<jupyter_start><jupyter_text>Problem 1: Nano XIX Sneakers<jupyter_code>import matplotlib.pyplot as plt
import numpy as np
from numpy import matrix<jupyter_output><empty_output><jupyter_text>Cost functionc(x) = 18x + 24000 Revenue functionr(x) = 120x  Breakeven point120x = 18x + 24000102x = 24000x = 24000/102r(x) = 120xr(24000/102) = 120(24000/102)y = 120*(24000/102) or approx. $28,235.29<jupyter_code>def graph(x_range, BE_X=None, BE_Y=None):  
    x = np.array(x_range)  
    rev_y = revenue_function(x)
    cost_y = cost_function(x)
    plt.plot(x, rev_y)
    plt.plot(x, cost_y)
    plt.legend (('r(x) = 120x','c(x) = 18x + 24000'),loc=4)
    plt.xlabel("Quantity of Nano XIX Sneakers")
    plt.ylabel("US Dollars $")
    plt.title("Nano XIX Cost and Revenue Functions")
    if BE_X == None:
        BE_X = round(24000/102, 2)
    if BE_Y == None:
        BE_Y = round(120*BE_X, 2)
    plt.annotate('breakeven point\n({}, {})'.format(BE_X, BE_Y), xy=(BE_X, BE_Y), xytext=(120, 85000),
                 arrowprops=dict(
                          connectionstyle="arc3,rad=0.",
                          shrinkA=0, shrinkB=10,
                          arrowstyle= '-|>', ls= '-', linewidth=2
                          ))
    plt.grid()
    plt.show()

def revenue_function(x):
    return 120*x

def cost_function(x):
    return 18*x + 24000

graph(range(0, 1000))<jupyter_output><empty_output><jupyter_text>Mathematical Breakeven Point is (235.29, 28,234.80)where cost and revenue are equal at $28,234.80 and number of sneakers produced is 235.29, mathematically.Since one cannot have a partial sneaker, assuming no hungry dogs were involved, we round up to 236.r(236) = 120*236 = $28,320  Practical Breakeven point (236, 28320)<jupyter_code>graph(range(0, 1000), BE_X=236, BE_Y=28320)<jupyter_output><empty_output><jupyter_text>Problem 2: Arielle's Investments The following equations apply to Arielle's investments, where a is mutual funds, b is government bonds, and c is CD's:a + b + c = 175001.11a + 1.07 + 1.05c = 18995a = 2cWritten as matrix:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;|&nbsp;&nbsp;175001.11&nbsp;&nbsp;1.07&nbsp;&nbsp;1.05&nbsp;&nbsp;&nbsp;|&nbsp;&nbsp;18995&nbsp;&nbsp;&nbsp;-1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;2&nbsp;&nbsp;&nbsp;&nbsp;|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0Using TI84 Calculator rref([A]) where A is the matrix above, the resulting matrix is below:1&nbsp;&nbsp;0&nbsp;&nbsp;0&nbsp;&nbsp;|&nbsp;&nbsp;90000&nbsp;&nbsp;1&nbsp;&nbsp;0&nbsp;&nbsp;|&nbsp;&nbsp;45000&nbsp;&nbsp;0&nbsp;&nbsp;1&nbsp;&nbsp;|&nbsp;&nbsp;4500a = 9000b = 4000c = 4500Plugging into the originals to double check:9000 + 4000 + 4500 = 175002(4500) = 90001.11(9000) + 1.07(4000) + 1.05(5400) = 18995 Therefore, Arielle invested 9000 USD in mutual funds, 4000 USD in government bonds, and 4500 USD in CDs.  Problem 3: Vandelay Industries a + b + c + d = 252a = 3bc = 2dwhere a is the first team, b is the second teamc is the third, and d is the fourth team.Augmented matrix:a&nbsp;&nbsp;&nbsp;b&nbsp;&nbsp;c&nbsp;&nbsp;d&nbsp;&nbsp;|1&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;1&nbsp;&nbsp;1&nbsp;&nbsp;|&nbsp;&nbsp;2521&nbsp;&nbsp;-3&nbsp;0&nbsp;&nbsp;0&nbsp;&nbsp;|&nbsp;&nbsp;00&nbsp;&nbsp;&nbsp;0&nbsp;1&nbsp;-2&nbsp;&nbsp;|&nbsp;&nbsp;0Using TI-84 graphing calculator with 3x5 matrix A and the amazing function rref([A]), the following is resulting matrix:a&nbsp;&nbsp;b&nbsp;&nbsp;c&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;d&nbsp;&nbsp;|1&nbsp;&nbsp;0&nbsp;&nbsp;0&nbsp;&nbsp;9/4&nbsp;&nbsp;|&nbsp;&nbsp;1890&nbsp;&nbsp;1&nbsp;&nbsp;0&nbsp;&nbsp;3/4&nbsp;&nbsp;|&nbsp;&nbsp;630&nbsp;&nbsp;0&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;-2&nbsp;&nbsp;|&nbsp;&nbsp;0Since there are four variables and only 3 equations, this is normal.The resulting matrix can tell us about the distribution in terms of the fourth team, d.a = 189 - 9/4db = 63 - 3/4dc = 2d (same as one of the original equations)We also still have the original equation a = 3b The third team is twice the fourth team. The first team is thrice the third team, or the first team is 189 less 9/4 of the fourth team. The second team is 63 less 3/4 the fourth team.  Problem 4: Phil's Candy<jupyter_code>candy = matrix([[6,8,1],[6,4,1],[5,7,1]])
print (candy)
K = matrix([[3],[5],[2]])
print(K)
cost_of_one_batch = np.dot(candy,K)
cost_of_100_batches = np.dot(cost_of_one_batch, 100)
print("cost of 100 batches using supplier K : \n{}".format(cost_of_100_batches))
print("\ntotal cost : {}".format(sum(cost_of_100_batches)))<jupyter_output>cost of 100 batches using supplier K : 
[[6000]
 [4000]
 [5200]]

