SOLUTIONS to Review Problems for Chapter 4. by Vladimir A. Dobrushkin

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1 Hughes-Hallett SOLUTIONS to Review Problems for Chapter 4 by Vladimir A. Dobrushkin Third Edition 4.1 The points: (1, 2) is local and global minimum, (3.5, 8) is local and global maximum, and (5, 4.5) is local minimum. 4.2 The points: (3, 4.5) and (9, 4.5) are global and local maximum, (6, 1.5) is local and global minimum, (1, 2.5) is local minimum, and (10, 3) is local minimum. 4.3 (a) f(1) is local minimum; f(0), f(2) are local maxima. (b) f(1) is global minimum; f(2) is global maximum. 4.4 (a) x = 1 is local minimum. (b) Since f(1) = 1, f(0.1) = 0.1 ln , and f(2) = 2 ln 2 1.3, the function f(x) = x ln x attains its global minimum on the interval [0.1, 2] at the point x = 1 and its global maximum at x = Local maximum, minimum, or neither. Price relatively constant around July (a) Time, t = 1/b, to reach the peak concentration does not depend on a. Although, as parameter a increases, the peak concentration increases. (b) t = 1/b, so t is inverse proportional to b. (c) C (t) = a e at (1 at). Time, t = 1/a, to reach the peak concentration is inverse proportional to a. 4.7 (a) Increasing for x > 0. Decreasing for x < 0. (b) Local and global minimum: f(0). 4.8 (a) The function is increasing everywhere except x = 0. (b) Neither maximum no minimum. 4.9 (a) Increasing for 0 < x < 4. Decreasing for x < 0 and x > 4. (b) Local maximum: f(4), local minimum: f(0) (a) Increasing for 1 < x < 0 and x > 1. Decreasing for x < 1 and 0 < x < 1. (b) x = 1 is local min; x = 0 is local max; x = 1 is local min (a) cm/week. (b) Growing at 1.6 cm/week in week (a) f (20) > f (36).

2 (b) After approximately 20 years, human fetus still grows, but its rate slow down (the function is concave down) (a) Week 14. (b) Point of fastest growth (a) f (20) 1 cm/week. (b) f (36) 1/3 cm/week. (c) The average rate of change is = 5 4 = a = 3e and b = The point θ = 1.15 is a local maximum; the point θ = 1.55 is a local minimum, and the point θ = 2.05 is a local maximum One (a) The derivative, f (x), has two critical points: x 2 and x 4 ; its graph is similar to the cubic function y = (x x 1 )(x x 3 )(x x 5 ). (b) The function f (x) changes its sign at x 1, x 3, and x 5. (c) The derivative, f (x), has a local maximum at the point x 2 and a local minimum at x The function f has local minima at points x 1 and x 5 and a local maximum at x 3. The graph of the function f changes concavity at points x 2 and x 4. The derivative f is negative for x < x 1 and x 3 < x < x 5. It is positive at x 1 < x < x 3. The derivative f has a local maximum at x 2 and a local minimum at x 5. { 3 + arctan(x 3), if x < 3; 4.20 y = 3 arctan(x 3), if x > (a) Polynomial; negative leading coefficient; degree 2. (b) Exponential. (c) Logistic. (d) Polynomial; positive leading coefficient; degree 2. (e) Exponential. (f) Surge. (g) Periodic. (h) Polynomial; negative leading coefficient; degree 3. 2

