BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Samosas

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 1132
Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members
Location: India
Thanked: 64 times
GMAT Score:760
by harsh.champ » Thu Feb 04, 2010 2:21 pm
Davji shop sells samosas in boxes of different sizes. The samosas are priced at Rs.2 per samosa upto 200 samosas. For every additional 20 samosas, the price of the whole lot goes down by 10 paisa per samosa. What should be the maximum size of the box that would maximize the revenue?

(1)240
(2)250
(3)300
(4)360
(5)400
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1275
Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members
Location: Arabian Sea
Thanked: 125 times

by ajith » Thu Feb 04, 2010 4:10 pm
harsh.champ wrote:Davji shop sells samosas in boxes of different sizes. The samosas are priced at Rs.2 per samosa upto 200 samosas. For every additional 20 samosas, the price of the whole lot goes down by 10 paisa per samosa. What should be the maximum size of the box that would maximize the revenue?

(1)240
(2)250
(3)300
(4)360
(5)400
At 200 samosas the revenue = 400
At 220 the revenue = 220*(2 -0.1) = 440 -22 = 418
At 240 = 240 *(2-0.2) = 480 -48 =432
at 300 =300*1.5 = 450
at 360 = 360 *(1+0.2) = 360+72 = 432
at 400 =400*1 =400

The revenue is maximized at 300
Always borrow money from a pessimist, he doesn't expect to be paid back.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 90
Joined: Fri Jan 22, 2010 1:10 pmFollowed by: 4 members
Location: New Jersey
Thanked: 13 times
GMAT Score:640

by Mom4MBA » Fri Feb 05, 2010 8:08 am
It can also be done using Maxima and Minima also;

The formula of revenue will be R = n[2 - 0.10(n-200)/20] , where n is number of samosas

R = n[2 - 0.10(n-200)/20] ....................eq� (i)
simplify it
R = 2n - (n²-200n)/200 ........................eq� (ii)
R' = 2 - (2n-200)/200 , first derivative of R
make R'=0, we will get n=300

R'' = -(2)/200 as this is negative the value of n at R'=0 is the maximum value

put n=300 in eq� (i) we get R=450

so maximum revenue is Rs 450 for a box of 300 samosas
Stay focused
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1275
Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members
Location: Arabian Sea
Thanked: 125 times

by ajith » Fri Feb 05, 2010 8:13 am
Mom4MBA wrote:It can also be done using Maxima and Minima also;

The formula of revenue will be R = n[2 - 0.10(n-200)/20] , where n is number of samosas

R = n[2 - 0.10(n-200)/20] ....................eq� (i)
simplify it
R = 2n - (n²-200n)/200 ........................eq� (ii)
R' = 2 - (2n-200)/200 , first derivative of R
make R'=0, we will get n=300

R'' = -(2)/200 as this is negative the value of n at R'=0 is the maximum value

put n=300 in eq� (i) we get R=450

so maximum revenue is Rs 450 for a box of 300 samosas
Strictly speaking, it is a step function and thus not continuous and maxima and minima cannot be applied.
In this case you have assumed that 10 paise decrease is gradual where in the actual decrease is discontinuous and happens only in intervals of 20
Always borrow money from a pessimist, he doesn't expect to be paid back.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1275
Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members
Location: Arabian Sea
Thanked: 125 times

by ajith » Fri Feb 05, 2010 8:48 am
Mom4MBA wrote:well isn't it a continuous step function.............
I think not, in your case the rates for 201 samosas and 219 samosas are different but, in actual case, it is same, the change occurs at 220 - hence called step function.

What would you do if your optimization technique gives a maxima at 291? [in this case it works perfectly and gives a maxima at a step which is 300]
Always borrow money from a pessimist, he doesn't expect to be paid back.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1132
Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members
Location: India
Thanked: 64 times
GMAT Score:760

by harsh.champ » Fri Feb 05, 2010 9:03 am
ajith wrote:
Mom4MBA wrote:It can also be done using Maxima and Minima also;

The formula of revenue will be R = n[2 - 0.10(n-200)/20] , where n is number of samosas

R = n[2 - 0.10(n-200)/20] ....................eq� (i)
simplify it
R = 2n - (n²-200n)/200 ........................eq� (ii)
R' = 2 - (2n-200)/200 , first derivative of R
make R'=0, we will get n=300

R'' = -(2)/200 as this is negative the value of n at R'=0 is the maximum value

put n=300 in eq� (i) we get R=450

so maximum revenue is Rs 450 for a box of 300 samosas
Strictly speaking, it is a step function and thus not continuous and maxima and minima cannot be applied.
In this case you have assumed that 10 paise decrease is gradual where in the actual decrease is discontinuous and happens only in intervals of 20
_________________________
Thanks ajith for pointing out the mistake.
Many a times people tend to make functions without even realizing its continuity and its limits.
Mom4MBA wrote:
well isn't it a continuous step function.............
I think not, in your case the rates for 201 samosas and 219 samosas are different but, in actual case, it is same, the change occurs at 220 - hence called step function.

What would you do if your optimization technique gives a maxima at 291? [in this case it works perfectly and gives a maxima at a step which is 300]

Also,we can see over here that by formulating the f:n and calculating the maxima and minima,the soln. process becomes quite cumbersome.

I would like to go by the hit-and-trial method in this case. What-say guys??
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 90
Joined: Fri Jan 22, 2010 1:10 pmFollowed by: 4 members
Location: New Jersey
Thanked: 13 times
GMAT Score:640

by Mom4MBA » Fri Feb 05, 2010 9:05 am
Sorry my mistake, even I would like to go for hit and trial.......Thanks ajith for pointing it out :)
Stay focused
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1