Math Assignment Help|Math Assignment Help Service Tutorhelpdesk David Luke Contact Us: Phone: (617) 807 0926 Web: http://www.tutorhelpdesk.com Email: - info@tutorhelpdesk.com Facebook: https://www.facebook.com/Tutorhelpdesk Twitter: http://twitter.com/tutorhelpdesk Blog: http://tutorhelpdesk.blogspot.com/ Tutorhelpdesk Copyright 2010-2015 Tutorhelpdesk.com Tutorhelpdesk Copyright 2010-2015 Tutorhelpdesk.com AboutMathAssignment:Mathematicsormathsasitis popularlycalledisagameof numbers. Thesenumberswork under a set of rules that set the rule for the number game i.e mathematics.Mathsisbyfartheeasiestsubjectinthe curriculum. Thestudentneedstoknowtherulesofthe number game and he is a winner if he plays it well. Due to its logicalstructuremathshasbegantobecountedasthebasis formanysubjects. Naturalsciences(physics, chemistryetc), engineering,medicine,financeandevensocialsciencelike economicsarebasedverystronglyonmathematics.Many students are heard moaning about mathematics being scary or is it just a myth.Well there canbenogreatmyththanthefactthatmathematicsistough.Itisindeedanapathyto termmathematicsaFrankenstein. Successin mathematics iseasyandpavesthewayfor successinothersubjects. Befriendnumbersandtheywillbefriendyouback.WeProvide math homework,mathassignmenthelpwhich helpyouto solveproblemsabout math and scoring good marks in exam. Sample Math Assignment Help Service Questions: Depreciation Sample Questions Question-1:Find the maximum and minimum values of 3 - 22 + x+ 6 Solution: Determination of the maximum and minimum values of the function Let y = 3 - 22 + x+ 6.

= 32 - 4x + 1 = (3x -1) (x -1) The function will have a turning value if

= 0. This means, (3x -1) (x-1) = 0. =x =

or x =1 Now

=

(32 - 4x + 4) = 6x -4 (a) When x =

,

=6

-4 = -2 < 0 i.e., - ve. Hence, the given function is at maximum when x = 13 and the maximum value is given by max.y =

2 - 2

2 +6 =

-

+

+ 6 = ++

=

Tutorhelpdesk Copyright 2010-2015 Tutorhelpdesk.com Tutorhelpdesk Copyright 2010-2015 Tutorhelpdesk.com (b) When x =1,

2

2 = 6(1) 4 = 2>0 i.e. +ve. Hence, the given function is at minimum when x =1 and the minimum value is given by Min.Y = 13 -2 (2)2 + 1+ 6 = 1 -2 + 1 + 6 =6. Question-2: Determine the minima and maxima of the following function. Y =

+

- 6x + 8. Solution: Given, Y =

+

- 6x + 8.

=

+

- 6

(x) +

(8) = 2x2 +x -6 + 0 = 2x2 + x -6 = (x+2) (2x -3) By the necessary condition,

= 0. =(x+2) (2x-3) = 0 = x = -2, or x = 32 Tutorhelpdesk Copyright 2010-2015 Tutorhelpdesk.com Now,

=

(22 + x -6) = 4x + 1 (i)When x = -2,

= 4(-2) +1 = -7 < 1 = -ve. Hence, the given function is at maximum when x = -2, and the maximum value of the function is given by, Max. y =

23 +

22 -6 (-2) + 8 =

+ 2 + 12 + 8 = 22 - 163 = 503 = 16 23 (ii)When x =

,

= 4

+ 1 = 7 > 0 = + ve Hence, the given function is minimum when x = 32, and the minimum value of the function is given by Min,Y = 32

3 +

2 -6

+ 8 =

+

-9 + 8 = 94 +

-1 = +

=

= 2

Tutorhelpdesk Copyright 2010-2015 Tutorhelpdesk.com Question-3: The cost and revenue functions of firm are set as under C = 100 + 0.0152 ; R = 3x Where, C = total cost, R = total revenue and x = number of units produced. Find the production rate, x that will maximize the profit of the firm, and ascertain such maximum profit, Find also, the profit when x = 120. Solution: We know that profit = Revenue cost. = P = R-C = P = (3x) (100 + 0.015x2) = P = 3x 100 0.015x2 Thus, the profit function or p = 3x 100 0.015x2 Differentiating P w. r. t x we get,

