## mathslover Group Title How to find the maximum and minimum value of the function : $$y = 25x^2 + 5 - 10x$$ one year ago one year ago

1. mathslover Group Title

Well I took : $$\cfrac{dy}{dt} = 0$$ And then: $$\cfrac{d(25x^2+5-10x)}{dt}= 0$$ $$\implies 50x - 10 = 0$$ $$\implies 50x = 10$$ $$\implies x = \cfrac{10}{50} = \cfrac{\cancel{10}^1}{\cancel{50}^5} = \cfrac{1}{5}$$

2. DLS Group Title

Find the second derivative now.

3. mathslover Group Title

therefore y has minimum value at x = 1/5 What about maximum.

4. DLS Group Title

You have to find 2nd derivative for maximum value

5. DLS Group Title

$\LARGE y=50x-10$ diff this.

6. DLS Group Title

You should get y=50.

7. DLS Group Title

That is the maximum value

8. DLS Group Title

As far as I remember,you can always check it though.

9. mathslover Group Title

$$\cfrac{d(50x-10)}{dt} \implies y = 50$$

10. mathslover Group Title

Thanks @DLS got it. Is their any condition for negative or positive?

11. DLS Group Title

I think its not an equal sign right there,its > or < I'll let you know.

12. mathslover Group Title

OH : $$\cfrac{d^2 y}{dt^2} = 50$$

13. DLS Group Title

yes :) I wrote, second derivative should be 0

14. mathslover Group Title

Should I substitute x = 1/5 in the equation? to find $$y_{min.}$$

15. DLS Group Title

should work

16. mathslover Group Title

$$y_{min.} = 25(\cfrac{1}{5})^2 + 5 - 10(\cfrac{1}{5})$$ $$y_{min.} =\cancel{25}^1(\cfrac{1}{\cancel{25}^1}) + 5 - \cancel{10}^2 (\cfrac{1}{\cancel{5}^1}$$ $$y_{min.} = 6-2 = 4$$

17. DLS Group Title

Seems correct,do you have the answer?

18. mathslover Group Title

Hmn ... let me check the answers.

19. mathslover Group Title

Only minimum value = 4 is given ... what would be maximum value?

20. mathslover Group Title

Max. value is not given , might be author forgot that.

21. mathslover Group Title

Though , can you help me finding max. value ?

22. DLS Group Title

isn't it x=50?

23. DLS Group Title

@agent0smith

24. Yahoo! Group Title

y = 25x^2 +5-10x y' = 50x - 10 y'' = 50 Since y'' > 0 The FUction has minimum value

25. Yahoo! Group Title

Minimum value is obstained bY: y' = 0

26. Yahoo! Group Title

y' = 50x - 10 = 0 x = 1/5 So...put x=1/5 in the equation....

27. Yahoo! Group Title

This Function has Only Minimum Value....

28. agent0smith Group Title

You can do this question without any calculus at all :P

29. Yahoo! Group Title

Maximum is infinite

30. Yahoo! Group Title

Yup..@agent0smith By using Some Rules in Quadratic Equation

31. DLS Group Title

i was thinking the same..maximum vallue can be anything actually..

32. Yahoo! Group Title

Function Attains Maximum or Minimum at x = -b/2a

33. DLS Group Title

@mathslover Maximum value of the function is x>50

34. mathslover Group Title

Oh k. Got it now.

35. DLS Group Title

Now it can be 51..51000..51000000000000000..anything

36. DLS Group Title

I told you i was confused,there isn't an equality sign,but a greater to one,just confirmed.

37. DLS Group Title

and minimum value x<1/5

38. agent0smith Group Title

Minimum value of the function with be when x=1/5, not when x is < 1/5

39. agent0smith Group Title

As @Yahoo! said, Function Attains Maximum or Minimum at x = -b/2a this is x = -(-10)/(2*25) = 1/5, no calculus needed :D

40. agent0smith Group Title

And since a is positive (in ax^2+bx+c) the function is concave up, so it's a minimum.

41. DLS Group Title

$\LARGE 6x^5+5x^4+4x^3+3x^2+2x+1$ @Yahoo! You got a formula for minimizing this?

42. Yahoo! Group Title

Here u have to use Calculus Since it is not a Quadratic Equation...

43. Yahoo! Group Title

find f'(x) first equate it to 0 and find the values of x then find f''(x) by inserting the values of x u got.. if f''(x) > 0 then function has minimum at that point of x and vice versa

44. agent0smith Group Title

@DLS there aren't general formulas for solving quintics or polynomials with order higher than that, this is the Abel-Ruffini theorem: http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Abel%E2%80%93Ruffini_theorem.html There are formulas for quadratics (obviously), cubics and quartics.

45. agent0smith Group Title

If you'd like to solve a quartic of the form $\large ax^4+bx^3+cx^2+dx +e = 0$ then here you go: http://planetmath.org/QuarticFormula Or watch wolfram derive it: http://www.wolframalpha.com/input/?i=a*x%5E4%2Bb*x%5E3%2Bc*x%5E2%2Bd*x+%2Be+%3D+0

46. DLS Group Title

For maxima, y=$\LARGE \frac{d^2(25x^2+5−10x)}{dx^2}<0$

47. DLS Group Title

@RnR @shubhamsrg

48. AravindG Group Title

yep @DLS

49. waterineyes Group Title

You are taking derivative with respect to x and I respect that too, but @mathslover I think you are writing dt instead of dx.. Right??

50. AravindG Group Title

just applying second derivative test here ..

51. AravindG Group Title

what c value did you get by the way ?

52. AravindG Group Title

oh 1/5 :)

53. waterineyes Group Title

Hey @yahoo, Congratulations for Green.. @AravindG, Congratulations for Ambassador.. And @mathslover, Congratulations for 98 + Green..

54. DLS Group Title

congratulate me too???

55. AravindG Group Title

for ?

56. DLS Group Title

nothing XD

57. waterineyes Group Title

@DLS, Congratulations for getting 1 medal in this post and now 2... :)

58. DLS Group Title

:")

59. klimenkov Group Title

Write $$y(x)=25x^2-10x+5$$ as $$(5x-1)^2+4$$ and use that this is the sum of two non-negative addends. So it has minimum, when $$5x-1=0\quad\Rightarrow\quad x=\frac15$$. To prove that there is no maximum try to compare $$y(x)$$ and $$y(x+1)$$ by subtracting: $$y(x+1)-y(x)=(25x^2+50x+25-10x-10+5)-(25x^2-10x+5)=50x+15$$. So for large $$x$$ this is not bounded. So it has no maximum.

60. mathslover Group Title

Thanks a lot everyone for your help

61. mathslover Group Title

Oh thanks @waterineyes , and sorry for using dt there... I meant dx. Actually I usually come up with problems having dt there na.