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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
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

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

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

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

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

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

DLS Group TitleBest ResponseYou've already chosen the best response.2
You should get y=50.
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.2
That is the maximum value
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

DLS Group TitleBest ResponseYou've already chosen the best response.2
isn't it x=50?
 one year ago

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

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

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

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

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

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
Maximum is infinite
 one year ago

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

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

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

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

mathslover Group TitleBest ResponseYou've already chosen the best response.1
Oh k. Got it now.
 one year ago

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

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

DLS Group TitleBest ResponseYou've already chosen the best response.2
and minimum value x<1/5
 one year ago

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

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

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

DLS Group TitleBest 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?
 one year ago

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

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

agent0smith Group TitleBest 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.
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

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

DLS Group TitleBest ResponseYou've already chosen the best response.2
@RnR @shubhamsrg
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
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??
 one year ago

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

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

AravindG Group TitleBest ResponseYou've already chosen the best response.0
oh 1/5 :)
 one year ago

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

DLS Group TitleBest ResponseYou've already chosen the best response.2
congratulate me too???
 one year ago

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

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Write \(y(x)=25x^210x+5\) 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.
 one year ago

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

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