=

(3x 100 0.0152) = 3-0.0105 (2x) = 3 0.030 x According to the necessary condition

= 0 = (3 0.030x) = 0 =x = 100 Now,

=

(3 0.030 x) = -0.030 < 0, = -ve This shows that the profit function is at the maximum when x = 100 (a) Thus, the maximum profit is given by, Max, P = 3(100) 100 0.015 (100)2 = 300 100 0.015. (10,000) = 300 100 150 = 50 or $ 50. (b) Profit when x = 120 Putting 120 in place of x in the profit function we get, P = 3(120) 100 0.015 1202 = 360 100 0.015 (14400) = 260 216 = 44 or $ 44. Tutorhelpdesk Copyright 2010-2015 Tutorhelpdesk.com Note. From the above two results it must be seen that under no level of sale other than its optimum level i.e., when x = 100, the profit will be greater than its optimum profit i.e., $ 50. Question-4:The total cost function for the inventory control of a firm is given by T = 10,000 + ,,

+

, where, T = total cost, Q = quantity ordered each time, 10,000 = the purchase price, ,,

= the ordering cost, and

16 = the carrying quantity and (ii) the cost corresponding to it. (iii) Also, find the total cost of the inventory, when each order is placed for 2,500 units. Solution: Given the cost function, T = 10,000 + ,,

+

16 Differentiating T, w.r.t. Q we have

=

(10,000 + 2,52,000 1 + 116 Q) Tutorhelpdesk Copyright 2010-2015 Tutorhelpdesk.com

= -1(2,50,000) 2 + 116 = ,,

+ 116 The function will have a turning value where

= 0 This implies that ,,

+ 116 = 0 = ,,

=- 116 =

=

,, = 2 = 16 2,50,000 = Q =

40,00,000 = 2,000 Q, being the quantity purchased can never be negative. Hence, Q = 2,000 units. Now

=

=

(-2,50,000 2 + 116) = (-2) (-2,50,000) 3 = 50,00,000

3 When, Q = 2,000,

= ,,,

=

, > 0 = +ve (i)EQQ, From the above it comes out that the given cost function attains the minimax, when Q = 2,000. This means that the economic ordering quantity, or EOQ = 2,000 units. (ii) Cost corresponding to EOQ. The minimum cost, or the cost corresponding to the ECQ is obtained by substituting 2,000 in place of Q in the given function Thus,Min. T = 10,000 + ,,, + ,

= 10,000 + 100 + 156.25 = $10,256.25 Tutorhelpdesk Copyright 2010-2015 Tutorhelpdesk.com Question-5: The demand function of a particular commodity is Y = / for 0 < x 8, where y is the price per unit and x is the number of units demanded. Determine the price, and the quantity for which revenue is maximum. Solution: Here, Revenue = Number of units demanded price. Where, x represents the units, and y the price Thus,R = xy GivenY =/ Thus, the revenue function or R = x. 15/3 Now, differentiating R (the revenue) w. r. t. x, (the units demanded) we have,

=

(/) = /

(x) + x

/ Tutorhelpdesk Copyright 2010-2015 Tutorhelpdesk.com = / (1) + x. 15

/ = / + /

= / + /

(x) = / + /

(1) = / + /

= / + / x = / (3 x) According to the necessary condition, the function will have a turning value when

= 0 =/ (3-x) = 0 =x = , orx =3 x = being absurd, x = 3 According to the sufficient condition,

=

=

(/ + / x) On putting x = 3 in the

, the value appears to be ve i.e. < 0 Hence, the revenue function is maximum, when x = 3 and the maximum revenue given by Max.R = . 15/3

= (153/3 = 451 = 45

= 452.72 = 16.54 (app.). Note 1. The reader is requested to differentiate

(/ + / x), and find its value after putting x = 3 therein, and see that the result is < 0 i.e., - ve. 2. e is an irrational number whose value lies between 2.7 and 2.8 i.e. 2.72 approx.