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Maximum value of a function

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by winnerhere » Sun Jul 04, 2010 5:18 am
Y = 25x - x^2

Find the value of X , for which Y attains maximum value.

How to proceed with this question

Thanks in Advance,
Sai
Last edited by winnerhere on Sun Jul 04, 2010 5:41 am, edited 1 time in total.
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by singhpreet1 » Sun Jul 04, 2010 5:34 am
winnerhere wrote:Y = x^2 - 25x

Find the value of X , for which Y attains maximum value.
Sai: if we take the value of x as negative we can get the maximum value of Y, though what are the answer choices?

Please post the same along with OA.

Preet
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by kvcpk » Sun Jul 04, 2010 5:38 am
winnerhere wrote:Y = x^2 - 25x

Find the value of X , for which Y attains maximum value.

How to proceed with this question

Thanks in Advance,
Sai
Y = x(x-25)
when x is less than -25 I beleive, Y will Attain its maximum and will keep increasing..

Are you sure about the question? What is the source!!
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by winnerhere » Sun Jul 04, 2010 5:39 am
Preet,

The answer choices are,

a) 14.5

b) 14

c) 13

d) 12.5

OA : d
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by winnerhere » Sun Jul 04, 2010 5:41 am
kvcpk and preet,

oops sorry,

its y = 25x - x^2
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by kvcpk » Sun Jul 04, 2010 6:28 am
winnerhere wrote:kvcpk and preet,

oops sorry,

its y = 25x - x^2
Ok, in this case, Y will attain a maximum value if x(25-x) os maximum.
let p = (25-x)
then x+p = 25
given their sum, the product of two number is maximum, when they are equal..
(I did not put this rule in the exact terms.. I dont remember the exact terminology.. I request some experts to restate it as per textbooks.)

So product of x and p is maximum when x=p
So, 2x=25 -> x= 12.5
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by singhpreet1 » Sun Jul 04, 2010 6:49 am
winnerhere wrote:kvcpk and preet,

oops sorry,

its y = 25x - x^2
Personally,

i wouldnt even calculate anything for this one, the answer stares in the face for me out of the options: least value of positive X will help Y be maximum.

out of the options given. 12.5 is the lowest.

eg: assume x=10

y= 25*10- 10^2= 150

assume x=20

y= 25*20-20^2= 100

assume y=30

25*30-30^2= -150

therefore the sequence clearly shows that the value of Y is decreasing as X is increasing.

thats my personal way..though please use what suits you.

Preet
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by Patrick_GMATFix » Sun Jul 04, 2010 7:10 am
winnerhere wrote:Y = 25x - x^2

Find the value of X , for which Y attains maximum value.
12.5. I'll explain below.
singhpreet1 wrote:i wouldnt even calculate anything for this one, the answer stares in the face for me out of the options: least value of positive X will help Y be maximum.

out of the options given. 12.5 is the lowest.
You are making a dangerous assumption that the least positive X will result in the maximum Y. This is not accurate. If X=1 then Y will not be greater than if X is 10.

Here is how I solved the question.

This is a quadratic function, meaning that the graph is a parabola. Furthermore, because the coefficient (the number in front) of X is negative (-1) the parabola opens down. The curve is shaped like a rainbow.

Parabolas are symmetrical around their "middle". The maximum value will be the value of X that is exactly halfway between the two points at which the parabola touches the horizontal x-axis (these are called the roots or solutions of the equation; the points where y=0). So let's find those points.

Y=25x - x^2. To solve any quadratic equation (to find its roots) you factor it ---> y = x (25-x). The solutions are the values of x that will make the equation 0. They are x=0 and x=25. This means that the parabola goes up, crossing the x-axis at 0, then reaches its maximum and comes back down, crossing the x-axis again at 25. The maximum is half-way between the two. The maximum value of Y is when X=25/2 --> 12.5

Note: although knowing how to handle parabolas in the xy-axis system can be useful on a few GMAT problems, this skill is not required for the GMAT and may not be worth investing a lot of time to learn if you have weaknesses in GMAT tested topics.

