how do you find x,y,z?

(1) If w is the number of widgets, g the number of gigets and s the number of sprockets we have then

(a) w+g+s=30 and

(b) 3w+4g+5s=96

From (a) we get w=30-g-s and, replacing in (b) and simplifying we get

g+2s=6

From where we get

s = (6-g)/2 = 3 - (g/2)

As w, g and s are integers >=0, it follows that g has to be even and the 4 possible solutions are:

g=0, s=3, w=30-0-3=27

g=2, s=2, w=26

g=4, s=1, w=25

g=6, s=0, w=24

(2) If we are still saying that we have to make use of the full 96 ounces and 30 items then the answer is 2, as those are the only answers amongst the ones above for each all variables are positive.

However, in the more interesting case where we are just interested in all possible ways to pack where the number of items is at most 30 and the weight is at most 96, then they can be counted (using a spreadsheet) as follows:

Consider 3 columns headed w, g and s.

On the first row, under the headings, make w(1)=1, g(1)= 1 and s(1) = INT ( (96-3*w(1)-4*g(1))/5 )

Then you will get s(1)=17 and this is the maximum number of sprockets you can get if you have 1 widget and 1 giget. Clearly you could have any number of sprockets from 1 to 17, so there are 17 possibilities with 1 widget and 1 sprocket

Now for the rows below make w(i+1) = w(i) if s(i)>1 and w(i+1)=w(i) + 1 if s(i)=1, and

g(i+1)=g(i) + 1 if s(i)>1 and 1 if s(i) = 1. Also make s(i)= INT ( (96-3*w(i)-4*g(i))/5 )

What this means is that for a fixed w(i) we will increase g(i) until s(i) becomes 1. Then we start on the next value of w(i) and bring s(i) back to 1 again.

The maximum possible value of w(i) is (96-4-5)/3 = 87/3=29 as this will give us g(i)=s(i)=1, so the process needs to be continued till w(i) reaches 29.

In all cases s(i) will represent the maximum value of combinations possible for the corresponding values of w(i) and g(i). Therefore, to determine the total number of possible ways to pack, all we have to do is add all the values of s(i). The result is 2,049 so that is the answer

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