Problem 1. Find the local optima of f (x1 , x2 ) = 8x13 − 12x1 x2 + x23 .
The gradient of f is
∇ f (x1 , x2 ) = [
24x12 − 12x2
]
−12x1 + 3x22
Solving the system of equations given by ∇ f (x1 , x2 ) = 0, we find that there are two critical points: (x1 , x2 ) =
(0, 0) and (x1 , x2 ) = (1, 2).
The Hessian matrix of f is
H(x1 , x2 ) = [
48x1 −12
]
−12 6x2
At the critical point (x1 , x2 ) = (0, 0), we have
0 −12
H(0, 0) = [
]
−12 0
∣H1 ∣ = 0 ∣H2 ∣ = −144
and so (0, 0) is a saddle point of f .
At the critical point (x1 , x2 ) = (1, 2), we have
48 −12
H(1, 2) = [
]
−12 12
∣H1 ∣ = 48 ∣H2 ∣ = 432
and so f (1, 2) = −8 is a local minimum of f .
1
Problem 2. Find the local optima of f (x1 , x2 , x3 ) = −x14 − 2x22 − x32 + 4x1 x2 + 2x3 .
The gradient of f is
⎡−4x 3 + 4x2 ⎤
⎥
⎢
1
⎥
⎢
∇ f (x1 , x2 , x3 ) = ⎢ 4x1 − 4x2 ⎥
⎥
⎢
⎢ −2x3 + 2 ⎥
⎦
⎣
Solving the system of equations given by ∇ f (x1 , x2 ) = 0, we find that there are three critical points: (x1 , x2 , x3 ) =
(−1, −1, 1), (x1 , x2 , x3 ) = (0, 0, 1), and (x1 , x2 , x3 ) = (1, 1, 1).
The Hessian matrix of f is
⎡−12x 2 4
0 ⎤⎥
⎢
1
⎢
⎥
−4 0 ⎥
H(x1 , x2 , x3 ) = ⎢ 4
⎢
⎥
⎢ 0
0 −2⎥⎦
⎣
At the critical point (x1 , x2 , x3 ) = (−1, −1, 1), we have
⎡−12 4
0 ⎤⎥
⎢
⎢
⎥
H(−1, −1, 1) = ⎢ 4 −4 0 ⎥
⎢
⎥
⎢ 0
0 −2⎥⎦
⎣
∣H1 ∣ = −12
∣H2 ∣ = 32 ∣H3 ∣ = −64
and so f (−1, −1, 1) = 2 is a local maximum.
At the critical point (x1 , x2 , x3 ) = (0, 0, 1), we have
⎡0 4
0 ⎤⎥
⎢
⎢
⎥
H(0, 0, 1) = ⎢4 −4 0 ⎥
⎢
⎥
⎢0 0 −2⎥
⎣
⎦
∣H1 ∣ = 0 ∣H2 ∣ = −16
∣H3 ∣ = 32
and so (−1, −1, 1) is saddle point of f .
At the critical point (x1 , x2 , x3 ) = (1, 1, 1), we have
⎡−12 4
0 ⎤⎥
⎢
⎢
⎥
H(1, 1, 1) = ⎢ 4 −4 0 ⎥
⎢
⎥
⎢ 0
0 −2⎥⎦
⎣
∣H1 ∣ = −12
and so f (1, 1, 1) = 2 is a local maximum.
2
∣H2 ∣ = 32 ∣H3 ∣ = −64
Problem 3. Is f (x1 , x2 , x3 ) = x12 +4x22 +2x32 + x1 x3 + x2 x3 +8 strictly convex or strictly concave? Why? Suppose
f (0, 0, 0) = 8 is a local minimum of f . Is f (0, 0, 0) = 8 also an absolute minimum? Why?
The gradient of f is
⎡ 2x1 + x3 ⎤
⎢
⎥
⎢
⎥
∇ f (x1 , x2 , x3 ) = ⎢ 8x2 + x3 ⎥
⎢
⎥
⎢x1 + x2 + 4x3 ⎥
⎣
⎦
The Hessian matrix of f is
⎡2 0 1 ⎤
⎢
⎥
⎥
⎢
H(x1 , x2 , x3 ) = ⎢0 8 1 ⎥
⎢
⎥
⎢ 1 1 4⎥
⎣
⎦
and so
∣H1 ∣ = 2
∣H2 ∣ = 16
∣H3 ∣ = 54
Therefore, f is strictly convex. As a result, f (0, 0, 0) = 8 is an absolute minimum, since for a strictly convex
function, every local minimum is also an absolute minimum.
