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

- mathslover

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

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- mathslover

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.

- mathslover

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

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## More 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.

- mathslover

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

- mathslover

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.

- mathslover

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

- DLS

yes :)
I wrote,
second derivative should be 0

- mathslover

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

- DLS

should work

- mathslover

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

- DLS

Seems correct,do you have the answer?

- mathslover

Hmn ... let me check the answers.

- mathslover

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

- mathslover

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

- mathslover

Though , can you help me finding max. value ?

- DLS

isn't it x=50?

- DLS

@agent0smith

- anonymous

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

- anonymous

Minimum value is obstained bY:
y' = 0

- anonymous

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

- anonymous

This Function has Only Minimum Value....

- agent0smith

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

- anonymous

Maximum is infinite

- anonymous

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

- DLS

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

- anonymous

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

- DLS

@mathslover
Maximum value of the function is
x>50

- mathslover

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

- agent0smith

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

- agent0smith

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

- agent0smith

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?

- anonymous

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

- anonymous

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

- agent0smith

@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.

- agent0smith

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

@RnR @shubhamsrg

- AravindG

yep @DLS

- anonymous

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??

- AravindG

just applying second derivative test here ..

- AravindG

what c value did you get by the way ?

- AravindG

oh 1/5 :)

- anonymous

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

- DLS

congratulate me too???

- AravindG

for ?

- DLS

nothing XD

- anonymous

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

- DLS

:")

- klimenkov

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.

- mathslover

Thanks a lot everyone for your help

- mathslover

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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