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 500,000 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:

If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink]

Show Tags

11 Feb 2013, 10:56

2

This post received KUDOS

10

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

52% (02:09) correct
48% (01:12) wrong based on 274 sessions

HideShow timer Statistics

If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is

A. Less than 10 B. Greater than or equal to 10 and less than 14 C. Greater than 14 and less than 19 D. Greater than 19 and less than 23 E. Greater than 23

Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink]

Show Tags

11 Feb 2013, 14:44

5

This post received KUDOS

Expert's post

6

This post was BOOKMARKED

emmak wrote:

If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is

A. Less than 10 B. Greater than or equal to 10 and less than 14 C. Greater than 14 and less than 19 D. Greater than 19 and less than 23 E. Greater than 23

We need to maximize \(x+y\), thus we need to maximize \(\sqrt{100+2xy}\). To maximize \(\sqrt{100+2xy}\) we need to maximize \(xy\).

Now, for given sum of two numbers, their product is maximized when they are equal, hence from \(x^2 + y^2 = 100\) we'll have that \(x^2y^2\) (or which is the same xy) is maximized when \(x^2=y^2\).

In this case \(x^2 + x^2 = 100\) --> \(x=\sqrt{50}\).

So, we have that \(\sqrt{100+2xy}\) is maximized when \(x=y=\sqrt{50}\), so the maximum value of \(\sqrt{100+2xy}\) is \(\sqrt{100+2*\sqrt{50}*\sqrt{50}}=\sqrt{200}\approx{14.1}\).

Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink]

Show Tags

11 Feb 2013, 15:05

2

This post received KUDOS

I tried a more crude method:

\(x^2 + y^2 = 100\)

We know \(6^2 + 8^2\) is one of the options which will lead to x+y = 14. So options A & B are out.

We also know \(7^2 + 7^2\) we get just \(98<100\). So something slightly more than \(7\) i.e. \(7.1\) or whatever that will lead to the answer of \(100\). So the maximum value of \(x+y\) is just over \(14\). Any other combination of x & y cannot be more than this value of just over 14. So answer is C.

Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink]

Show Tags

13 Feb 2013, 09:31

1

This post received KUDOS

An elegant solution is looking at this problem geometrically.

x^2 + y^2 = 100 is a circumference with radius 10 and center in (0,0).

The equation x + y = k (where we want to maximize k) is a set of infinite parallel negative slope (-45 deg) lines.

Now note that k is the Y INTERSECT of all such lines. In order to maximize this y intersect (and therefore maximize k) we need to find the line of the set that is tangent to the circunference in the 1st quadrant.

Now this becomes a (fairly easy) geometry problem.

Drawing it out you'll have no problem seeing that Y intersect = k = RADIUS / Sin(45 deg) = 10.sqrt(2) = approx 14,1

Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink]

Show Tags

11 Mar 2013, 18:17

Dear Bunuel,

Can you please explain this statement,"for given sum of two numbers, their product is maximized when they are equal". Can you give some theory and examples for this statement to be true ? I would be happy to understand this. Thank you. _________________

A Ship in port is safe but that is not what Ships are built for !

Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink]

Show Tags

11 Mar 2013, 20:43

3

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

emmak wrote:

If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is

A. Less than 10 B. Greater than or equal to 10 and less than 14 C. Greater than 14 and less than 19 D. Greater than 19 and less than 23 E. Greater than 23

You can use the logical approach to get the answer within seconds.

First you need to understand how squares work - As you go to higher numbers, the squares rise exponentially (obviously since they are squares!) What I mean is 2^2 = 4 3^2 = 9 4^2 = 16 (as we increase the number by 1, the square increases by more than the previous increase. From 2^2 to 3^2, the increase is 9-4= 5 but from 3^2 to 4^2, the increase is 16 - 9 = 7... As we go to higher numbers, the squares will keep increasing more and more.)

If we want to keep the square at 100 but maximize the sum of the numbers, we should try and make the numbers as small as possible so that their contribution in the square doesn't make the other number very small i.e. if you take one number almost 10, the other number will become very very small and the sum will not be maximized. If one number is made small, the other will become large, hence both numbers should be equal to maximize the sum. So the square of each number should 50 i.e. each number should be a little more than 7. Both numbers together will give us something more than 14.

Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink]

Show Tags

12 Mar 2013, 06:50

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

Backbencher wrote:

Dear Bunuel,

Can you please explain this statement,"for given sum of two numbers, their product is maximized when they are equal". Can you give some theory and examples for this statement to be true ? I would be happy to understand this. Thank you.

If a+c=k, then the product ab is maximized when a=b.

For example, if given that a+b=10, then ab is maximized when a=b=5 --> ab=25.

Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink]

Show Tags

05 Jul 2014, 15:00

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink]

Show Tags

04 Jun 2015, 06:50

Expert's post

VeritasPrepKarishma wrote:

emmak wrote:

If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is

A. Less than 10 B. Greater than or equal to 10 and less than 14 C. Greater than 14 and less than 19 D. Greater than 19 and less than 23 E. Greater than 23

You can use the logical approach to get the answer within seconds.

First you need to understand how squares work - As you go to higher numbers, the squares rise exponentially (obviously since they are squares!) What I mean is 2^2 = 4 3^2 = 9 4^2 = 16 (as we increase the number by 1, the square increases by more than the previous increase. From 2^2 to 3^2, the increase is 9-4= 5 but from 3^2 to 4^2, the increase is 16 - 9 = 7... As we go to higher numbers, the squares will keep increasing more and more.)

If we want to keep the square at 100 but maximize the sum of the numbers, we should try and make the numbers as small as possible so that their contribution in the square doesn't make the other number very small i.e. if you take one number almost 10, the other number will become very very small and the sum will not be maximized. If one number is made small, the other will become large, hence both numbers should be equal to maximize the sum. So the square of each number should 50 i.e. each number should be a little more than 7. Both numbers together will give us something more than 14.

Answer (C)

Sorry to open this post after so long time again but i got a bit confused reg the solution. I agree with the solution that maximizing sum is by making both the numbers x and y equal resulting in value greater than 14. but what about checking whether the value is less than 19 or not. Can you please explain. _________________

The only time you can lose is when you give up. Try hard and you will suceed. Thanks = Kudos. Kudos are appreciated

Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink]

Show Tags

04 Jun 2015, 07:03

Expert's post

Mechmeera wrote:

VeritasPrepKarishma wrote:

emmak wrote:

If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is

A. Less than 10 B. Greater than or equal to 10 and less than 14 C. Greater than 14 and less than 19 D. Greater than 19 and less than 23 E. Greater than 23

You can use the logical approach to get the answer within seconds.

First you need to understand how squares work - As you go to higher numbers, the squares rise exponentially (obviously since they are squares!) What I mean is 2^2 = 4 3^2 = 9 4^2 = 16 (as we increase the number by 1, the square increases by more than the previous increase. From 2^2 to 3^2, the increase is 9-4= 5 but from 3^2 to 4^2, the increase is 16 - 9 = 7... As we go to higher numbers, the squares will keep increasing more and more.)

If we want to keep the square at 100 but maximize the sum of the numbers, we should try and make the numbers as small as possible so that their contribution in the square doesn't make the other number very small i.e. if you take one number almost 10, the other number will become very very small and the sum will not be maximized. If one number is made small, the other will become large, hence both numbers should be equal to maximize the sum. So the square of each number should 50 i.e. each number should be a little more than 7. Both numbers together will give us something more than 14.

Answer (C)

Sorry to open this post after so long time again but i got a bit confused reg the solution. I agree with the solution that maximizing sum is by making both the numbers x and y equal resulting in value greater than 14. but what about checking whether the value is less than 19 or not. Can you please explain.

Hi Mechmeera,

The maximum value has been calculated when x = y = \(5\sqrt{2}\)

i.e. x+y will have maximum value of \(5\sqrt{2}\) + \(5\sqrt{2}\) = \(10\sqrt{2}\) = 14.14

Since the value of x+y can't exceed beyond the calculated value therefore

we can conclude that it is less that 19 and greater than 14. _________________

READ:http://gmatclub.com/forum/620-to-760-getting-reborn-161230.html Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!

gmatclubot

Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x
[#permalink]
04 Jun 2015, 07:03

http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...