Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

2 Part question- Max., & Min. value of function! [#permalink]
09 Sep 2012, 06:39

1

This post received KUDOS

1

This post was BOOKMARKED

Hello Friends!

Try this question...

The function f(x,y) is such that f(x,y) = x^2.y^3 & x+y =25. Where x, & y are non-negative integers. Select one minimum possible, & one maximum possible value for f(x,y). Make only two selections, one in each column.

Attachments

2 Part Qn.JPG [ 26.61 KiB | Viewed 1066 times ]

_________________

Shalabh Jain, e-GMAT Instructor

Last edited by ShalabhsQuants on 09 Sep 2012, 19:01, edited 2 times in total.

Re: 2 Part question- Max., & Min. value of function! [#permalink]
09 Sep 2012, 10:36

ShalabhsQuants wrote:

Hello Friends!

Try this question...

I checked the OG13 and QuantReview: they don't use neither the term whole number, nor that of natural number. I guess, in order to avoid the lack of consensus regarding 0. And in fact, why have so many different definitions? It is much easier to have just integers, which are positive, negative or zero. Further, you can have non-negative integers if you want to include 0,...

Oh, and I don't think the symbol for infinity should be known on the GMAT. I would say, not a must to solve such questions for a future test taker. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: 2 Part question- Max., & Min. value of function! [#permalink]
09 Sep 2012, 18:43

EvaJager wrote:

ShalabhsQuants wrote:

Hello Friends!

Try this question...

I checked the OG13 and QuantReview: they don't use neither the term whole number, nor that of natural number. I guess, in order to avoid the lack of consensus regarding 0. And in fact, why have so many different definitions? It is much easier to have just integers, which are positive, negative or zero. Further, you can have non-negative integers if you want to include 0,...

Oh, and I don't think the symbol for infinity should be known on the GMAT. I would say, not a must to solve such questions for a future test taker.

Re: 2 Part question- Max., & Min. value of function! [#permalink]
09 Sep 2012, 19:04

Min value of f(x,y) = 0, If we choose either x or y as 0 and the other as 25. The product x^2y^3 becomes 0. Max value of f(x,y) = Infinity. Since x+y =25, and both are integers, we can choose y as a +ve number such as 10000xxxxxx000 and x as -ve no such as 999999xxxxx75. If we add we will get 25, but if we square x it will be a positive expression and the value of f(x,y) can go up to infinity. _________________

Re: 2 Part question- Max., & Min. value of function! [#permalink]
09 Sep 2012, 21:25

SOURH7WK wrote:

Min value of f(x,y) = 0, If we choose either x or y as 0 and the other as 25. The product x^2y^3 becomes 0. Max value of f(x,y) = Infinity. Since x+y =25, and both are integers, we can choose y as a +ve number such as 10000xxxxxx000 and x as -ve no such as 999999xxxxx75. If we add we will get 25, but if we square x it will be a positive expression and the value of f(x,y) can go up to infinity.

As far as Minimum value of f(x,y) is concerned, your answer is correct. But Maximum value is not infinity. Remember x, & y both are non-negative integers. Minimum values of x, & y would be 0 only, whereas maximum as 25. _________________

Re: 2 Part question- Max., & Min. value of function! [#permalink]
09 Sep 2012, 23:04

ShalabhsQuants wrote:

SOURH7WK wrote:

Min value of f(x,y) = 0, If we choose either x or y as 0 and the other as 25. The product x^2y^3 becomes 0. Max value of f(x,y) = Infinity. Since x+y =25, and both are integers, we can choose y as a +ve number such as 10000xxxxxx000 and x as -ve no such as 999999xxxxx75. If we add we will get 25, but if we square x it will be a positive expression and the value of f(x,y) can go up to infinity.

As far as Minimum value of f(x,y) is concerned, your answer is correct. But Maximum value is not infinity. Remember x, & y both are non-negative integers. Minimum values of x, & y would be 0 only, whereas maximum as 25.