3 4.22 The profit function π(q) = R(q) C(q), where R(q) = 788q is the revenue and C(q) = q 3 60q q is the total cost, attains its maximum at the point q 0 where C (q 0 ) = 788; so we get the equation 3q 2 120q = 788 or q 2 40q = 0. Solving it, we obtain two possible values: q 0 = 34 and q = 6. Since q = 6 is a local minimum π(6) = $2728, the remaining point q 0 = 34 is its maximum: π(34) = $8248. The total cost: C(34) = $18544, the revenue: R(34) = $26792, and the profit: π(34) = $ (a) The fixed cost, which is incurred even if nothing is produced, is $0. (b) The maximum profit: $ (c) Raise the price by $ Since the revenue is defined as R = pq, we have the profit, π(q), to be π(q) = R(q) C(q) = b 1 q a 1 q 2 b 2 a 2 q. Its critical point we obtain from the equation: b 1 2a 1 q a 2 = 0, so the point of maximum is q = (b 1 a 2 )/2a (a) The profit π(q) attains its maximum when R(q) > C(q) and R and Q are farthest apart. (b) C (q 0 ) = R (q 0 ) = p. (c) Graph is close to x 2 + c 4.26 (a) The average cost is a(q) = C(q)/q = 0.04q 2 3q /q. (b) Near q = 0 the graph is close to 96/q. When q grows, the graph of a(q) approaches 0.04q 2. (c) The function a(q) is decreasing for q < and is increasing for q > , where q = is the root of the equation 0.08q 3 = 96/q 2. (d) Graph of a cost function should be tangent to the straight line that goes through the origin and the point (15000, (a) F is the point of y-intercept. (b) The break-even point is the point where the profit is zero and the revenue equals cost (there are two such points on the graph). (c) Marginal cost, C (q) attains its minimum when the tangent line is almost horizontal (its slope is close to zero). (d) Critical points of average cost occur when marginal cost equals average cost: C (q) = C(q)/q, which has the general solution C(q) = kq for some constant k. Therefore the minimum average cost occurs at the point where the straight line C = kq touches the graph of C(q). 3

4 4.29 C (2) (e) The profit π(q) = R(q) C(q) attains its maximum when R and Q are farthest apart. C(5) C(3) (C(75) C(50))/ (C(75) C(25))/ C(3)/ C(10)/ E = 0.05, demand is inelastic Since the revenue, R, is R = pq, its derivative is dr dp = d(pq) = q + p dq. Therefore the dp dp elasticity of demand is E = p q dq ( dp = 1 q + p dq ) 1 q dp = 1 dr q dp 1 > 1 because dr/dp < Similar to the previous problem, the elasticity of demand is E = 1 q dr/dp > 0. dr dp 1 < 1 because 4.38 (a) A logistic model is reasonable because it grows exponentially at the beginning and after a while its growth rate slows down. (b) Taking the difference in the number of households with cable television, we get Year P Year P Hence we see that after 1982 the increase in growth slows down and we expect the curve to be concave down. This corresponds to the half of the limiting value, L/2, to be somewhere between 28.3 and 35. (c) (d) The limiting value looks close to 69%. ( 4.39 (a) d2 P dt = k 1 2P ) dp 2 L dt. (b) From part (a), d 2 P/dt 2 = 0 when 2P/L = 1. 4

5 4.40 Since the concentration of the triamterene with the bemetizide reaches zero faster than bemetizide along, it is wise to use it in the cases when the drug should be eliminated from the body in shorter period of time. Or it would be better for somebody who can t handle strong concentration Global minimum = 3 at z = 1/2. No global maximum Global minimum =0 is attained only when t. Global maximum at t = Global maximum = 1/2 at x = 1. No global minimum The only one point of y-intercept is y = 0 when x = 0. The points of x-intercept: x 0 = 0, x 1 = π , x 2 = 2π Critical points: x = 0, x = π/ , x = 3π/ Inflection points: x = , x = , x = are solutions of the equation cos(x 2 ) 2x 2 sin(x 2 ) = 0 because f (x) = 2x cos(x 2 ) and f (x) = 2 cos(x 2 ) 4x 2 sin(x 2 ) < a < < a < (a) The slope is the ratio of rise over run. (b) 2 hours. (c) Equal (a) The straight line connecting the point P (starting point) with the point on the curve. (b) Approximately 7 worms because this is the point where the tangent line touches the curve and goes through the starting point, P. (c) The optimal load increases as the traveling time increases (a) The capacity is 200, just before 8:30. (b) (c) About 7 : 30. (d) t 5 : 00 5 : 30 6 : 00 6 : 30 7 : 00 7 : 30 8 : 00 8 : 30 C At x = 1/3e The slope is rise/run or y(x)/x = x e 3x. Denoting this function by m(x) = x e 3x, we need to find its maximum, which is attained at the points where m (x) = e 3x (1 3x) = 0. Therefore, the maximum slop of such line is at the point x = 1/3 and its value is m(1/3) = 1/3e (a) 1/(2e). (b) (ln 2)

6 4.52 (a) At θ = arccos(1/ 3) = radians or (b) It is almost optimal (a) Vertical intercept: W = A e rb. (b) No critical points, inflection point at t = b/c, W = A e 1. (c) For bigger c, the graph reaches its maximum faster. (d) Yes. 6