3
Problem 4. Find the local optima of f (x1 , x2 ) = −x12 + x22 subject to the constraint x12 + 4x22 = 4.
The Lagrangian function is
Z(x1 , x2 , λ) = −x12 + x22 + λ[4 − x12 − 4x22 ]
The gradient of Z is
⎡ −2x1 − 2x1 λ ⎤
⎥
⎢
⎥
⎢
∇Z(x1 , x2 , λ) = ⎢ 2x2 − 8x2 λ ⎥
⎥
⎢
⎢4 − x12 − 4x22 ⎥
⎦
⎣
Solving the system of equations given by ∇Z(x1 , x2 , λ) = 0, we find that there are four critical points:
(x1 , x2 , λ) = (−2, 0, −1), (x1 , x2 , λ) = (2, 0, −1), (x1 , x2 , λ) = (0, −1, 1/4), (x1 , x2 , λ) = (0, 1, 1/4).
The bordered Hessian matrix is
⎡ 0
2x1
8x2 ⎤⎥
⎢
⎥
⎢
0 ⎥
H(x1 , x2 , λ) = ⎢ 2x1 −2 − 2λ
⎥
⎢
⎢8x2
0
2 − 8λ⎥⎦
⎣
At the critical point (x1 , x2 , λ) = (−2, 0, −1), we have
⎡ 0 −4 0 ⎤
⎥
⎢
⎥
⎢
H(−2, 0, −1) = ⎢−4 −4 0 ⎥
⎥
⎢
⎢0
0 −6⎥⎦
⎣
∣H 2 ∣ = −160
Therefore, f (−2, 0) = −4 is a local minimum.
At the critical point (x1 , x2 , λ) = (2, 0, −1), we have
⎡0 4 0 ⎤
⎥
⎢
⎥
⎢
H(2, 0, −1) = ⎢4 0 0 ⎥
⎥
⎢
⎢0 0 10⎥
⎦
⎣
∣H 2 ∣ = −160
Therefore, f (2, 0) = −4 is a local minimum.
At the critical point (x1 , x2 , λ) = (0, −1, 1/4), we have
⎡0
0
−8⎤⎥
⎢
⎢
⎥
H(0, −1, 1/4) = ⎢ 0 −5/2 0 ⎥
⎢
⎥
⎢−8
0
0 ⎥⎦
⎣
∣H 2 ∣ = 160
Therefore, f (0, −1) = 1 is a local maximum.
At the critical point (x1 , x2 , λ) = (0, 1, 1/4), we have
⎡0
0
8⎤⎥
⎢
⎢
⎥
H(0, 1, 1/4) = ⎢0 −5/2 0⎥
⎢
⎥
⎢8
0
0⎥⎦
⎣
Therefore, f (0, 1) = 1 is a local maximum.
4
∣H 2 ∣ = 160
Problem 5. Find the local optima of f (x1 , x2 , x3 ) = x1 x2 x3 subject to the constraint x1 + 2x2 + 3x3 = 6.
The Lagrangian function is
Z(x1 , x2 , x3 , λ) = x1 x2 x3 + λ[6 − x1 − 2x2 − 3x3 ]
The gradient of Z is
⎤
⎡
x2 x3 − λ
⎥
⎢
⎢
⎥
x1 x3 − 2λ
⎥
⎢
⎥
∇Z(x1 , x2 , x3 , λ) = ⎢
⎥
⎢
x
x
−
3λ
1
2
⎥
⎢
⎢6 − x1 − 2x2 − 3x3 ⎥
⎦
⎣
Solving the system of equations given by ∇Z(x1 , x2 , x3 , λ) = 0, we find that there are four critical points:
(x1 , x2 , x3 , λ) = (0, 0, 2, 0), (x1 , x2 , x3 , λ) = (0, 3, 0, 0), (x1 , x2 , x3 , λ) = (6, 0, 0, 0), (x1 , x2 , x3 , λ) = (2, 1, 2/3, 2/3).
