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

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mathslover
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2Find the second derivative now.

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

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

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

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

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

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

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

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

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

mathslover
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2Seems correct,do you have the answer?

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

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y = 25x^2 +510x y' = 50x  10 y'' = 50 Since y'' > 0 The FUction has minimum value

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Minimum value is obstained bY: y' = 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y' = 50x  10 = 0 x = 1/5 So...put x=1/5 in the equation....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This Function has Only Minimum Value....

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yup..@agent0smith By using Some Rules in Quadratic Equation

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Function Attains Maximum or Minimum at x = b/2a

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

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

DLS
 3 years 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
 3 years 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
 3 years 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
 3 years 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
 3 years 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here u have to use Calculus Since it is not a Quadratic Equation...

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

anonymous
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0just applying second derivative test here ..

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

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

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

klimenkov
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1Thanks a lot everyone for your help

mathslover
 3 years 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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