total cost : [[15200]]<jupyter_text>cost of 100 batches using supplier K in $| chocolate bar type | cost ||:------------------:|:----:|| cherry | 6000 || almond | 4000 || raisen | 5200 |  Problem 5: Welsh-Ryan Arena Where c is courtside, f is first level, and u is upper deck:    c + f + y = 150008c + 6f + 4u = 760004c + 6f + 2u = 44000Written as an Augmented matrix:1&nbsp;&nbsp;1&nbsp;&nbsp;1&nbsp;&nbsp;|&nbsp;&nbsp;150008&nbsp;&nbsp;6&nbsp;&nbsp;4&nbsp;&nbsp;|&nbsp;&nbsp;760004&nbsp;&nbsp;6&nbsp;&nbsp;2&nbsp;&nbsp;|&nbsp;&nbsp;44000Using a Ti-84 calculator rref([A]) where A is the above matrix, the resulting matrix is below:1&nbsp;&nbsp;0&nbsp;&nbsp;0&nbsp;&nbsp;|&nbsp;&nbsp;&nbsp;30000&nbsp;&nbsp;1&nbsp;&nbsp;0&nbsp;&nbsp;|&nbsp;&nbsp;&nbsp;20000&nbsp;&nbsp;0&nbsp;&nbsp;1&nbsp;&nbsp;|&nbsp;&nbsp;10000c = 3000f = 2000u = 10000This makes sense from a glance that upper deck seats are the most plentiful. We can be sure by plugging these values into the original equations:3000 + (2000) + 10000 = 15000 is correct8(3000)+ 6(2000) + 4(10000) = 76000 is correct4(3000) + 6(2000) + 2(10000) = 44000 is also correct Therefore, there are 3000 courtside seats, 2000 first level, and 10,000 upper deck seats in the stadium.  Problem 6: Hoch Industries |                       |  old  |  new  ||----------------------:|-------|-------|| quantity of processes |   x   |   y   ||    sulphur            |  6x   |  2y   ||      lead             |  3x   |  4y   |6x + 2y <= 180003x + y <= 9000z = 25x + 16yGraphing y <= -3x + 9000 for sulphur and y <= -3/4x + 3000 for lead:<jupyter_code>x = np.arange(0, 100000, 10)

y_sulphur = -3*x + 9000

y_lead = -3/4*x + 3000

y3 = np.minimum(y_sulphur, y_lead) # the smaller of sulphur and lead functions, since both have y<=
plt.xlim(0, 10000)
plt.ylim(0, 10000)
plt.xlabel('x-axis')
plt.ylabel('y-axis')
plt.title('Shaded Area Shows the Feasible Region')
plt.plot(x, y_sulphur, color='b', label='sulphur y <= -3x + 9000')
plt.plot(x, y_lead, color='r', label='lead y <= -3/4x + 3000')
plt.legend()
plt.fill_between(x, 0, y3, color='grey',alpha=0.2)
plt.show()<jupyter_output><empty_output><jupyter_text>corner points are where lead function hits the y-axis, where lead and sulphur functions meet, and where sulphur function hits the x-axis.lead&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;y <= -3/4x + 3000sulphur&nbsp;&nbsp;y <=&nbsp;&nbsp;&nbsp;&nbsp;-3x + 9000-3/4x + 3000 = -3x + 9000x = 6000/2.25 ~= 2667y = -3(6000/2.25) + 9000 = 1000| Corner Points | z = 25x+16y|--------------:|-------------:||(0, 0)|0|(0, 3000)|48000|(6000/2.25, 1000)|82667|(3000, 0)|75000The point where the two functions meet give the maximum value.   Hoch Industries should run 2667 of the old and 1000 of the new.  Problem 7: Northwestern<jupyter_code>x = np.arange(0, 1000, 1)

