Solution 1

The following shows commands typed to LABVIEW MATHSCRIPT (at the >> command window) and the output produced by LABVIEW MATHSCRIPT:

>>d = 1.54
d =

           1.54     


>>b = 8
b =

           8     


>>h = 14
h =

           14     


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

           703.72     


>>vcube = d^3
vcube =

           3.6523     


>>nl = vcyl/vcube
nl =

           192.68     

Solution 2

We used the LABVIEW MATHSCRIPT editor (or any text editor) to create 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 LABVIEW MATHSCRIPT hold on command after the plot statement to plot each point individually in the for loop.

clear all
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*')
    hold on
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 LABVIEW MATHSCRIPT's vector capabilities 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