The bordered Hessian matrix is
⎡0 1 2 3 ⎤
⎢
⎥
⎢1 0 x x ⎥
⎢
3
2⎥
⎥
H(x1 , x2 , x3 , λ) = ⎢
⎢2 x 3 0 x 1 ⎥
⎢
⎥
⎢3 x2 x1 0 ⎥
⎣
⎦
At the critical point (x1 , x2 , x3 , λ) = (0, 0, 2, 0), we have
⎡0
⎢
⎢1
⎢
H(0, 0, 2, 0) = ⎢
⎢2
⎢
⎢3
⎣
1
0
2
0
2
2
0
0
3⎤⎥
0⎥⎥
⎥
0⎥⎥
0⎥⎦
∣H 2 ∣ = 8
∣H 3 ∣ = 36
Therefore, there is neither a local minimum nor a local maximum at (x1 , x2 , x3 ) = (0, 0, 2).
At the critical point (x1 , x2 , x3 , λ) = (0, 3, 0, 0), we have
⎡0
⎢
⎢1
⎢
H(0, 3, 0, 0) = ⎢
⎢2
⎢
⎢3
⎣
1
0
0
3
2
0
0
0
3⎤⎥
3⎥⎥
⎥
0⎥⎥
0⎥⎦
∣H 2 ∣ = 0
∣H 3 ∣ = 36
Therefore, there is neither a local minimum nor a local maximum at (x1 , x2 , x3 ) = (0, 3, 0).
At the critical point (x1 , x2 , x3 , λ) = (6, 0, 0, 0), we have
⎡0
⎢
⎢1
⎢
H(6, 0, 0, 0) = ⎢
⎢2
⎢
⎢3
⎣
1
0
0
3
2
0
0
6
3⎤⎥
0⎥⎥
⎥
6⎥⎥
0⎥⎦
∣H 2 ∣ = 0 ∣H 3 ∣ = 36
Therefore, there is neither a local minimum nor a local maximum at (x1 , x2 , x3 ) = (6, 0, 0).
At the critical point (x1 , x2 , x3 , λ) = (2, 1, 2/3, 2/3), we have
⎡0 1
2 3⎤⎥
⎢
⎢ 1 0 2/3 1 ⎥
⎢
⎥
⎥
H(2, 1, 2/3, 2/3) = ⎢
⎢2 2/3 0 2⎥
⎢
⎥
⎢3 1
2 0⎥⎦
⎣
Therefore, f (2, 1, 2/3) = 4/3 is a local maximum.
5
∣H 2 ∣ = 8/3 ∣H 3 ∣ = −12
1/2
1/2
Problem 6. Find the local optima of f (x1 , x2 , x3 ) = x1 subject to the constraints x1 − 4x2 − 8x3 = 0 and
2x2 + 4x3 = 6.
The Lagrangian function is
1/2
1/2
Z(x1 , x2 , x3 , λ1 , λ2 ) = x1 + λ1 (−x1 + 4x2 + 8x3 ) + λ2 (6 − 2x2 − 4x3 )
The gradient of Z is
⎡
⎤
1 − λ1
⎢
⎥
⎢
⎥
−1/2
⎢ 2x
⎥
λ
−
2λ
1
2 ⎥
⎢
2
⎢
⎥
−1/2
∇Z(x1 , x2 , x3 , λ1 , λ2 ) = ⎢ 4x3 λ1 − 4λ2 ⎥
⎢
⎥
⎢−x + 4x 1/2 + 8x 1/2 ⎥
⎢ 1
⎥
2
3
⎢
⎥
⎢ 6 − 2x2 − 4x3 ⎥
⎣
⎦
Solving the system of equations given by ∇Z(x1 , x2 , x3 , λ1 , λ2 ) = 0, we find that there is one critical point:
(x1 , x2 , x3 , λ1 , λ2 ) = (12, 1, 1, 1, 1).
