Nice answers guys. I am in a rush today (hope to return pm tomorrow. Take good care of the place) so here are the answers; Pick the ones you want.
Pizza
The area of a circle is proportional to the square of its radius. It follows that we are comparing 16/3, 25/4, 36/6 and 49/8
Mark:
PLUMS
m/4 + n/3 + m/4 + m/6 = 10
m/2 + n/2 = 10
m + n = 20
Plums
The total expenditure is 10 = m/4 + n/3 + m/4 + n/6 = (m + n)/2.
Mark:
TABLES & CHAIRS
revenue = $1200
chairs = 24
tables = 48
TABLE & CHAIRS
The problem is to maximize 10C + 20T subject to:
10C + 5T <= 10*60
5C + 5T <= 6*60
5C + 20T <= 18*60
The solution space is a pentagon bounded by:
C = 0
T = 0
and the three lines from above.
The optimal answer will be at the intersection of two of the lines. You just need to compute the revenue function at the four non-trivial intersections ((0,0) can't be optimal).
For complex problems of this sort, there is an algorithm called the simplex method.
PLUMS
m/4 + n/3 + m/4 + m/6 = 10
m/2 + n/2 = 10
m + n = 20
Raprap:
By varying the production rate you can get better machine utilization
The optimal production rate curve is
# Tables # Chairs Prod ($)
54 0 1080
53 4 1100
52 8 1120
51 12 1140
50 16 1160
49 20 1180
48 24 1200
47 25 1190
46 26 1180
45 27 1170
44 28 1160
43 29 1150
42 30 1140
41 31 1130
As you can see it maximizes at 48 tables per hour and 24 chairs per hour
This is also accompanied by 100% usage of two of the machines.
A factory
The maximum profit per hour, of $1200, comes when 24 chairs and 48 tables are produced per hour.
The 10 saws can produce 600 minutes of work per hour (10 saws * 60 minutes).
The 6 lathes can produce 360 minutes of work per hour (6 lathes * 60 minutes).
The 18 sanding machines can produce 1080 minutes of work per hour (18 sanding machines * 60 minutes).
Let c be the number of chairs produced per hour and t the number of tables produced per hour.
The number of saws limit the combination of chairs and tables to 600=10c+5t.
The number of lathes limit the combination of chairs and tables to 360=5c+5t.
The number of sanding machines limit the combination of chairs and tables to 1080=5c+20t.
Next graph these three lines. It should be expected that the answer will lie on the intersection of two of these lines or to make all chairs or all tables. The intersection of the saw and sanding machine line occurs outside of how many chairs the lathe can make so this combination is not a viable answer. The saw and lathe lines cross at 48 chairs and 24 tables. The lathe and sanding machine lines cross at 24 chairs and 48 tables.
Next determine the revenue at all points of intersection.
Chairs Tables Revenue
60
. 0
. 600
48
24
.. ..960
24
48
.. 1200
0
.. 54
.. 1080
So the optimal answer is to make 24 chairs and 48 tables for revenue of $1200 per hour.
To check it will take 24*10 + 48*5 = 480 minutes of saw time. There are 600 minutes available so the saws will be idle 20% of the time.
It will take 24*5 + 48*5 = 360 minutes of lathe time, which is exactly what we have.
It will take 24*5 + 48*20 = 1080 minutes of sanding machine time, which is exactly what we have.
Mark:
DOWRY
The strategy is to note the maximum of the dowries of the first N women, then pick the first woman after that with a dowry that is greater.
Rap:
This is an extension of Baysean probabilities, and if memory serves me from a class I once took on quality assurance sampling let a third of the sample pass and pick the next one with a value greater (or less) than any of the first third.
I haven't worked this out so this is just be a memory induced inductive strategy and not an analytical solution.
I am glad to see you were paying attention. Great answers.
You should let 37 ladies go by and select the first one with a dowry greater than the maximum of the first 37 dowry's.
