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
Maximum value of a function

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Last edited by winnerhere on Sun Jul 04, 2010 5:41 am, edited 1 time in total.

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Sai: if we take the value of x as negative we can get the maximum value of Y, though what are the answer choices?winnerhere wrote:Y = x^2  25x
Find the value of X , for which Y attains maximum value.
Please post the same along with OA.
Preet
Y = x(x25)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
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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Ok, in this case, Y will attain a maximum value if x(25x) os maximum.winnerhere wrote:kvcpk and preet,
oops sorry,
its y = 25x  x^2
let p = (25x)
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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Personally,winnerhere wrote:kvcpk and preet,
oops sorry,
its y = 25x  x^2
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*2020^2= 100
assume y=30
25*3030^2= 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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12.5. I'll explain below.winnerhere wrote:Y = 25x  x^2
Find the value of X , for which Y attains maximum value.
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.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.
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 xaxis (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 (25x). 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 xaxis at 0, then reaches its maximum and comes back down, crossing the xaxis again at 25. The maximum is halfway between the two. The maximum value of Y is when X=25/2 > 12.5
Note: although knowing how to handle parabolas in the xyaxis 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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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.
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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Y = 25x  x^2winnerhere 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=(x^225x)
y=(x^22*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
Ideation without execution is delusion
Can any expert restate this for me in exact terms? Thank you!!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.)
 amising6
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u r rite kvcpkkvcpk 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!!
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
Ideation without execution is delusion

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Praveen bhai,kvcpk wrote:Can any expert restate this for me in exact terms? Thank you!!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.)
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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Hi!kvcpk wrote:Can any expert restate this for me in exact terms? Thank you!!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.)
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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