The bordered Hessian matrix is
⎡
⎢ 0
⎢
⎢ 0
⎢
⎢
H(x1 , x2 , x3 , λ1 , λ2 ) = ⎢ 1
⎢
⎢−2x −1/2
⎢
2
⎢
⎢−4x3−1/2
⎣
0
0
0
2
4
−1/2
−1/2 ⎤
⎥
1 −2x2
−4x3
⎥
⎥
0
2
4
⎥
⎥
0
0
0
⎥
⎥
−3/2
⎥
0 −x2 λ1
0
⎥
−3/2 ⎥
0
0
−2x3 λ1 ⎥⎦
At the critical point (x1 , x2 , x3 , λ1 , λ2 ) = (12, 1, 1, 1, 1), we have
⎡0
⎢
⎢0
⎢
⎢
H(12, 1, 1, 1, 1) = ⎢ 1
⎢
⎢−2
⎢
⎢−4
⎣
0
0
0
2
4
1 −2 −4⎤⎥
0 2
4 ⎥⎥
⎥
0 0
0⎥
⎥
0 −1 0 ⎥⎥
0 0 −2⎥⎦
Therefore, f (12, 1, 1) = 12 is a local maximum.
6
∣H 3 ∣ = −24
Problem 7. Consider a firm that produces and sells two products. Below is a model that represents the firm’s
profit maximization problem.
● Variables:
Q1 = quantity of product 1 produced and sold
P1 = unit price of product 1
Q2 = quantity of product 2 produced and sold
P2 = unit price of product 2
● Model:
maximize
P1 Q1 + P2 Q2 − 12 − 4Q1 − 8Q2
subject to
P1 = 46 − 3Q1
P2 = 32 − 2Q2
a. Describe the objective function and the constraints of the model. (e.g. What are the prices of the
products? What are the costs of production? Do the prices depend on demand?)
b. By substituting the constraints into the objective function, find the local maximum values of profit.
c. Using the Lagrange multiplier method, find the local maximum values of profit.
d. What do your answers from parts b and c tell you about what the firm should do?
a. In this model, prices depend on the quantity produced and sold, as given in the two equality constraints.
There is a fixed cost of 12; each unit of quantity 1 costs 4, while each unit of quantity costs 8.
b.
Let π be the profit function. By substitution, we have that
π(Q1 , Q2 ) = 42Q1 − 3Q12 + 24Q2 − 2Q22 − 12
The gradient of π is
42 − 6Q1
∇π(Q1 , Q2 ) = [
]
24 − 4Q2
Solving the system of equations given by ∇π(Q1 , Q2 ) = 0, we find that there is one critical point: (Q1 , Q2 ) =
(7, 6). The Hessian of π is
−6 0
H(Q1 , Q2 ) = [
]
0 −4
At the critical point (Q1 , Q2 ) = (7, 6), we have
−6 0
H(7, 6) = [
]
0 −4
∣H1 ∣ = −6 ∣H2 ∣ = 24
Therefore π(7, 6) = 207 is a local maximum.
c. Let f (Q1 , Q2 , P1 , P2 ) = P1 Q1 +P2 Q2 −12−4Q1 −8Q2 , g1 (Q1 , Q2 , P1 , P2 ) = P1 +3Q1 , c1 = 46, g2 (Q1 , Q2 , P1 , P2 ) =
P2 + 2Q2 , c2 = 32. The Lagrangian function is
Z(Q1 , Q2 , P1 , P2 , λ1 , λ2 ) = P1 Q1 + P2 Q2 − 12 − 4Q1 − 8Q2 + λ1 (46 − P1 − 3Q1 ) + λ2 (32 − P2 − 2Q2 )
7
The gradient of Z is
⎡ P1 − 4 − 3λ1 ⎤
⎥
⎢
⎥
⎢
⎢ P2 − 8 − 2λ2 ⎥
⎥
⎢
⎢ Q1 − λ1 ⎥
⎥
⎢
∇Z(Q1 , Q2 , P1 , P2 , λ1 , λ2 ) = ⎢
⎥
⎢ Q2 − λ2 ⎥
⎥
⎢
⎢ 46 − P1 − 3Q1 ⎥
⎥
⎢
⎢32 − P2 − 2Q2 ⎥
⎦
⎣
Solving the system of equations given by ∇Z(Q1 , Q2 , P1 , P2 , λ1 , λ2 ) = 0, we find that there is one critical point:
(Q1 , Q2 , P1 , P2 , λ1 , λ2 ) = (7, 6, 25, 20, 7, 6).
