Presentation on Maxima and Minima (2nd Derivative Test)

A Presentation on Maxima and Minima

1. Maxima and Minima (2 nd derivative test) Prepared by Group 4 Afsana Parvin Mim (24230225008) Suvojit Kumar Debnath (24230225021) Md. Zahid Hasan (24230225022) Shariful Mamun (24230225023) Md. Nazmus Sakib (24230225028) Prepared for Md. Ashraful Alam Associate Professor Bangladesh University of Professionals

2. INTRODUCTION

3. INTRODUCTION Derivative is the Rate of change of a function Maxima and minima are largest and smallest values taken by a function At Stationary point function’s first derivative is zero At Inflection point second derivative is zero and third derivative is non zero

4. Problem 1 Determine all local optimum points and all inflection points of y= by the second derivative test. Solution: Given, f(x)= - .˙. f’(x)= -3 -24x-45 .˙. f’’(x)= -6x-24 .˙. f’’’(x)= -6  

5. Problem 1 To get the stationary points, Let, f’(x)= 0 ⇒ -3x 2 -24x-45= 0 ⇒ x 2 +8x+15=0 ⇒ x 2 +5x+3x+15= 0 ⇒ (x+3)(x+5)= 0 .˙. x= -3, -5

6. Problem 1 For x= -3, f’’(x)= f’’(-3) = -6.(-3)-24= -6<0, which is negative number. So, f(x) has a maximum value at x = -3, and the maximum value is- f max = (-3) 3 -12(-3) 2- 45.(-3)+2= 2

7. Problem 1 For x= -5, f’’(x)= f’’(-5) = -6.(-5)-24= 6 > 0, which is positive number. So, f(x) has a minimum value at x = -5, and the minimum value is- f min = (-5) 3 -12(-5) 2- 45.(-5)+2= -198

8. Problem 1 Inflection points: Let, f”(x)= 0 ⇒ -6x-24= 0 .˙. x= -4 (Here, f’’’(x)= f’’’(-4)= -6≠ 0 )

9. Problem 1 Coordinates of inflection point: f(x)= f(-4)= -(-4) 3 -12(-4) 2 -45.(-4)+2= -74 So, the inflection point is (-4,-74). Answer: Maximum point(-3,2) Minimum Point(-5,-198) Inflection point(-4,-74).

10. Problem 2 2 . Determine all local optimum points and all inflection points of 𝒇(𝒙) = 𝟏𝟓𝒙 𝟒 + 𝟖𝒙 𝟑 − 𝟏𝟖𝒙 𝟐 + 𝟏 by second derivative test. Solution: Given, 𝑓(𝑥) = 15 𝑥 4 + 8 𝑥 3 − 18 𝑥 2 + 1 𝑓’( 𝑥) = 60 𝑥 3 + 24 𝑥 2 − 36𝑥 𝑓’’( 𝑥) = 180 𝑥 2 + 48𝑥 − 36 𝑓’’’( 𝑥) = 360𝑥 + 48

11. Problem 2 To get stationary points, put, 𝑓’( 𝑥) = 0 60 𝑥 3 + 24 𝑥 2 − 36𝑥 = 0 5 𝑥 3 + 2 𝑥 2 − 3𝑥 = 0 𝑥(5 𝑥 2 + 2𝑥 − 3) = 0 𝑥(5 𝑥 2 + 5𝑥 − 3𝑥 − 3) = 0 𝑥{5𝑥(𝑥 + 1) − 3(𝑥 + 1)} = 𝑥 (𝑥 + 1)(5𝑥 − 3) =

12. Problem 2 For 𝒙 = 𝟎 𝑓’’( 𝑥) = 𝑓’’(0 ) = −36 < So, 𝑓(𝑥) has a maximum value at 𝑥 = 0 and the maximum value is 𝑓 max (𝑥 ) = 𝑓(0) = 1 So, maximum point is (0, 1).

13. Problem 2 For 𝒙 = −𝟏 𝑓’’( 𝑥) = 𝑓’’( −1) = 180. (− 1) 2 + 48. (−1) − 36 = 180 − 48 − 36 = 96 > 0. So , 𝑓(𝑥) has a minimum value at 𝑥 = −1, and the minimum value is 𝑓 min ( 𝑥) = 𝑓(−1) = 15. (− 1) 4 + 8. (− 1) 3 − 18. (− 1) 2 + 1 = 15 − 8 − 18 + 1 = −10 So, minimum point is (−1,−10).

