b) 3 (c) 1 (d) 17 (e) 15 8.(6 pts) Find the x-coordinate of all points where the graph of y = 3x 2 + 6x + 3 has slope 2. (a) x = 3 5 and x = 1 (b) There are no points where the slope is 2. (c) x = and x = 1 2 (d) x = 2 3 (e) x = 0 and x = 1

5 9.(6 pts) A restaurant owner studied the sales of an octopus dish and determined that its average number of orders q each night is given by p = 72, where p is the price in dollar of q + 2 an order of the dish. Supposing each appetizer costs the restaurant $4 to make, what price should the owner charge to maximize profit from the appetizer? (a) 12 (b) 8 (c) 3 (d) 14 (e) The more he charges the greater the profit. 10.(6 pts) Consider the function f(x) = 1 + ln x. On which interval below is f decreasing? x (a) (0.5, 3) (b) ( 1, 3) (c) (2, 5) (d) (0.5, 5) (e) f is increasing on its domain.

6 11.(12 pts) A cylindrical tank is filling with water at a rate of 4t cubic meters per minute at any given time t. Assume that the radius of the tank is 2 meters. (1) Determine the rate at which the height of the water is changing at any given time t. (2) Determine the rate at which the height of the water is changing at time t = (14 pts) A certain company can produce up to 500 units of a product it wants to sell. The demand function for this product is p = 1000e q/100, where p is the price in dollars and q is the quantity being produced. Find the production level (i.e., how many units) that maximizes the profit of this company, knowing that it has a fixed cost of $700. This problem is badly phrased. From the solution we see that it was assumed that $700 is the cost no matter what the level of production. This is not usually what fixed cost means.

7 13.(14 pts) NASA is tracking an asteroid on a parabolic orbit passing close to earth. In their coordinate system the earth is at the point (0, 10) and the orbit is the graph of y = x 2. How close does the asteroid come to the earth? Be sure to explain why your distance is a minimum and not just a local minimum (or even worse, a local maximum). Hint: Minimizing the distance squared makes your calculations easier.

8 1. Solution. The critical points are solutions to f (x) = 3x 2 4x = 0, which are x = 0 and x = 4/3. Since f (x) = 6x 4, we have f (0) = 4 < 0 and f (4/3) = 4 > 0, so by the second derivative test, one is a local maximum and the other a local minimum. 2. Solution. The conditions given imply that f ( 2) = f ( 1) = f (0) = f (1) = 0, but we cannot say what happens to f (2); it could be negative or zero, so we could have up to 5 critical points. This information alone determines the right answer. By looking at the signs of f (x) on the given intervals, we can easily see that twice it changes from positive to negative, and twice it changes from negative to positive, so we have two local maxima, two local minima. 3. Solution. First, find the critical points on the interior of [0, 2] by solving the equation (1) f (x) = 1(x2 + 1) x(2x) (x 2 + 1) 2 = 1 x2 (x 2 + 1) 2 (2) f (x) = 0 1 x2 (x 2 + 1) 2 = 0 x = ±1. In [0, 2], x = 1 is the only critical. Now, comparing the values of f(x) at the critical point x = 1 with the values at the endpoints, we see that f(1) = 1/2 is a global maximum. Thus, m = 1 and 1/2 < m < 3/2.

9 4. Solution. Taking the first derivative of R(p) = 20pe 0.5p, we have Setting this equal quantity equal to zero, R (p) = 20e 0.5p 10pe 0.5p. 20e 0.5p 10pe 0.5p = 0 = p = 2. Choosing values to the left and right of p = 2, in this case p = 0 and p = 100, we have R (0) positive and R (100) negative. Hence we see that p = 2 is a maximum. Plugging this value back into our function gives our maximal revenue as R(2) = 40e 1 5. Solution. Using implicit differentiation, we see that d [ x 2 + y 3 + 2xy + xy 2 ] = d [ ] 6 dx dx = 2x + 3y 2 dy + 2y + 2xdy dx dx + y2 + 2xy dy dx = 0 = 2(1) + 3(2) 2 dy + 2(2) + 2(1)dy dx dx + (2)2 + 2(1)(2) dy dx = 0 = 18 dy dx = 10 = dy dx = 5 9

