1. Maxima and Minima

2. Maximum: Let f(x) be a function with domain DC IR then f(x) is said to attain the maximum value at a point a є D if f(x)<f(a) for all values of x. Also ‘a’ is called the point of maxima and f(a) is called the maximum value or the absolute maximum value of f(x). Minimum: Let f(x) be a function defined on the domain DC IR. Then f(x) is said to have a minimum value at a point a є D if f(x) > f(a) for all x є D.

3. Moreover, ‘a’ is called the point of minimum and f(a) is known as the minimum value or the absolute minimum value of f(x). Local Maxima: A function f(x) is said to attain a local maximum at x=a if there exist a neighborhood (a-δ, a+ δ) of a such that f(x) < f(a) for all xє (a-δ, a+ δ), x≠a. In such a case, f(a) is called the local maximum value of f(x) at x=a.

4. Local Minima: A function f(x) is said to attain a local minima at x=a if there exist a neighborhood (a-δ, a+ δ) of a such that f(x) > f(a) for all xє (a-δ, a+ δ), x ≠ a. In such a case, f(a) is called the local maximum value of f(x) at x=a.

5. Second Derivative Test for Local Maxima and Local Minima Let f(x) be a derivable function on an interval I 1)Find 2)Put and solve the equation for x. Let be the roots of this equation. These are the possible point or local maxima or minima. 3)Find at x =

6. 4) (a) If at some Ci then f(x) has local minima at x = Ci (b) IF at some Ci then f(x) has local maxima a at x = Ci

7. Example: Find all the points of local maxima and minima and corresponding maximum and minimum values of the function (Critical values or stationary values of x) Now

10. Example: The total profit y in rupees of a drug company from the manufacturer and sale of x drug bottles is given by, (a) How many drug bottles must the company sell to achieve the maximum profit? (b) What is the profit per drug bottle when this maximum is achieved?

11. Solution: Given. Therefore The first order condition for maximum value of y is,, i.e., or x =400 Since, therefore the company must sell x=400 drug bottles to achieve the maximum profit and it is equal to

12. Example: The efficiency E of a small manufacturing concern depends on the workers W and is given by. Find the strength of the workers which give maximum efficiency. Solution: Given, or. Therefore, The 1 st order condition for maximum value of E is i.e. Also

14. Maximization of total revenue function Step 1. Let R be the revenue function then put =0 and solve for x. Step 2. Find at the value of x obtained in step 1. Step 3. If >0 then revenue is minimum and if. < 0 then revenue is maximum at that value of x.

15. Example Given below is the revenue function where R is revenue and x is the quantity sold. Find the revenue maximizing level of output and the total revenue at this level of output. Solution :

17. Maximization of Profit ∏(x) = R(x) – C(x) where, ∏(x) =profit function R(x) = Revenue function C(x) = Cost function To maximize the profit, follow the steps given below 1. Put

19. Example A manufacturer can sell x items at a price of Rs (300-x) each. The cost of producing x items is Rs.. How many items should the manufacturer sell to make maximum profits. Also determine total profit.

20. Solution:

22. Homogenous function Let f(x,y) = ----(1). Here degree of each term in (1) is n, So f(x,y) is a homogeneous function Euler’s theorem on Homogenous function If Z=f(x,y) is a homogeneous function of degree n and possess continuous partial derivatives then

23. Euler’s theorem on Homogenous function… (Corollary) If Z=f(x,y) is a homogeneous function of degree n and possess continuous partial derivatives then

24. Example Solution

25. Maxima and Minima for a function of two variables Let f(x,y) be a function of 2 variables x and y, let (a, b) be any point of domain f(x,y). Then a point (a,b) is said to be critical of f if f x (a,b)=0=f y (a,b). And if (a,b) is the critical point of f(x,y) then f(a,b) is called the stationary value of f. Necessary and Sufficient conditions for Finding Maxima and Minima for a function of Two varaibles.

26. Maxima and Minima for a function of two variables …. Necessary and Sufficient conditions for Finding Maxima and Minima for a function of Two variables. Necessary condition f x (a,b)=0, f y (a,b)=0 Sufficient Condition Let A = f x,x (a,b), B = f x,y (a,b), C = f y,y (a,b) Calculate then

27. Maxima and Minima for a function of two variables …. 1.IF >0 and A < 0 then f(a,b) is the maximum value. 2.IF >0 and A > 0 then f(a,b) is the minimum value. 3.IF <0 then neither maximum nor minimum. 4.IF =0 then test fails and needs further investigation.

31. Stationary points Maximum point point of inflection Minimum point

32. Stationary points At a maximum point + - 0 = 0 is negative + - At a minimum point 0 = 0 is positive

33. Stationary points At some stationary points + - - + These are: points of inflection 0 = 0 and 0 Note is also 0 at some maximum and minimum points.

34. To sketch y = 5 + 4x - x 2 = 4- 2x- 2x At stationary points = 0 2x = 4 The stationary point is a maximum y = 5 + 4  2 - 2 2 = 9 y = 5 + 4x - x 2 x = 2 = - 2 There is a maximum point at (2, 9) When x = 0, y = 5 When y = 0, 5 + 4x - x 2 = 0 5 1 ( - x)( + x) = 0 x = 5 or x = -1 x Y 0 5 5 2 9

35. To sketch y = 2 + 3x 2 - x 3 = 6x- 3x2- 3x2 y x 0 y = 2 + 3x 2 - x 3 At stationary points this is 0 6x - 3x 2 = 0 3x3x = 0 (2 - x) x = 0 or x = 2 and y = 2 and y = 2 + 12 - 8 = 6 Minimum point (0, 2) Maximum point (2, 6) When x = 0, y = 2 When y = 0, 2 + 3x 2 - x 3 = 0 = 6- 6x- 6x When x = 0 this is positive When x = 2, is negative maximum (2,6) minimum (0, 2)

36. To sketch y = x 4 - 4 = 4x 3 At stationary points = 0 4x 3 = 0 Gradient before x = 0 is y = - 4 negative y = x 4 - 4 x = 0 = 12x 2 There is a minimum point at (0, - 4) When x = 0, y = - 4 When y = 0, x 4 - 4 = 0 x 2 = 2 x 4 = 4 x = ±   2 x Y 0 = 0 Gradient after x = 0 is positive 22 -4 -  2