Ok, I missed that "non" part in the stem. Now looking at the choices (i,e ending with zeros) I tried few combination and got 10^2x15^3 = 33750. So is Option C is the maximum value? _________________

Re: 2 Part question- Max., & Min. value of function! [#permalink]
10 Sep 2012, 00:54

SOURH7WK wrote:

ShalabhsQuants wrote:

SOURH7WK wrote:

Min value of f(x,y) = 0, If we choose either x or y as 0 and the other as 25. The product x^2y^3 becomes 0. Max value of f(x,y) = Infinity. Since x+y =25, and both are integers, we can choose y as a +ve number such as 10000xxxxxx000 and x as -ve no such as 999999xxxxx75. If we add we will get 25, but if we square x it will be a positive expression and the value of f(x,y) can go up to infinity.

As far as Minimum value of f(x,y) is concerned, your answer is correct. But Maximum value is not infinity. Remember x, & y both are non-negative integers. Minimum values of x, & y would be 0 only, whereas maximum as 25.

Ok, I missed that "non" part in the stem. Now looking at the choices (i,e ending with zeros) I tried few combination and got 10^2x15^3 = 33750. So is Option C is the maximum value?

You are right! Let's understand the concept. This will eliminate the hit & trial approach for these kinds of questions.

If a function is such that f(a,b)=a^m.b^n, & a+b= constant, then f(a,b) would be maximum when a/m=b/n.

Coming to this question....

Given is x+y=25 =constant. for f(x,y)=x^2.y^3 to be max., x/2 should be equal to y/3 =>x/2=y/3 or x=2y/3.

By plugging in this value in x+y=25, we get 2y/3+y=25 => y=15, & x=10.

So Maximum of f(x,y)= 10^2.15^3 = 337500. _________________

Re: 2 Part question- Max., & Min. value of function! [#permalink]
10 Sep 2012, 01:47

ShalabhsQuants wrote:

If a function is such that f(a,b)=a^m.b^n, & a+b= constant, then f(a,b) would be maximum when a/m=b/n.

Coming to this question....

Given is x+y=25 =constant. for f(x,y)=x^2.y^3 to be max., x/2 should be equal to y/3 =>x/2=y/3 or x=2y/3.

By plugging in this value in x+y=25, we get 2y/3+y=25 => y=15, & x=10.

So Maximum of f(x,y)= 10^2.15^3 = 337500.

How you have derived that formula. The formula seems very conditional with a+b constant & only for maximum value. Is there any partial derivative involved?? _________________

Re: 2 Part question- Max., & Min. value of function! [#permalink]
10 Sep 2012, 04:13

1

This post received KUDOS

SOURH7WK wrote:

ShalabhsQuants wrote:

If a function is such that f(a,b)=a^m.b^n, & a+b= constant, then f(a,b) would be maximum when a/m=b/n.

Coming to this question....

Given is x+y=25 =constant. for f(x,y)=x^2.y^3 to be max., x/2 should be equal to y/3 =>x/2=y/3 or x=2y/3.

By plugging in this value in x+y=25, we get 2y/3+y=25 => y=15, & x=10.

So Maximum of f(x,y)= 10^2.15^3 = 337500.

How you have derived that formula. The formula seems very conditional with a+b constant & only for maximum value. Is there any partial derivative involved??

Formally, yes, it is by partial derivatives, looking for extremum point...

A sort of justification without partial derivatives: For any real numbers \(x\) and \(y\), \(\, (x + y)^2 \geq{4xy}\). Equality holds if and only if \(x=y\) (the given inequality is equivalent to \((x-y)^2\geq{0}\). In words: when the sum of two real numbers is constant, the maximum product of the two numbers is obtained when they are each equal to half of the sum.

In our case, the sum is constant, but in the product we have two different powers, 2 and 3. Intuitively, the maximum will be obtained for a weighted average between \(x\) and \(y\), \(y\) being closer to 25 as in the product it has a greater power, but still not "too far away" from the half of the sum.

But, since here we have integers and in addition, it is a GMAT multiple choice question, we can use some number properties. The possible answers (after we eliminate infinity) are all multiples of 5, and since the sum\(x+y=25\) is a multiple of 5, if one of the numbers is multiple of 5, then the other one is also. And of course, \(y\) should be greater than \(x.\) Therefore, we only have to check \(x=5, \, y=20\) and \(x=10, \, y=15.\) _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.