Question

How to find the maximum and minimum value of the function :

\(y = 25x^2 + 5 - 10x\)

Answer 1

Well I took : \(\cfrac{dy}{dt} = 0\)

And then: \(\cfrac{d(25x^2+5-10x)}{dt}= 0\)
\(\implies 50x - 10 = 0\)
\(\implies 50x = 10\)
\(\implies x = \cfrac{10}{50} = \cfrac{\cancel{10}^1}{\cancel{50}^5} = \cfrac{1}{5} \)

Answer 2

Find the second derivative now.

Answer 3

therefore y has minimum value at x = 1/5

What about maximum.

Answer 4

You have to find 2nd derivative for maximum value

Answer 5

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

Answer 6

You should get y=50.

Answer 7

That is the maximum value

Answer 8

As far as I remember,you can always check it though.

Answer 9

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

Answer 10

Thanks @DLS got it.

Is their any condition for negative or positive?

Answer 11

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

Answer 12

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

Answer 13

yes :)
I wrote,
second derivative should be 0

Answer 14

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

Answer 15

should work

Answer 16

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

Answer 17

Seems correct,do you have the answer?

Answer 18

Hmn ... let me check the answers.

Answer 19

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

Answer 20

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

Answer 21

Though , can you help me finding max. value ?

Answer 22

isn't it x=50?

Answer 23

@agent0smith

Answer 24

y = 25x^2 +5-10x

y' = 50x - 10

y'' = 50

Since y'' > 0 The FUction has minimum value

Answer 25

Minimum value is obstained bY:

y' = 0

Answer 26

y' = 50x - 10 = 0

x = 1/5


So...put x=1/5 in the equation....

Answer 27

This Function has Only Minimum Value....

Answer 28

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

Answer 29

Maximum is infinite

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

@mathslover
Maximum value of the function is
x>50

Answer 34

Oh k. Got it now.

Answer 35

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

Answer 36

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

Answer 37

and minimum value
x<1/5

Answer 38

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

Answer 39

As @Yahoo! said, Function Attains Maximum or Minimum at x = -b/2a

this is x = -(-10)/(2*25) = 1/5, no calculus needed :D

Answer 40

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

Answer 41

\[\LARGE 6x^5+5x^4+4x^3+3x^2+2x+1\]
@Yahoo! You got a formula for minimizing this?

Answer 42

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

Answer 43

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

Answer 44

@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.

Answer 45

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

Answer 46

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

Answer 47

@RnR @shubhamsrg

Answer 48

yep @DLS

Answer 49

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??

Answer 50

just applying second derivative test here ..

Answer 51

what c value did you get by the way ?

Answer 52

oh 1/5 :)

Answer 53

Hey @yahoo, Congratulations for Green..

@AravindG, Congratulations for Ambassador..

And @mathslover, Congratulations for 98 + Green..

Answer 54

congratulate me too???

Answer 55

for ?

Answer 56

nothing XD

Answer 57

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

Answer 58

:")

Answer 59

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.

Answer 60

Thanks a lot everyone for your help

Answer 61

Oh thanks @waterineyes , and sorry for using dt there... I meant dx. Actually I usually come up with problems having dt there na.