Let x be the number of ladies you pass before choosing the next lady with a dowry greater than the first x ladies. Let max x be the maximum dowry in the first x ladies.
The probability of winning is sum for i=x+1 to 100 of the probability that the highest dowry is in the ith position multiplied by the probability that you choose it if it is in that position.
This is based on the condition probability formula: Pr(A)=Sum for i=1 to n of Pr( A | Bi ) * Pr( Bi ), where the sum for i=1 to n of Pr(Bi)=1.
For all i the probability that the highest dowry is in that position is 1/100.
The probability that you will choose the dowry if it is in the ith position is equal to the probability that the highest of the first i-1 dowries belongs to one of the first x ladies. This equals x/(i-1). If the highest dowry of the first i-1 were not in the first x ladies you would choose it before getting to the largest dowry, thus losing the game.
For example if you let 30 ladies pass the probability that the highest dowry is in the 75th postion and that you will choose it equals 1/100 * 30/74.
Let A denote the overall probability of winning. The proability of winning given x is the sum for i=x+1 to 100 of x/(i-1) * 1/n. This equals:
Pr(1/n * [ 1 + x/(x+1) + x/(x+2) + ... + x/99 ] ).
Some value of x must be optimal to maximize this formula. This value of x must the first such that Pr(A|x+1) - Pr(A|x) < 0.
Pr(A|x+1) - Pr(A|x) = 1/n * [ 1/(x+2) + 1/(x+3) + 1/(x+4) + ... + 1/99 - x/(x+1) ].
let y=x+1.
Pr(A|x+1) - Pr(A|x) = 1/n * [ 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99 - (y-1)/y ]. =
Pr(A|x+1) - Pr(A|x) = 1/n * [ 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99 - 1 + 1/y ]. =
Pr(A|x+1) - Pr(A|x) = 1/n * [ 1/y + 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99 - 1 ].
If Pr(A|x+1) - Pr(A|x) = 0, then 1/y + 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99 = 1.
Theorem: The integral from r to n of 1/x = log(n) - log(r) = log(n/r).
The equation above the theorm is an approximation of the integral in the theorem. By applying the theorem log(100/y) = 1.
Taking e to the power of both sides: 100/y = e.
y=100/e.
Since e=~2.7182818 y=~36.79, since x=y-1 x=~35.79.
Since the integral is an underestimate of the series look at some specific values of Pr(A) for values of x close to 35.79:
x A
--- ---
35 37.0709%
36 37.1015%
37 37.1043%
38 37.0801%
So the optimal strategy is to let 37 ladies pass and then choose the first lady with a dowry larger than the greatest of the first 37.
In general the answer is going to be n/e, where n is the total number of ladies. Because the answer must be an exact integer and error in applying the theorem mentioned above you should check the few integers just above the optimal value. As n increases the error decreases. The probability of winning turns out to approach 1/e =~ 36.79% as n approaches infinity (I'll leave this proof up to you).
Here are some optimal values of x for various values of n and the probability of winning given x:
Code:
n x Pr(winning)
---- --- -----------
3 2 50.00%
4 2 43.83%
5 3 43.33%
6 3 42.78%
7 3 41.43%
8 4 40.98%
9 4 40.60%
10 4 39.87%
15 6 38.94%
20 8 38.42%
30 12 37.87%
50 19 37.42%
100 38 37.10%
1000 369 36.82%
For more information I suggest section 2.5 of Probability and Statistics, second edition by Morris H. DeGroot.
At McRonalds you can order Chicken Nuggets in boxes of 6, 9, and 20. What is the largest number such that you cannot order any combination of the above to achieve exactly the number you want
You need to warm milk in a baby bottle from its initial temperature of 15 degrees centigrade to 25 degrees. You put the bottle in a pot of boiling water which stays at constant temperature of 100 degrees.
The thickness and conductivity of the bottle are such that the initial rate of heat transfer is 85 degrees per minute. However, heat transfer is proportional to the difference between the temperature of the milk and the water.
How many minutes will it take to heat the bottle to 25 degrees