y1 = -x + 45
y2 = -x + 30
y3 = [10 for i in range(0, 1000)]
y4 = (2/3)*x

plt.xlim(0, 50)
plt.ylim(0, 50)
plt.xlabel('x-axis')
plt.ylabel('y-axis')
plt.title('Shaded Area Shows the Feasible Region')
plt.plot(x, y1, color='b', label='y1 <= -x + 45 ')
plt.plot(x, y2, color='purple', label='y2 >= -x + 30')
plt.hlines(10, 0, 50, color='magenta', label='y3 >= 10')
plt.plot(x, y4, color='cyan', label='y4 <= 2/3x')
plt.legend()
x_fill = [20, 18, 27, 35]
y_fill = [10, 12, 18, 10]
plt.fill_between(x_fill, y_fill, 10, alpha=0.3)
plt.show()<jupyter_output><empty_output><jupyter_text>|          | z = 2400x + 1100y|---------:|------------------|| (20,10)  |   59,000| (18,12)  |   56,400| (27,18)  |   84,600| (35,10)  | 95,000 Northwestern should staff 18 teachers and 12 TA's for a minimized cost of $56,400.  Problem 8: Roger the Athlete |               | pill 1 | pill 2 | pill 3 | mins|--------------:|--------|--------|--------|------|| qty of pills  |   y1   |  y2    |   y3    |  -|    A          |  4y1   |  1y2   | 10y3   | >=10|    B          |  3y1   |  2y2   |  1y3   | >=12|    C          |   0y1  |  4y2   |  5y3   | >=20Mimimize cost: z = 0.06y1 + 0.08y2 + 0.01y34y1 + y2 + 10y3 >= 103y1 + 2y2 + y3 &nbsp;&nbsp;>= 12&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;4y2 + 5y3 &nbsp;>= 20y1 >= 0y2 >= 0y3 >= 0We cannot graph these... but we can attempt a matrix. Since there are 3 equations, there will be 3 slack variables. A minimization problem is also just a maximization problem with the objection function z = -wz = -0.06y1 - 0.08y2 - 0.01y3y1&nbsp;&nbsp;&nbsp;y2&nbsp;&nbsp;&nbsp;y3&nbsp;&nbsp;&nbsp;s1&nbsp;&nbsp;&nbsp;s2&nbsp;&nbsp;&nbsp;s3&nbsp;&nbsp;&nbsp;z&nbsp;|4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;3&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;| 0.061&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;4&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;| 0.0810&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;5&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;| 0.01-------------------------------------------------------------------------------------10&nbsp;-12&nbsp;-20&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;| 0-10 is the least negative so will be the pivot column.<jupyter_code>print(.06/4)
print(.08/1)
print(.01/10)<jupyter_output>0.015
0.08
0.001<jupyter_text>10 will be the pivot because it's the smallest quotient.Resulting matrix comes from the following row transformations:-4/10R3 + R1 -> R1-1/10R3 + R2 -> R2R3 + R4 -> R4on a Ti-84 calculator, where [B] is the Matrix shown above:  *row+(-2/5, [B], 3, 1) -> [C]*row+(-1/10, [C], 3, 2) -> [D]*row+(1, [D], 3, 4) -> [E]y1&nbsp;&nbsp;&nbsp;y2&nbsp;&nbsp;&nbsp;y3&nbsp;&nbsp;&nbsp;s1&nbsp;&nbsp;&nbsp;s2&nbsp;&nbsp;&nbsp;s3&nbsp;&nbsp;&nbsp;z&nbsp;|0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;2.6&nbsp;&nbsp;-2&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;-0.4&nbsp;&nbsp;0&nbsp;| 0.0560&nbsp;&nbsp;&nbsp;&nbsp;1.9&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;-11&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;| 0.0810&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;5&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;| 0.01------------------------------------------------------------------------------------0&nbsp;&nbsp;-11&nbsp;-15&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;| 0.01-11 is the least negative so will be the next pivot column.<jupyter_code>print(0.056/2.6)
print(0.08/1.9)
print(0.01/1)<jupyter_output>0.021538461538461538
0.04210526315789474
0.01<jupyter_text>1 will be the pivot, in the third row.Resulting matrix comes from the following row transformations:-2.6/1R3 + R1 -> R1-1.9/1R3 + R2 -> R211/1R3 + R4 -> R4on a Ti-84 calculator, where [B] is the Matrix shown above:  *row+(-2.6, [E], 3, 1) -> [F]*row+(-1.9, [F], 3, 2) -> [G]*row+(11, [G], 3, 4) -> [H]y1&nbsp;&nbsp;&nbsp;y2&nbsp;&nbsp;&nbsp;y3&nbsp;&nbsp;&nbsp;s1&nbsp;&nbsp;&nbsp;s2&nbsp;&nbsp;&nbsp;s3&nbsp;&nbsp;&nbsp;z&nbsp;|-26&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;-15&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;-3&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;| 0.03-19&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;-6&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;-2&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;| 0.0610&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;5&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;| 0.01------------------------------------------------------------------------------------110&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;40&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;12&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;| 0.12There are no more negative indicators, so the optimum solution has been achieved. y1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;y2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;y3&nbsp;&nbsp;&nbsp;&nbsp;s1&nbsp;&nbsp;&nbsp;s2&nbsp;&nbsp;&nbsp;s3&nbsp;&nbsp;&nbsp;&nbsp;z&nbsp;&nbsp;&nbsp;&nbsp;|-2600&nbsp;0&nbsp;&nbsp;-1500&nbsp;&nbsp;100&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;-300&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;| 3-1900&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;-600&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;100&nbsp;-200&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;| 61000&nbsp;&nbsp;100&nbsp;&nbsp;500&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;100&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;| 1------------------------------------------------------------------------------------11000&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;4000&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;&nbsp;&nbsp;0&nbsp;&nbsp;1200&nbsp;100&nbsp;| 12 Basic variables from this Tableau are y2, s1, and s2.y1 = 0y2 = 1y3 = 0s1 = 3s2 = 6s3 = 0and z = 12  Problem 9: Better Buy Inventory Better Buy has a capacity of 210 items. The fact that they don't take the size of the items into account is something I wouldn't point out if I was their consultant, because it makes this problem simpler.x1: DVD playersx2: surround sound stereo speakersx3: smart tvsx1 + x2 + x3 = 2102x2 = x1x3 >= 30z = 450x1 + 2000x2 + 750x3I notice x3 has to be greater than or equal to 30, but it's also kind of cheap.Based on x2's relationship with x1, and taking a shot at x3 = 30 since it's less than half the price of x2, we can rewrite the first equation:2x2 + x2 + 30 = 2103x2 = 180x2 = 60 implying x1 = 120z = 450(120) + 2000(60) + 750(30) = $196,500if x3 is a little larger, it makes the numbers not round out at 31 and 32, until it gets to 33. So if x3 = 33, x2 = 59, and x1 = 118, and profit is lower at $195,850.I wouldn't use this kind of deductive reasoning in every situation, but 210 is not a very high number, so I don't mind shooting darts until I reach the answer, kind of like neural networks do.I did some iterations on the calculator and it looked linear that the higher x3 got, the lower the profit, but just to be safe, I decided to write a short python program (below), finding it is safe to conclude the following:  120 DVD players, 2000 surround sound stereo systems, and 30 smart TV's yields the maximum profit of $196,500<jupyter_code>profits = []
all_the_info = []