Hope that made sense,
-Patrick
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by mj78ind » Sun Jul 04, 2010 7:44 am
Given the 2 mins timeline,

Approach 1
engineers have an advantage. Calculate the first differential wrt x of the quadratic, and that is your answer since the second differential is negative. First differential of the equation is 25/2 = 12.5

Approach 2
Plug in the numbers and you will see that 12.5 gives the maxima.
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by amising6 » Sun Jul 04, 2010 7:49 am
winnerhere wrote:Y = 25x - x^2

Find the value of X , for which Y attains maximum value.

How to proceed with this question

Thanks in Advance,
Sai
Y = 25x - x^2
y=-(x^2-25x)

y=-(x^2-2*x*25/2)
y=-(x^2-(2*x*25/2 )+(25/2)^2 - (25/2)^2) ( (adding (25/2)^2) and subtracting (25/2)^2)))
y=-((x+25/2)^2-(25/2)^2) (a^2+b^2+2ab)
y=(25/2)^2)=625/4 max values
since square of a number is always gretaer than or equal to zero
(x+25/2)^2>=0
so x=-25/2
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by kvcpk » Sun Jul 04, 2010 8:13 am
kvcpk wrote:given their sum, the product of two number is maximum, when they are equal..
(I did not put this rule in the exact terms.. I dont remember the exact terminology.. I request some experts to restate it as per textbooks.)
Can any expert restate this for me in exact terms? Thank you!!
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by amising6 » Sun Jul 04, 2010 8:31 am
kvcpk wrote:
kvcpk wrote:given their sum, the product of two number is maximum, when they are equal..
(I did not put this rule in the exact terms.. I dont remember the exact terminology.. I request some experts to restate it as per textbooks.)


Can any expert restate this for me in exact terms? Thank you!!
u r rite kvcpk

it says if sum of gven number is contant then therir product will be maximum if numbers as close as possible
if you have a+b=12
then ab i.e product will be maximum when and b are equal or close to each other
you can check if a=b=6
ab=36 i.e maximu m
next if a=7 b=5 ab=35 next maximum so on and so far
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by gmatmachoman » Sun Jul 04, 2010 8:45 am
kvcpk wrote:
kvcpk wrote:given their sum, the product of two number is maximum, when they are equal..
(I did not put this rule in the exact terms.. I dont remember the exact terminology.. I request some experts to restate it as per textbooks.)
Can any expert restate this for me in exact terms? Thank you!!
Praveen bhai,
Probably this one should help u!!

Y = 25x - x^2

step 1 :find dy/dx

dy/dx = 25 - 2x

step 2: set dy/dx= 0

25 = 2x

x= 25/2 = 12.5

For max values d^2y/dx^2 (second degree differential) <0

For minumim values d^2y/dx^2 >0

So cross check by putting value of x in d^2y/dx^2

d^2y/dx^2 = -2 which is anyway less than zero.

Pick D.

On a side note dont worry GMAC never tests maxima & minima...
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by Stuart@KaplanGMAT » Sun Jul 04, 2010 9:14 am
kvcpk wrote:
kvcpk wrote:given their sum, the product of two number is maximum, when they are equal..
(I did not put this rule in the exact terms.. I dont remember the exact terminology.. I request some experts to restate it as per textbooks.)
Can any expert restate this for me in exact terms? Thank you!!
Hi!

This rule is easy to remember if you think of rectangles and squares.

For rectangles, given a fixed perimeter, you maximize the area of the shape by making it a square.

In algebraic terms, given a fixed value x, the maximum product that can be derived by two numbers that sum to x is:

(x/2)^2 = (x^2)/4.
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by kvcpk » Sun Jul 04, 2010 9:56 am
Thanks for all your replies guys..

I wanted to confirm the rule that I was using.. I am answered now..

Thank you!!
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