M230 Test III Fall 1998 Solutions

1. The Feasible set is shaded:
tabular24
The maximum value of x - 3y is 9 at the point (9, 0). The maximum value of y-3x is 0 at the point (0,0). So the points are not the same.

2.
tex2html_wrap182

tex2html_wrap183

So, tex2html_wrap184

3. Let
tex2html_wrap185

tex2html_wrap186 R1-R2 tex2html_wrap_inline212 tex2html_wrap187
tex2html_wrap_inline214 tex2html_wrap188 tex2html_wrap_inline216 tex2html_wrap189

Thus tex2html_wrap190

4. Let
a = # of airplanes produced
b = # of boats produced
c = # of cars produced

Then tex2html_wrap_inline224, tex2html_wrap_inline226 and tex2html_wrap_inline228, since it is not possible to produce a negative number of anything. Also the plastic used cannot exceed 10,500 grams, so :
tex2html_wrap_inline230

The total wood usage cannot exceed 1500 inches, so :
tex2html_wrap_inline232

Finally the steel usage cannot be greater than the 25,500 grams available, so:
tex2html_wrap_inline234

The total profit for a given production schedule is:
3a + 5b + 4c ,
so this function needs to be maximized subject to the given six constraints.

5.
(a) Transition diagram:

(b) Transition matrix: tex2html_wrap191 (States in order 1-2-3 for rows and columns.)
OR tex2html_wrap192

BONUS tex2html_wrap193

So the probability of state # 2 after 3 years (after starting in state # 2) is tex2html_wrap_inline238. (underlined in tex2html_wrap_inline240).

OR using the tree diagram below: The probability a person infected now is also infected 3 years from now is:
tex2html_wrap_inline242


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Author: Rebecca Sullivan
Wed Feb 3 19:56:40 EST 1999