14. Problem 2 For 𝒙 = f ’’(x)= f min (x)= - So the minimum point is ( , - )  

15. Problem 2 Absolute minimum point is (-1,-10) Inflection point: Let, 𝑓’’( 𝑥) = 0 180 x 2 + 48𝑥 − 36 = 0 ⇒ 15 x 2 + 4𝑥 − 3 = 0 ⇒ 15 x 2 − 5𝑥 + 9𝑥 − 3 = 0 ⇒ 5 𝑥(3𝑥 − 1) + 3(3𝑥 − 1) = 0 ⇒ (3 𝑥 − 1)(5𝑥 + 3) = 0 .˙. 𝑥 = - ,   Here, 𝑓’’’( - )=− 168 ≠ 0 , and 𝑓’’’( = 168 ≠ 0  

16. Problem 2 Coordinates of inflection points: 𝑓( )= 15 .( ) 4 + 8 . ( ) 3 − 18 . .( ) 2 + 1 = − So, an inflection point is ( , − ) 𝑓( )= 15 .( - ) 4 + 8 .( - ) 3 − 18 . .( - ) 2 + 1 = − So, another inflection point is ( − ) Hence, Maximum point is (0, 1). Minimum points (-1,-10), ( , - ) Inflection points ( , − ) , ( − )  

17. Problem 3 3.Determine all local optimum points and all inflection points of y=x 3 -6x 2 +12x-5 by the second derivative test. Solution: Given, f(x)= x 3 -6x 2 +12x-5 f’(x)= 3x 2 -12x+12 f’’(x)= 6x-12 f’’’(x)= 6

18. Problem 3 To get the stationary points, Let, f’(x)= 0 ⇒ 3x 2 -12x+12= 0 ⇒ x 2 -4x+4=0 ⇒ (x-2) 2 = 0 .˙. x= 2

19. Problem 3 For x= 2, f’’(x)= f’’(2) = 6*2-12= 0 So, at x=2, the test is inconclusive for the given function. The use of other methods, such as the first derivative test or higher-order derivatives are needed to determine the nature of the critical point.

20. Problem 4 4. When x gallons of olive oil are produced, the average cost per barrel is 𝑨 ( 𝒙 ), where 𝑨 ( 𝒙 )= 0.25x, 𝒙 > 𝟎 . (a) Find the value of x that minimizes average cost per barrel. (b) Compute the minimum average cost per barrel.  

21. Problem 4 Solution: (a) Given, 𝑨(𝒙)= + 0.25x, 𝒙 > 𝟎. = 4000(0.1𝑥 + 20) -1 + 0.25𝑥 Now, 𝐴’(𝑥) = 4000 (−1)(0.1𝑥 + 20) -2 (0.1) + 0.25 = - + 0.25  

22. Problem 4 ∴ 𝐴’’(𝑥) = −400 (−2)(0.1𝑥 + 20) -3 (0.1) = 80(0.1𝑥 + 20) -3 For stationary points, Let, 𝐴’(𝑥) = 0 ⇒ - + 0.25 = 0 ⇒ = 0.25 ⇒ x = 200  

23. Problem 4 Now, 𝐴’’(200) = = > 0, which is positive So, the given function has a minimum value at 𝑥 = 200. (Answer)  

24. Problem 4 (b) Since, for 𝑥 = 200, 𝐴(𝑥) has minimum value. [From (a)] So, the minimum average cost per barrel will be, 𝐴 min (200) = + {(0.25)×200} = [4000/40] + 50 = $150 (Answer)  

25. Problem 5 5. A manufacturer sells x units of a product at a dollar price of p = 6565 −10x − 0.1x 2 per unit. The cost of manufacturing the product is C(x)= 0.05x 3 -5x 2 + 20x+ 250000, 0 ≤ x ≤ 150. How many units should be produced and sold to maximize the resulting profit? Solution: Total revenue = (Price per unit) × (Number of units sold)

26. Problem 5 R (x)= (6565−10x − 0.1x 2 ) × x = 6565x −10x 2 − 0.1 x 3 Profit, P(x) = Revenue – Cost =(6565x- 10x 2 -0.1x 3 ) - (0.05x 3 -5x 2 +20x+250000) = -0.15x 3 -5x 2 +6545x-250000 To determine the quantity x that maximizes the profit function, we first find, = -0.45x 2 -10x +6545  

27. Problem 5 Setting this first derivative to be equal to zero, we solve for critical values of x and x = 110 (Other root is negative). Now, = −0.9x-10 = -(0.9)(110)− 10 =− 109 < 0 ; indicating that this critical value does indeed represent a maximum.  

28. Problem 5 P(110) = - 0.15(110) 3 –5(110) 2 + 6545(110) – 250000 = $209800. (Answer)

29. Conclusion A fundamental tool Combined with first derivative test, offers a robust framework Extension to higher derivative test provides additional precision Relevance extends beyond theoretical mathematics

30. Thank You