Solution 1

The following shows the commands typed at the >> prompt and the output produced:

>> d = 1.54

d =

    1.5400

>> b = 8

b =

     8

>> h = 14

h =

    14

>> vcyl = h*pi*(b/2)^2

vcyl =

  703.7168

>> vcube = d^3

vcube =

    3.6523

>> nl = vcyl/vcube

nl =

  192.6796

Solution 2

We created the following file named PingPong.m:

% PingPong.m - computes a lower bound on the number of 
%   ping pong balls that fit into a cylinder
% Note that most lines end with a ";", so they don't print
%   intermediate results

d = 1.54;
h = 14;
b = 8;
vcyl = h*pi*(b/2)^2;
vcube = d^3;
nl = vcyl/vcube

>> PingPong

nl =

  192.6796

Solution 3

This solution is by BrieAnne Davis.

for d=1.0:.05:2.0
    b=8;
    h=14;
    vcyl=h*pi*(b/2)^2
    vcube=d^3
    nl=vcyl/vcube
end

Solution 4

This solution is by Wade Stevens. Note that it uses the command hold on to plot each point individually in the for loop.

clear all
hold on
for d=1.0:0.1:2.0;
    b=8;
    h=14;
    C=h*pi*(b/2)^2; %volume of cylinder
    c=d^3; %volume of cube
    N=C/c; %Lower bound
    floor(N)
    plot (d,N,'g*')
end

Figure 1: Plot of N L N L as a function of d d ; each point plotted individually.

This different solution is by Christopher Embrey. It uses the index variable j to step through the dv array to compute elements of the nlv array; the complete nlv array is computed, and then plotted outside the for loop.

clear
dv=1.0:.05:2.0;
 
[junk,dvsize] = size(dv)
for j=1:dvsize
    d=dv(j)
    b=8; %in
    h=14; %in
    vcyl=h*pi*(b/2)^2;
    vcube=d^3;
    nl=vcyl/vcube; 
    nlv(j)=nl;
end
plot (dv,nlv)

Figure 2: Plot of N L N L as a function of d d ; points plotted as vectors.

And finally, this solution by Travis Venson uses vector computations to perform the computation without a for loop.

%creates a vector for diameter
dv=1:.02:2;
b=5.5;
h=12;
 
%computes volume of cylinder
vcyl=h*pi*(b/2)^2;
 
%computes volume of cube
vcube=dv.^3;
 
%computes lower bound
lowerboundv=vcyl./vcube;
 
%plots results
plot(dv,lowerboundv)

Solution 5

This solution is by AJ Smith. The height, h, ranges from 12 to 15 and the base, b, ranges from 8 to 12.

for h=12:15; %ranges of height
    for b=8:12; %ranges of the base
        d=1.54; %diameter of ping pong ball.
        Vcyl=h*pi*(b/2)^2; %Volume of cylinder
        Vcube=d^3; %volume of a cube that encloses a single ball
        Nl=Vcyl/Vcube %lower bound on the number of balls that fit in the cylinder
    end
end