Question

4. (a) A function f has first derivative f'(r) and second derivative It is also known that the function f has r-intercept at (-3,0), and a y-intercept at (0, 0) i) Find all critical points, and use them to identify the intervals over which you will examine the behaviour of the first derivative 3 marks (İİ) Úse the f,(x), and the First Derivative Test to classify each critical point. [3 marks] Iİİ) Úse the second derivative to examine the concavity around critical points that are not stationary points. 3 marks (iv) Now consider the graphs of f'() and f"(x) (You can use Desmos or any other graphing package here). Based on these and your investigation of critical points and concavity sketch a possible graph of the function f() showing all critical values 3 marks (b) An engineering firm has hired a Human Resources consultant to see how they can optimize the hourly productivity of its engineers. It is found through 'experiment' that in a large shared office the productive hours H that n number of engineers work each day, can be modelled by What is the optimal number of engineers the firm should allocate to one of its large offices to ensure maximum hourly productivity? 3 marks

Answer 1

N^2 = 130 n = plusminus squareroot 30 n = plusminus 11.40 n = 11 approximately [since engineers cannot be negative] at point after 11 (suppose B) H�(n) < 0 at point before 11 (suppose A) H�(n) > 0 therefore h = 11 approximately is the optimal engine. \r\n (b) H(n) = 8(n - n^3/n^2 + beta_0) Here n rightarrow number of engines For optional no. of engineers we have to differentiate H(n) with espect to n. H�(n) = 8[1 - (n^2 + 130)3n^2 - n^3 times 2n/(n^2 + 130)^2] and equating H�(n) = 0 8[1 - 3n^4 + 390n^2 - 2n^4/(n^2 + 130)^2] = 0 1 - (3n^4 + 390n^2 - 2n^4)/(n^2 + 130)^2 = 0 1/1 = n^4 + 390n^2/(n^2 + 130)^2 (n^2 + 130)^2 = n^4 + 390n^2 n^4 + 16900 + 260n^2 = n^4 + 390n^2 16900 = 390n^2 - 260n^2 16900 = 130n^2 130 n^2 = 1690 n^2 = 1690/13