ANSWERS TO HOMEWORK SET 1
Problem 1.5
(a) Step 1: Ask the question.
Variables:
x =population (whales)
E =level of effort (boat-days)
g =growth rate (whales per year)
h =harvest rate (whale per year).
Assumptions:
g =0.08x(1 − x/400, 000)
h =0.00001Ex
g =h, x ≥ 0, E ≥ 0.
Objective: Maximize h.
Step 2: Select the modeling approach.
We will model this problem as a one variable optimization problem.
Step 3: Formulate the model.
Let y = h, and write
y = f (x) = 0.08x(1 − x/400, 000).
Our goal is to maximize f (x) over the internal x ≥ 0.
Step 4: Solve the model.
Solving f 0 (x) = 0 for x we get x = 200, 000. f (200, 000) = 8000. Since the graph
of f (x) is a parabola, we know this is the global maximum. Also, the correspond
value of E equals
8000/(0.00001 × 200, 000) = 4000.
Step 5: Answer the question.
According to this model, the optimal policy is to harvest 8000 whales per year,
which requires controlling the level of effort at 4000 boat-days per year. This will
maintain the population of whales at 200,000 which is higher than the current
population. This indicates that in the past the harvesting rate has exceeded the
optimum level according to this model.
(b) Generalize the model in part (a) to obtain the assumption
g = rx(1 − x/400, 000)
where currently r = 0.08. Then we have
y = f (x) = rx(1 − x/400, 000)
and the optimum is still at x = 200, 000 but now
f (x) = 100, 000r
which leads to
E = 50, 000r.
1
2
ANSWERS TO HOMEWORK SET 1
Then
S(x, r) = 0, S(E, r) = 1,
because x does not depend on r and E is proportional to r.
(c) Generalize the model in part (a) to obtain the assumption
g = 0.08x(1 − x/K)
where currently K = 400, 000. Then we have
y = f (x) = 0.08x(1 − x/K)
and the optimum is at x = K/2, f (x) = 0.02K which lead to E = 4000. Then
S(x, K) = 1,
S(E, K) = 0,
because E does not depend on K, and x is proportional to K.
Problem 2.5
(a) Step 1: Ask the question.
Variables:
s =sales of 19 inch sets (sets/year)
t =sales of 21 inch sets (sets/year)
p =price for 19 inch sets ($/set)
q =price for 21 inch sets ($/set)
T =tariff ($/set)
C =manufacturing cost ($/year)
R =revenue ($/year)
P =profit ($/year)
Assumptions:
p =339 − 0.01s − 0.003t
q =399 − 0.004s − 0.01t
T =25(s + t)
R =ps + qt
C =400, 000 + 195s + 225t + T
P =R − C
s ≥0
t ≥0
Objective: Maximize P .
Step 2: Select the modeling approach.
We will model this problem as a multivariable unconstrained optimization problem.
Step 3: Formulate the model.
Let x = s, y = t, and z = P , and write
z = f (x, y) =(339 − 0.01x − 0.003y)x + (399 − 0.004x − 0.01y)y
− (400, 000 + 195x + 225y + 25(x + y))
ANSWERS TO HOMEWORK SET 1
3
Our goal is to maximize f (x, y) over the set of all (x, y) for which x ≥ 0 and y ≥ 0.
Figure 1. Graph of z = f (x, y)
Figure 2. Level curves of f (x, y)
Step 4: Solve the model.
Solving the system
∂z/∂x = −0.02x + 119 − 0.007y = 0,
∂z/∂y = −0.007x − 0.02y + 149 = 0
yields
x = 3809.12, y = 6116.81.
The corresponding value of z is 282344.73. The graph indicates that the maximum
occurs at this unique local extreme point.
Step 5: The optimal strategy is to produce 3809 of the 19” LCD TV sets and
6117 of the 21” LCD TV sets per year. This will result in about $282,000 profit
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ANSWERS TO HOMEWORK SET 1
per year. In the no-tariff case we made $554,000 per year, so the tariff has cost the
company $272,000 in profits, almost half. Of this amount $248,000 is paid to the
government as a tariff and the remaining $24,000 is due to lost sales.
(b) If we move to the US, net operating costs will increase by $350,000 per year.
The optimal production policy in this case is the same as the non-tariff case, i.e.
we would make 4735 of the 19 inch and 7043 of the 21 inch sets. We expect profits
of 553, 641 − 350, 000 = 203, 641 dollars per year, and so it would not be better to
move to the US.
(c) We will use sensitivity analysis to estimate the minimum increase in tariffs
that will make it optimal for the company to relocate to the US. Let r denote the
tariff amount, where we currently assume r = 25($/set). We have
z = f (x, y) =(339 − 0.01x − 0.003y)x + (399 − 0.004x − 0.01y)y
− (400, 000 + 195x + 225y + r(x + y))
∂z/∂x = −0.02x + 144 − 0.007y − r,
By solving
∂z
∂x
=
∂z
∂x
∂z/∂y = −0.007x − 0.02y + 174 − r.
= 0, we get
x = 4735.04 − 37.04r,
y = 7042.74 − 37.04r.
We also have
dz/dr = −11777.78 + 74.07r.
Then S(z, r) = −0.88 and so if the tariff goes up by 10% the company’s expected
profits decrease by 8.8%. To motivate the company to relocate we have to decrease
the expected profit form 282,000 to 204,000 which is a decline of 28%. This requires
a tariff increase of 32% to $33 per set. Since this is just an estimate, and since we
want the US relocation to be a better option, not just an equally good option, we
recommend setting the tariff at $35 per set or more.
(d) We have already calculated S(z, r) = −0.88. In part (c) we know that
dx/dr = dy/dr = −37.04. Then S(x, r) = −0.24 while S(y, r) = −0.15. Given that
the only motivation of the tariff is to induce the company to relocate to the US,
we should set the tariff as low as possible, given that the tariff is large enough to
motivate the move. So we recommend a tariff of $35 per set. The only significant
result of a tariff higher than $35 is to lower company profits.
Problem 2.6
(a) Step 1: Ask the question.
Variables:
p =price ($/computer)
s =sales (computers/month)
a =advertising budget ($/month)
C =manufacturing cost ($/month)
R =revenue ($/month)
P =profit ($/month).
ANSWERS TO HOMEWORK SET 1
5
Assumptions:
C =700s + a
s =10000 + 5000(950 − p)/100 + 200(a − 50000)/10000
R =ps
P =R − C
p ≥0, 0 ≤ a ≤ 100000.
Objective: Maximize P .
Step 2: Select the modeling approach.
We will model this problem as a multivariable contained optimization problem.
Step 3: Formulate the model.
Let x = p, y = a and z = P , and write
z =f (x, y)
h
i
=(x − 700) 10000 + 5000(950 − x)/100 + 200(y − 50000)/10000 − y
y
− y.
=(x − 700) 56500 − 50x +
50
Our goal is to maximize f (x, y) over the set
S = {(x, y) : 0 ≤ x, 0 ≤ y ≤ 10000}.
Figure 3. Graph of z = f (x, y)
Step 4: Solve the model.
First, we look for possible maxima in the interior of the feasible region S. We
have
dz
y
dz
x
= 91500 − 100x + ,
=
− 15.
dx
50
dy
50
dz
dz
Solving the system dx
= dy
= 0, we obtain
x = 750,
y = −825000,
which is outside the feasible region. Therefore, there are no critical point in the
interior.
Now look along the constraint line
g(x, y) = y = 100000.
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ANSWERS TO HOMEWORK SET 1
Figure 4. Level curves of f (x, y)
Note that ∇g = h0, 1i. Solving the system