The bordered Hessian is
⎡0
⎢
⎢
⎢0
⎢
⎢3
H(Q1 , Q2 , P1 , P2 , λ1 , λ2 ) = ⎢⎢
⎢0
⎢
⎢1
⎢
⎢0
⎣
0
0
0
2
0
1
3
0
0
0
1
0
0
2
0
0
0
1
1
0
1
0
0
0
0⎤⎥
⎥
1⎥
⎥
0⎥⎥
1 ⎥⎥
⎥
0⎥⎥
0⎥⎦
At the critical point (Q1 , Q2 , P1 , P2 , λ1 , λ2 ) = (7, 6, 25, 20, 7, 6), we have
⎡0
⎢
⎢
⎢0
⎢
⎢3
H(7, 6, 25, 20, 7, 6) = ⎢⎢
⎢0
⎢
⎢1
⎢
⎢0
⎣
0
0
0
2
0
1
3
0
0
0
1
0
0
2
0
0
0
1
1
0
1
0
0
0
0⎤⎥
⎥
1⎥
⎥
0⎥⎥
1 ⎥⎥
⎥
0⎥⎥
0⎥⎦
∣H 3 ∣ = −24
∣H 4 ∣ = 24
Therefore, f (7, 6, 25, 20) = 207 is a local maximum.
d. Assuming a local maximum is sufficient, the firm should set the price of product 1 to 25 and the price of
product 2 to 20, resulting in 7 units of product 1 and 6 units of product 2 sold.
8
Problem 8. Consider a firm that produces a good that requires two inputs to produce. In particular, x1 units
1/3 1/3
of input 1 and x2 units of input 2 yield 3x1 x2 units of the good. Each unit of input 1 costs $20, and each unit
of input 2 costs $160.
a. Using the variables x1 and x2 defined above, write cost as a function of x1 and x2 : c(x1 , x2 ) = . . .
b. The firm needs to produce 24 units of the product. Using the variables x1 and x2 defined above, write an
equality constraint that models this.
c. Using the Lagrange multiplier method, find the local minimum values of cost.
d. What does your answer from part c tell you about what the firm should do?
a.
c(x1 , x2 ) = 20x1 + 160x2
b.
3x1 x2 = 24
c.
The Lagrangian function is
1/3 1/3
1/3 1/3
Z(x1 , x2 , λ) = 20x1 + 160x2 + λ(24 − 3x1 x2 )
The gradient of Z is
⎡ 20 − λx −2/3 x 1/3 ⎤
⎢
1
2 ⎥
⎢
⎥
1/3 −2/3 ⎥
⎢
∇Z(x1 , x2 , λ) = ⎢160 − λx1 x2 ⎥
⎢
⎥
⎢ 24 − 3x 1/3 x 1/3 ⎥
⎣
⎦
1
2
Solving the system of equations given by ∇Z(x1 , x2 , λ) = 0, we find that there is one critical point: (x1 , x2 , λ) =
(64, 8, 160).
The bordered Hessian is
−2/3 1/3
1/3 −2/3
⎡
⎤
0
x1 x2
x1 x2
⎢
⎥
⎢ −2/3 1/3
⎥
−5/3
1/3
−2/3
−2/3
2
1
⎢
⎥
H(x1 , x2 , λ) = ⎢x1 x2
x
x
λ
−
x
x
λ
⎥
2
2
3 1
3 1
⎢ 1/3 −2/3
⎥
−2/3
−2/3
1/3
−5/3
1
2
⎢x x
− 3 x1 x2 λ
λ ⎥⎦
⎣ 1 2
3 x1 x2
At the critical point (x1 , x2 , λ) = (64, 8, 160), we have
⎡0
1/8
1 ⎤⎥
⎢
⎢
⎥
H(64, 8, 160) = ⎢1/8 5/24 −5/6⎥
⎢
⎥
⎢ 1 −5/6 40/3 ⎥
⎣
⎦
∣H 3 ∣ = −
5
8
Therefore, c(64, 8) = 2560 is a local minimum.
d.
Assuming a local minimum is sufficient, the firm should use 64 units of input 1 and 8 units of input 2.
9