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Thus . Or .12. Same procedure as in the question no. 10.20. (a) Given. So . Thus multiplying the two derivatives we get.(b) Similar as part (a).24. (a) Given. Differentiating on both sides with respect to x, we get, which implies. So at given point (1, 2), the slope of the tangent to the curve is given by.(b) Given. Differentiating on both sides with respect to x, we get, which implies. So at given point (0, 3), the slope of tangent to the curve is given by.50. Take the road towards east as x-axis and that towards north as y-axis.
So their intersection is the origin. Set the clock at 0 at noon. After t hours the truck will be at point A (0, 70t) and the car will be at B (105(t-1), 0). So the distance S = AB is given by, which is equal to or simply. Thus . So required rate of change of distance at t = 2 is.52. The quantity produced by the worker is given by.(a) So R(t) = Q'(t) =(b) Setting clock at 0 at 8:00 am., the time at 9:00 am is represented by t = 1. So R(1) =. (c) The time 9:06 am is represented by t = 1.1. So Q(1.1) - Q(1)R(1).(1.1 - 1) = (27)(0.1) = 2.7.(d) Q(1.1) - Q(1) = Q(1.1)- 27 can be calculated by actually substituting t = 1.
1 in the expression for Q(t).56. Given no. of units manufactured by the relation. The cost is given by. Now . So at t = 2, it is equal to 1663.2.58. Letbe the error in the measurement of radius r. And be the error in the measurement of surface area. Here. So . That is, % error in S = 2(% error in r). As % error in S is no greater than 8%. So the largest % error in r is 4%.Chapter 318. Given. For critical values we solve or. This gives. Also, which is +ve for and is -ve for . So at there is a relative minimum and at there is a relative maximum.22. Given. So . This gives the critical.
Equating 6t and 90 + 0.5t, we can get t. Set the clock at 0 at noon. After t hours the truck will be at point A (0, 70t) and the car will be at B (105(t-1), 0). So the distance S = AB is given by, which is equal to or simply. Thus . So required rate of change of distance at t = 2 is. 58. Letbe the error in the measurement of radius r. And be the error in the measurement of surface area. Here. So . That is, % error in S = 2(% error in r). As % error in S is no greater than 8%. So the largest % error in r is 4%. (a) The fence is on all the four sides of the pasture.
So 2(x + y)= 320. Thus y = 160 - x. Therefore the area A = xy = x(160 - x) =. . On equating it with 0, we get x = 80. So y = 80. These dimensions give maximum area. 28. The traffic speed is given by. Where t represents the time in hours measured 0 at noon. So . The critical values are given by. Or . Thus t = 1 and t = 5. Also . It is -ve for t = 1 and is +ve for t = 5. Therefore, the traffic speed is fastest at 1 hour past noon that is at 1:00 pm. It is slowest at 5 hours past noon that is at 5:00 pm. 30. Let the breadth = x in.
So the length = 2x in. If height = y in, then the . Thus . Now the area of the four walls and the bottom equals. The area of the top is. Therefore cost of the material used is. Its derivative can be equated with zero to get the value of x. 40. Let the coffee break occurs after x hours.
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