y


 91500 − 100x + 50 = 0
x
− 15 = λ

50


y = 10000,
we obtain
The maximum occurs when
Along the constraint line








 x = 935
y = 100000

λ = 3.7.
x = 935. At this point z = 2661250.
y = 0, we similarly solve the system
y
91500 − 100x +
= 0
50
x
− 15 = λ
50
y = 0,
we obtain

 x = 915
y = 0

λ = 3.3.
The maximum occurs when x = 915. At this point z = 2311250.
Along the constraint line x = 0,
f (x, y) = f (0, y) = −5(7910000 + 3y).
It is linear, and for 0 ≤ y ≤ 100000, the maximum occurs at the origin, where
z < 0.
Therefore, the maximum over the entire feasible region is at the point (935, 100000),
and the maximum is 2661250.
Step 5: Answer the question.
ANSWERS TO HOMEWORK SET 1
7
We recommend lowering the price slightly to $935 and spending $100,000 per
month on advertising. This should result in a monthly profit of around $2,660,000.
(b) Let e denote the price elasticity, where currently e = 50%. Let α = 100e.
In other words, every $1 drop in price is assumed to generate 50 more sales per
month. Now,
z =f (x, y)
h
i
=(x − 700) 10000 + α(950 − x) + 200(y − 50000)/10000 − y
h
yi
− y.
=(x − 700) 9000 + α(950 − x) +
50
Direct computation leads to
dz
= 9000 + α(950 − x) + y/50 − (x − 700)α,
dx
dz
= x/50 − 15.
dy
The key point is that for value of α near 50, the maximum should still occur on
the constraint line y = 100000. Therefore, by Lagrange multiplier method, we solve
the system

