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mathslover
 one year ago
How to find the maximum and minimum value of the function :
\(y = 25x^2 + 5  10x\)
mathslover
 one year ago
How to find the maximum and minimum value of the function : \(y = 25x^2 + 5  10x\)

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mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Well I took : \(\cfrac{dy}{dt} = 0\) And then: \(\cfrac{d(25x^2+510x)}{dt}= 0\) \(\implies 50x  10 = 0\) \(\implies 50x = 10\) \(\implies x = \cfrac{10}{50} = \cfrac{\cancel{10}^1}{\cancel{50}^5} = \cfrac{1}{5} \)

DLS
 one year ago
Best ResponseYou've already chosen the best response.2Find the second derivative now.

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1therefore y has minimum value at x = 1/5 What about maximum.

DLS
 one year ago
Best ResponseYou've already chosen the best response.2You have to find 2nd derivative for maximum value

DLS
 one year ago
Best ResponseYou've already chosen the best response.2\[\LARGE y=50x10\] diff this.

DLS
 one year ago
Best ResponseYou've already chosen the best response.2As far as I remember,you can always check it though.

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1\(\cfrac{d(50x10)}{dt} \implies y = 50\)

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Thanks @DLS got it. Is their any condition for negative or positive?

DLS
 one year ago
Best ResponseYou've already chosen the best response.2I think its not an equal sign right there,its > or < I'll let you know.

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1OH : \(\cfrac{d^2 y}{dt^2} = 50\)

DLS
 one year ago
Best ResponseYou've already chosen the best response.2yes :) I wrote, second derivative should be 0

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Should I substitute x = 1/5 in the equation? to find \(y_{min.}\)

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1\(y_{min.} = 25(\cfrac{1}{5})^2 + 5  10(\cfrac{1}{5})\) \(y_{min.} =\cancel{25}^1(\cfrac{1}{\cancel{25}^1}) + 5  \cancel{10}^2 (\cfrac{1}{\cancel{5}^1}\) \(y_{min.} = 62 = 4\)

DLS
 one year ago
Best ResponseYou've already chosen the best response.2Seems correct,do you have the answer?

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Hmn ... let me check the answers.

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Only minimum value = 4 is given ... what would be maximum value?

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Max. value is not given , might be author forgot that.

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Though , can you help me finding max. value ?

Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.1y = 25x^2 +510x y' = 50x  10 y'' = 50 Since y'' > 0 The FUction has minimum value

Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.1Minimum value is obstained bY: y' = 0

Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.1y' = 50x  10 = 0 x = 1/5 So...put x=1/5 in the equation....

Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.1This Function has Only Minimum Value....

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2You can do this question without any calculus at all :P

Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.1Yup..@agent0smith By using Some Rules in Quadratic Equation

DLS
 one year ago
Best ResponseYou've already chosen the best response.2i was thinking the same..maximum vallue can be anything actually..

Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.1Function Attains Maximum or Minimum at x = b/2a

DLS
 one year ago
Best ResponseYou've already chosen the best response.2@mathslover Maximum value of the function is x>50

DLS
 one year ago
Best ResponseYou've already chosen the best response.2Now it can be 51..51000..51000000000000000..anything

DLS
 one year ago
Best ResponseYou've already chosen the best response.2I told you i was confused,there isn't an equality sign,but a greater to one,just confirmed.

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2Minimum value of the function with be when x=1/5, not when x is < 1/5

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2As @Yahoo! said, Function Attains Maximum or Minimum at x = b/2a this is x = (10)/(2*25) = 1/5, no calculus needed :D

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2And since a is positive (in ax^2+bx+c) the function is concave up, so it's a minimum.

DLS
 one year ago
Best ResponseYou've already chosen the best response.2\[\LARGE 6x^5+5x^4+4x^3+3x^2+2x+1\] @Yahoo! You got a formula for minimizing this?

Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.1Here u have to use Calculus Since it is not a Quadratic Equation...

Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.1find 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

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2@DLS there aren't general formulas for solving quintics or polynomials with order higher than that, this is the AbelRuffini theorem: http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Abel%E2%80%93Ruffini_theorem.html There are formulas for quadratics (obviously), cubics and quartics.

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2If 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

DLS
 one year ago
Best ResponseYou've already chosen the best response.2For maxima, y=\[\LARGE \frac{d^2(25x^2+5−10x)}{dx^2}<0\]

waterineyes
 one year ago
Best ResponseYou've already chosen the best response.0You are taking derivative with respect to x and I respect that too, but @mathslover I think you are writing dt instead of dx.. Right??

AravindG
 one year ago
Best ResponseYou've already chosen the best response.0just applying second derivative test here ..

AravindG
 one year ago
Best ResponseYou've already chosen the best response.0what c value did you get by the way ?

waterineyes
 one year ago
Best ResponseYou've already chosen the best response.0Hey @yahoo, Congratulations for Green.. @AravindG, Congratulations for Ambassador.. And @mathslover, Congratulations for 98 + Green..

waterineyes
 one year ago
Best ResponseYou've already chosen the best response.0@DLS, Congratulations for getting 1 medal in this post and now 2... :)

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Write \(y(x)=25x^210x+5\) as \((5x1)^2+4\) and use that this is the sum of two nonnegative addends. So it has minimum, when \(5x1=0\quad\Rightarrow\quad x=\frac15 \). To prove that there is no maximum try to compare \(y(x) \) and \(y(x+1)\) by subtracting: \(y(x+1)y(x)=(25x^2+50x+2510x10+5)(25x^210x+5)=50x+15\). So for large \(x\) this is not bounded. So it has no maximum.

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Thanks a lot everyone for your help

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Oh thanks @waterineyes , and sorry for using dt there... I meant dx. Actually I usually come up with problems having dt there na.
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