def find_x2(x3):
    return (210 - x3)/3

def find_x1(x2):
    return 2*x2

def find_z(x1, x2, x3):
    return 450*x1 + 2000*x2 + 750*x3

for x3 in range(30, 211):
    x2 = find_x2(x3)
    x1 = find_x1(x2)
    z = find_z(x1, x2, x3)
    profits.append(z)
    all_the_info.append({z:[x1,x2,x3]})

max(profits)
plt.hist(profits, bins=len(profits))
plt.title("Does max profit claim of $196,500 occur more than once in list profits?")
plt.show()<jupyter_output><empty_output><jupyter_text>Since 196,500 only occurs once in the profits list, this conclusion is valid.  Problem 10: McDowell's Sweepstakes P(A|winner) = P(A) * P(winner|A) / (P(A) * P(winner|A) + P(B) * P(winner|B))P(A) = 0.27P(A') = P(B) = 1 - 0.27 = 0.73P(winner|A) = 0.04P(winner|B) = 0.05P(A|winner) = [(0.27) * (0.04)] / [(0.27) * (0.04) + (0.73) * (0.05)] The probability of a winner coming from box A is 22.83%<jupyter_code>num = 0.27*0.04
b = 0.73*0.05

print("{} %".format(round(num/(num+b)*100, 2)))<jupyter_output>22.83 %
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