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Finding Minimum value of radius

Asked by Suman Koirala

Suman Koirala (view profile)

on 26 Mar 2013
Problem 1: The volume V and paper surface area of a conical paper cup are given by: 
V=1/3*pi*r^2*h 
A =pi*r*sqrt(r^2+h^2)
For V = 10 in 3 , compute the value of the radius, r that minimizes the area A. What is the corresponding value of the height, h? What is the minimum amount that r can vary from its optimal value before the area increases by 10%. 

6 Comments

Walter Roberson

Walter Roberson (view profile)

on 26 Mar 2013

You have asked fminbnd() to invoke your function 'Untitled3', which then will invoke fminbnd() which will then invoke Untitled3, which will then invoke fminbnd()...

Walter Roberson

Walter Roberson (view profile)

on 26 Mar 2013

I wonder if "10 in 3" is intended to mean "10 cubic inches" ?

Suman Koirala

Suman Koirala (view profile)

on 26 Mar 2013

Hey guys, sorry for that..it was 10 cubic inches.

Suman Koirala

Suman Koirala (view profile)

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3 Answers

Answer by Youssef Khmou

Youssef Khmou (view profile)

on 26 Mar 2013
Edited by Youssef Khmou

Youssef Khmou (view profile)

on 26 Mar 2013
Accepted answer

hi Suman Koirla, try this :

The Volume is given by : V=(1/3)*r²*h, and the surface A=pi*r*sqrt(r²+h²)

for V=10 m^3, we search for r that minimizes the Surface , :

 Min(A)   ,  SUBject to  V=10

we have : h=3*V/pi*r² then : A=pi*r*sqrt(r²+90/pi²*r^4) .

Min(A) means the dA/dr=0=......=4*pi*r^3-180 /(2*sqrt(pi*r^4+90/r²))=0

Fast way to find R :

 syms r
 A=(pi^2*r^2+90/r^2)^1/2
 ezplot(A)
 S=subs(A,-6:0.1:6); % AXIS based on the first graph
 min(S)

1)so the minimum value for S=29.83 meter is R=1.89 ( FROM THE GRapH )

2) The corresponding value for h=3*10/(pi*1.89)=5.0525 meter .

4 Comments

Walter Roberson

Walter Roberson (view profile)

on 26 Mar 2013

No, not h=3*V/pi*r² -- h=3*V/(pi*r²)

The actual minimum value for r is 1.890102955

Youssef Khmou

Youssef Khmou (view profile)

on 26 Mar 2013

YES true i made mistake its S=29 m², corresponding r~1.9 meter .

Suman Koirala

Suman Koirala (view profile)

on 26 Mar 2013

How to do the third part where it says "What is the minimum amount that r can vary from its optimal value before the area increases by 10%." I had no idea on that one. Thanks for any inputs.

Youssef  Khmou

Youssef Khmou (view profile)

Answer by Walter Roberson

Walter Roberson (view profile)

on 26 Mar 2013

Are you required to use a minimizer? The question can be solved analytically with a tiny amount of algebra together with some small calculus.

1 Comment

Suman Koirala

Suman Koirala (view profile)

on 26 Mar 2013

Not required to use minimizer. Intro Matlab course.

Walter Roberson

Walter Roberson (view profile)

Answer by Youssef Khmou

Youssef Khmou (view profile)

on 27 Mar 2013
Edited by Youssef Khmou

Youssef Khmou (view profile)

on 27 Mar 2013

3)What is the minimum amount that r can vary from its optimal value before the area increases by 10% ( with fixed h ) :

Given S=29.83 m² and h=5.05 m, we have the new surface S2 :

                            __________
 S2=S+0.1*S=32.81 m²=pi*r*\/ r²+h²      .
 S2²=pi².r^4 + pi²r²h² , make it as equation of 4th order :
 r^4 + r² . h² -S2²/pi² = 0 ==>   r^4 + 25.50 *r² - 109.7 = 0

We use the command "root" :

the Polynomial is a*r^4 + b*r^3 + c*r^2 + b*r + d = 0

a=1; b=0; c=25.50; d=-109.7

 R_amount = roots([1 0 25.50 0 -109.7]) 
 R_amount =
   0.0000 + 5.4084i
   0.0000 - 5.4084i
   1.9366          
  -1.9366     

The reasonable answer is the third one, R=1.9366 the amount change is

DELTA_R=1.9366-1.89=0.04 meter .

0 Comments

Youssef  Khmou

Youssef Khmou (view profile)

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