MATLAB: Volume/Surface Area of Paper Cup

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Homework Help Overview

The problem involves determining the dimensions of a paper cup shaped as a frustum of a cone, specifically finding the radius R_1 and height h given a volume constraint of 240 cm³ and a relationship between R_2 and R_1. The context is MATLAB programming, focusing on optimization techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of equations for volume and surface area, with some attempting to express the surface area in terms of a single variable. Questions arise regarding the correctness of initial assumptions and definitions, particularly the relationship between R_1 and R_2.

Discussion Status

There is an ongoing exploration of the problem with some participants providing MATLAB code snippets and others questioning the validity of the initial equations and relationships. Guidance has been offered regarding the need for clarity in the definitions used.

Contextual Notes

Participants note discrepancies in the definitions of R_2, with one instance stating R_2 = 1.3R_1 and another stating R_2 = R_1/3. There is also mention of the requirement to use specific MATLAB commands and functions.

malindenmoyer
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A paper cup shpaed as a frustum of a cone with [tex]R_2=1.3R_1[/tex] is designed to have a volume of 240 cm3. Determine [tex]R_1[/tex] and height [tex]h[/tex]of the cup such that the least amount of paper will be used for making the cup.

The problem must be done in MATLAB, so please keep that in mind. We have learned commands such as feval, fminbnd, syms, and function handles...all of which I think could be applicable to the problem.
 
Last edited:
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Show us the MATLAB code you have written. You won't get much help until you try something first.
 
I am not particularly sure if this is even how you start the problem, but this is what I have done so far.

Code:
syms R1 h;
R2=1.3*R1;
V=(pi*h)/3*(R1^2+R2^2+R1*R2)-240;
S=pi*(R1+R2)*sqrt((R2-R1)^2+h^2)+pi*R1^2;

I know that I could solve for [tex]h[/tex] using the Volume equation, and then have the Surface Area equation in terms of one variable essentially, as R2 is defined by R1; my thought process is such that I would take the new Surface Area equation (in terms of only R1) and take its derivative to find the critical points. This would give me R2 and then I could solve for h. Is this the right process?

Per LCKurtz's request, the two formulas I used to write the above code were the following:
[tex]V=\frac{1}{3}\pi h(\left R_1^2+R_2^2+R_1R_2)\right[/tex]
[tex]S=\pi(R_1+R_2)\sqrt{(\left R_2-R_1)\right^2+h^2}+\pi R_1^2[/tex]
 
Last edited:
malindenmoyer said:
I am not particularly sure if this is even how you start the problem, but this is what I have done so far.

Code:
syms R1 h;
R2=R1/3;
V=(pi*h)/3*(R1^2+R2^2+R1*R2)-240;
S=pi*(R1+R2)*sqrt((R2-R1)^2+h^2)+pi*R1^2;

I know that I could solve for [tex]h[/tex] using the Volume equation, and then have the Surface Area equation in terms of one variable essentially, as R2 is defined by R1; my thought process is such that I would take the new Surface Area equation (in terms of only R1) and take its derivative to find the critical points. This would give me R2 and then I could solve for h. Is this the right process?

I don't use Matlab, but still...in your statement of the problem you give R2 = 1.3R1. Then in your code you put R2 = R1/3. Which is it? You need to be more careful.

For starters, your equation for volume needs to be just that, an equation, not a function.
 

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