Finding Minimum value of radius

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%.

[Suman Koirala]

I have done this so far:

function [A,a] =Untitled3(r)
    global  V;
    V=10;
    a=pi*r.*sqrt(r.^2 +((9*V^2)/pi^2*r.^4))
    [A,r]= fminbnd('Untitled3', 0,50)
end

[Image Analyst]

What does "10 in 3" mean?

[Youssef Khmou]

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

[Walter Roberson]

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]

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

[Suman Koirala]

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

Answer 1

[Youssef Khmou]

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 .

[Suman Koirala]

How did you get A=pi*r*sqrt(r²+90/pi²*r^4) ? where did you get the 90 from?

[Walter Roberson]

No, not h=3*V/pi*r² -- h=3*V/(pi*r²)

The actual minimum value for r is 1.890102955

[Youssef Khmou]

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

[Suman Koirala]

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 2

[Walter Roberson]

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

[Suman Koirala]

Not required to use minimizer. Intro Matlab course.

Answer 3

[Youssef Khmou]

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 .