This document provides solutions for various statistics problems for economists. It includes topics such as differentiation, integration, cost and profit maximization, and solving linear equations. The solutions are explained step by step.
Contribute Materials
Your contribution can guide someone’s learning journey. Share your
documents today.
Statistics for Economists
Secure Best Marks with AI Grader
Need help grading? Try our AI Grader for instant feedback on your assignments.
(c) (i)∫(5x – 3x2) dx = 5x2– 3x3 23 =5x2– x3 2 2 (ii)∫1(x2– 2x) dx x3– x22 31 8–1- 1 + 22 33 7– 3 = -2/3 3 (iii)∫4x exp.(x2) .dx Put exp.(x2) = u 2x. exp(x2) dx = du Then,∫2.u. du = u2 = [exp.(x2)]2 3
Ques 2. Solution Given, f(x) = x3– 3x2– x + 3 (a) If given function can divide with x – 3 then, f(3) will equal to zero f(3) = (3)3– 3(3)2– 3 + 3 = 27 – 27 – 3 + 3 = 0 Hence, f(x) is divisible by x-3 (b) x3– 3x2– x + 3 (x – 3) (x2– 1) (x + 1) (x – 3) (x – 1) So, at x = 1, -1 and 3 given f(x) = 0 while, -1 ≤ x ≤ 3 (c) f(x) = x3– 3x2– x + 3 f’(x) = 3x2– 6x – 1 f’(2) = 3(2)2– 6 (2) – 1 = 12 – 12 – 1 = -1 So, function is decreasing at x = 2 4
Ques. 4 Solution: Given, Total cost of producing x units of an item as – K (x) = -0.25x2+ 75x + 1000 where,0 ≤ x ≤ 100 (a) Expression of boundary cost - -0.25x2+ 75x + 1000 = 0 (b)How many units are produced when the limit cost is NOK 60? When limit cost is 60 then, K(60) = -0.25(60)2+ 75(60) + 1000 = -0.25 x 3600 + 75 x 60 + 1000 = -900 + 4500 + 1000 = 4600 units (c) Expression of average cost – Average cost =Total production cost Quantity of unit produced =-0.25x2+ 75x + 1000 x = -0.25x + 75 + 1000/x (d)What is the average profit per unit sold when producing 50 units and each unit in this case sold for 100 kroner? Average profit per unit at x = 50 Average cost = -0.25(50) + 75 + 1000/50 = 107.5 kroner per unit Then, average profit per unit will be 7.5 kroner. 5
Secure Best Marks with AI Grader
Need help grading? Try our AI Grader for instant feedback on your assignments.
Ques. 5 Solution – P = 210 – 3x – y Q = 180 – x – 2y Total cost of production c (x.y) = 1200 + 100x + 80y Total profit = Total selling cost – Total production cost f(x,y) = Px + Qy – c(x.y) = x (210 – 3x – y) + y (180 – x – 2y) – (1200 + 100x + 80y) = 210 x – 3x2– xy + 180y – xy – 2y2– 1200 – 100x – 80y = -3x2– 2xy – 2y2+ 110x +100y – 1200 (b) For maximise the profit – f(x) = -3x2– 2xy – 2y2+ 110x +100y – 1200 By differentiating with respect to x and y individually f(x) = -6x -2y + 110 f(y) = -2x – 4y + 100 y = 24 and x = 3 (approx.) (c) Proof – Use D-test for (3,24) to check maximum value of given function D = fxx(3,24) fyy(3,24) – [fxy(3,24)]2 = (-6) (-4) – 1200 = 24 – 1200 = -1176 so, (3,24) gives maximum value of given function 6
Ques 7 Given x + y – 2z = 9 2x – y – z =6 -x + 5y + z = 5 11-2x9 2-1-1y=6 -151z5 solving using Guass elimination method – x = 49/15 y = 29/15 z = 7/5 7