Forum Home > GMAT > Quantitative > Problem Solving (PS)
Events & Promotions
|
|
GMAT Club Daily Prep
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Dec 1310:00 AM EST
-12:00 PM EST
Have you ever wondered how to score a PERFECT 805 on the GMAT? Meet Julia, a banking professional who used the Target Test Prep course to achieve this incredible feat. Julia's story is nothing short of an inspiration.
Dec 1410:00 AM EST
-12:00 PM EST
Think a 100% GMAT Verbal score is out of your reach? Target Test Prep will make you think again! Our course uses techniques such as topical study and spaced repetition to maximize knowledge retention and make studying simple and fun.
Dec 1411:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.- Dec 15
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!
11 Jul 2003, 05:59
Kudos
Bookmarks
I have a long rectangular metal sheet of 12' wide and I need to construct a rain gutter by turning up two sides so that they are perpendicular to the sheet. Help me, all wonderful mathematicians out there, to find out how many inches should be turned up to give my rain gutter to hold its greatest capacity ?
:help2
:help2
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
11 Jul 2003, 06:15
Kudos
Bookmarks
try out the combinations
2*5 =10
4*4 =16
5*3.5 =17.5
6*3 = 18
7*2.5 = 17.5
8*2 =16
maximum is at 18, so it should be turned up 3"
2*5 =10
4*4 =16
5*3.5 =17.5
6*3 = 18
7*2.5 = 17.5
8*2 =16
maximum is at 18, so it should be turned up 3"
11 Jul 2003, 06:31
Kudos
Bookmarks
No problem, Sir
You have to bend your metal sheet in some way, to have a П-shaped object. Since, the lenght is not important, concentrate on a П-section.
A section of maximal area will give you maximal volume.
So, how to bend a 12' line in 3-side, П-object to cover the maximal area?
Imagine a rectangular with sides: X, X, 12–2X, 12–2X. Let X be a vertical side, and 12–2X be a horizontal one.
X+X+12–2X=given 12 and one side is virtual
An area of the section is X*(12–2X)= 12X–2X^2
A derivative is 12–4X=0, where X=3
so your gutter should have two 3' vertical sides and one 6' horizontal.
Such section gives you 18 square inch area
All others are smaller (10+1+1) gives 10
(8+2+2) gives 16
(4+4+4) gives 16
With the same perimetr a circle has the smallest area, but a square -- the greatest.
You have to bend your metal sheet in some way, to have a П-shaped object. Since, the lenght is not important, concentrate on a П-section.
A section of maximal area will give you maximal volume.
So, how to bend a 12' line in 3-side, П-object to cover the maximal area?
Imagine a rectangular with sides: X, X, 12–2X, 12–2X. Let X be a vertical side, and 12–2X be a horizontal one.
X+X+12–2X=given 12 and one side is virtual
An area of the section is X*(12–2X)= 12X–2X^2
A derivative is 12–4X=0, where X=3
so your gutter should have two 3' vertical sides and one 6' horizontal.
Such section gives you 18 square inch area
All others are smaller (10+1+1) gives 10
(8+2+2) gives 16
(4+4+4) gives 16
With the same perimetr a circle has the smallest area, but a square -- the greatest.
