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

Mathematics
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Well I took : \(\cfrac{dy}{dt} = 0\) And then: \(\cfrac{d(25x^2+5-10x)}{dt}= 0\) \(\implies 50x - 10 = 0\) \(\implies 50x = 10\) \(\implies x = \cfrac{10}{50} = \cfrac{\cancel{10}^1}{\cancel{50}^5} = \cfrac{1}{5} \)
  • DLS
Find the second derivative now.
therefore y has minimum value at x = 1/5 What about maximum.

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Other answers:

  • DLS
You have to find 2nd derivative for maximum value
  • DLS
\[\LARGE y=50x-10\] diff this.
  • DLS
You should get y=50.
  • DLS
That is the maximum value
  • DLS
As far as I remember,you can always check it though.
\(\cfrac{d(50x-10)}{dt} \implies y = 50\)
Thanks @DLS got it. Is their any condition for negative or positive?
  • DLS
I think its not an equal sign right there,its > or < I'll let you know.
OH : \(\cfrac{d^2 y}{dt^2} = 50\)
  • DLS
yes :) I wrote, second derivative should be 0
Should I substitute x = 1/5 in the equation? to find \(y_{min.}\)
  • DLS
should work
\(y_{min.} = 25(\cfrac{1}{5})^2 + 5 - 10(\cfrac{1}{5})\) \(y_{min.} =\cancel{25}^1(\cfrac{1}{\cancel{25}^1}) + 5 - \cancel{10}^2 (\cfrac{1}{\cancel{5}^1}\) \(y_{min.} = 6-2 = 4\)
  • DLS
Seems correct,do you have the answer?
Hmn ... let me check the answers.
Only minimum value = 4 is given ... what would be maximum value?
Max. value is not given , might be author forgot that.
Though , can you help me finding max. value ?
  • DLS
isn't it x=50?
  • DLS
y = 25x^2 +5-10x y' = 50x - 10 y'' = 50 Since y'' > 0 The FUction has minimum value
Minimum value is obstained bY: y' = 0
y' = 50x - 10 = 0 x = 1/5 So...put x=1/5 in the equation....
This Function has Only Minimum Value....
You can do this question without any calculus at all :P
Maximum is infinite
Yup..@agent0smith By using Some Rules in Quadratic Equation
  • DLS
i was thinking the same..maximum vallue can be anything actually..
Function Attains Maximum or Minimum at x = -b/2a
  • DLS
@mathslover Maximum value of the function is x>50
Oh k. Got it now.
  • DLS
Now it can be 51..51000..51000000000000000..anything
  • DLS
I told you i was confused,there isn't an equality sign,but a greater to one,just confirmed.
  • DLS
and minimum value x<1/5
Minimum value of the function with be when x=1/5, not when x is < 1/5
As @Yahoo! said, Function Attains Maximum or Minimum at x = -b/2a this is x = -(-10)/(2*25) = 1/5, no calculus needed :D
And since a is positive (in ax^2+bx+c) the function is concave up, so it's a minimum.
  • DLS
\[\LARGE 6x^5+5x^4+4x^3+3x^2+2x+1\] @Yahoo! You got a formula for minimizing this?
Here u have to use Calculus Since it is not a Quadratic Equation...
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
@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.
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
  • DLS
For maxima, y=\[\LARGE \frac{d^2(25x^2+5−10x)}{dx^2}<0\]
  • DLS
yep @DLS
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??
just applying second derivative test here ..
what c value did you get by the way ?
oh 1/5 :)
Hey @yahoo, Congratulations for Green.. @AravindG, Congratulations for Ambassador.. And @mathslover, Congratulations for 98 + Green..
  • DLS
congratulate me too???
for ?
  • DLS
nothing XD
@DLS, Congratulations for getting 1 medal in this post and now 2... :)
  • DLS
:")
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.
Thanks a lot everyone for your help
Oh 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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