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Asked by Suman Koirala 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%.

*No products are associated with this question.*

Answer by Youssef KHMOU on 26 Mar 2013

Edited by Youssef KHMOU 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 .

Show 1 older comment

Walter Roberson 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

Suman Koirala 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.

Answer by Walter Roberson 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.

Answer by Youssef KHMOU on 27 Mar 2013

Edited by Youssef KHMOU 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 .

## 6 Comments

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/68627#comment_138948

I have done this so far:

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/68627#comment_138955

What does "10 in 3" mean?

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/68627#comment_138959

i think, it means for V=10 in "equation 3" , maybe

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/68627#comment_138960

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

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/68627#comment_138962

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

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/68627#comment_138964

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