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Find the second derivative now.

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

You have to find 2nd derivative for maximum value

\[\LARGE y=50x-10\]
diff this.

You should get y=50.

That is the maximum value

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?

I think its not an equal sign right there,its > or <
I'll let you know.

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

yes :)
I wrote,
second derivative should be 0

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

should work

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 ?

isn't it x=50?

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

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

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

@mathslover
Maximum value of the function is
x>50

Oh k. Got it now.

Now it can be 51..51000..51000000000000000..anything

I told you i was confused,there isn't an equality sign,but a greater to one,just confirmed.

and minimum value
x<1/5

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

And since a is positive (in ax^2+bx+c) the function is concave up, so it's a minimum.

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

For maxima,
y=\[\LARGE \frac{d^2(25x^2+5−10x)}{dx^2}<0\]

just applying second derivative test here ..

what c value did you get by the way ?

oh 1/5 :)

congratulate me too???

for ?

nothing XD

:")

Thanks a lot everyone for your help