10 6. Solution. We see that the first derivative for f is and the second derivative is Exam A f (x) = 4x 3 6x 2 24x f (x) = 12x 2 12x 24 Setting the second derivative equal to zero we find 12x 2 12x 24 = 0 = x 2 x 2 = 0 = (x 2)(x + 1) = 0 = x = 2, x = 1 Checking the three values x = 10, f ( 10) = (12)( 10 2)( ) = (12)( 12)( 9) > 0. x = 0, f (0) = (12)(0 2)(0 + 1) = (12)( 2)(1) < 0. x = 10, f (10) = (12)(10 2)(10 + 1) = (12)(8)(11) > We see that the sign switches at each zero so that x = 2 and x = 1 are both inflection points. 7. Solution. Find critical points: y = 3x 2 3; y = 0 has solutions ±1. y(0) = 1; y(1) = 3; y(3) = Solution. y = 6x + 6. Solve 6x + 6 = 2, 6x = 4, x = 2 3

11 9. Solution. Revenue R = q p = 72q 72q. Cost C = 4q so profit P = q + 2 q + 2 4q = ( ) 72q 4q(q + 2) = 72q 4q2 8q 64q (64 8q)(q + 2) (64q 4q 2 )(1) 4q2 =. P = = q + 2 q + 2 q + 2 (q + 2) 2 64q q 2 16q 64q + 4q q 4q2 32 4q q2 (8 + q)(4 q) = = 4 = 4. (q + 2) 2 (q + 2) 2 (q + 2) 2 (q + 2) 2 Solutions are q = 8 and q = 4 and we can t charge a negative price so q = 4 and hence p = = Solution. f (x) = ( 1 ) x (1 + ln x) (1) x x 2 = 1 1 ln x x 2 = ln x. Solve f (x) = 0 so ln x = 0 so x = 1 is the only critical point. If x > 1, ln x > 0 so f < 0; x < 1, ln x < 0 so f > 0. x so decreasing on any subinterval of (1, )

12 11. Solution. (1) We see that V = πr 2 h = V = 4πh since r = 2 = dv dt = 4π dh dt = 4t = 4π dh dt (2) so dh dt = t π dh dt = 3 π 12. Solution. The profit function, in terms of the quantity q being produced, will be P (q) = R(q) C(q) = pq 700 = 1000qe q/ To maximize it, first find its critical points: ( ) P (q) = 1000 e q/100 q + e q/100 = 0 q = Comparing the values at the endpoints with the value at the critical point, we have (3) P (0) = 700 (4) 100, 000 P (100) = 700 e (5) 500, 000 P (250) = 700 < 100, e 5 Thus, the global maximum is achieved when q = 100.

13 13. Solution. Distance squared is D = x 2 +(x 2 10) 2 = x 2 +x 4 20x = x 4 19x D = 4x 3 38x so x = 0 and x = ± ±3.08 are the critical points. 2 ) 19 Since D = 12x 2 38, D (0) = 38, local max.; D (± = = > 0, 2 2 local minimum s. You can also use the first derivative test. Since = the asteroid orbit is eventually arbitrarily far away from the earth. lim x ± D Hence as the asteroid its distance to the earth is steadily decreasing until it gets to x = Then its distance starts to increase until x = 0 where it is as far away from the earth as it has been since x = Then it starts to get closer to the earth again until in gets to x = 3.08, after which the distance begins to increase again and gets steadily bigger thereafter. Just the full number line would get you full marks ( ) 19 D ± = 2 ( 19 2 so the minimum distance is ) = = = Another approach. D = x 2 + (y 10) 2 = y + (y 10) 2 dd. = 1 + 2(y 10) = 2y 19. dy There is one critical point at y = d2 D = 1 + 2(y 10) = 2 > 0 so local minimum. Since dy 2 y = x 2, y takes values in [0, ). + D 0 19/2 Hence D decreases from y = 0 to y = 19/2 and increases thereafter. Hence 19/2 is a global minimum and the distance is D = 19 + ( 19 10) = 19 + ( ) 2 = = Yet another approach. D = x 2 + (y 10) 2 dd. = 2x + 2(y 10)dy dx dx. Since y = x2, dy dd = 2x so dx dx = 2x + 2(x2 10)2x = 2x(1 + 2x 2 20) = 2x(2x 2 19). Proceed from here as in the first solution. =