 9000 + α(950 − x) + y/50 − (x − 700)α = 0
x/50 − 15 = λ

y = 1000000,
and obtain
x = 275(3α + 20)/α,
y = 100000,
λ = (220 + 3α)/(2α).
It follow that
dx
= 825/α − 275(3α + 20)/α2 ,
dα
and at α = 50, i.e. e = 50%,
S(x, e) = S(x, α) =
dx α
· = −2/17 ≈ −0.118.
dα x
Since we are on the constraint line y = 100000 (or notice that
dy
dα
= 0), we have
S(y, e) = S(y, α) = 0
at α = 50, i.e. e = 50%.
Both the optimal selling price and the optimal advertising expenditure are quite
insensitive to the price elasticity. If the effect of lowering price is 10% greater
than expected (i.e. if lowering the price by $100 results in a 55% increase in sales)
then the optimal selling price is about 1.2% lower than expected, and the optimal
advertising budget stays the same.
(c) Let n denote the number of new sales for each $10,000 increase in the ad
budget. Currently, we assume n = 200. Now
z = f (x, y) = (x − 700)(10, 000 + 50(950 − x) + n(y − 50000)/10000) − y
and we need to maximize f (x, y) over the set of all (x, y) for which x ≥ 0 and
0 ≤ y ≤ 10000. For values of n near 200 the maximum should still occur on the
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ANSWERS TO HOMEWORK SET 1
constraint line y = 100, 000. Solving the system
∂z/∂x = 92500 − 100x + n(y − 50000)/10000 =0
∂z/∂y = (x − 700)n/10000 − 1 =λ
y =100000
we obtain
x = 925 + n/20, y = 100000, λ = −1 + 9n/400 + n2 /200000.
Therefore, S(x, n) = 2/187 ≈ 0.01069518717. Since we are on the constraint line
y = 100, 000, we have S(y, n) = 0. Both the optimal selling price and the optimal
advertising expenditure are quite insensitive to n.
(d) The value of the Lagrange multiplier was 3.7 which means that every 1$
increase in the advertising budget should result in about $3.70 in additional profits.
Management should probably consider a larger advertising budget, assuming that
the effect of advertising is as great as the ad agency claims.
Problem 3.1 (a) Notice that
dp
= 0.65 · (−0.01/0.65)e−(0.01/0.65)t = 0.01e−(0.01/0.65)t .
dt
We have
dp = 0.01.
dt t=0
0
It is also not hard to see that p (t) is decreasing, which means the rate of change
of the price decreases as t increases.
(b) The profit function here is
P (t) = 0.65e−(0.01/0.65)t (200 + 5t) − 0.45t
Figure 5. Graph of P (t), 0 ≤ t ≤ 20
By graphing method (see Figures 5,6,7) or Newton’s method, we obtain that
the optimal time to sell the pig is t = 13.86 days, and the corresponding profit is
P = 135.194 dollars.
(c) Similarly to part (b), for
P (t) = 0.65e−(0.0101/0.65)t (200 + 5t) − 0.45t,
ANSWERS TO HOMEWORK SET 1
9
Figure 6. Graph of P (t), 12 ≤ t ≤ 16
Figure 7. Graph of P (t), 13.5 ≤ t ≤ 14
applying either the graphing method or Newton’s method, we obtain the optimal
values
t = 13.39, P = 134.90
Comparing to the values obtained in part (b), we see that the best time to sell
is down by 3.4%, so that S(x, a) = −3.4, and the profit is down by 0.22%, so
S(P, a) = −0.22.
(d) We have replaced our linear price function with one which is nearly linear
over the range of t which is critical to the problem (0 to 20 days). Even so the
best time to sell the pig changed from 8 to 14 days. Our model is not robust with
regard to the linearity assumption.