ECON 11: Problem Set 1 Solutions

Due Thursday October 8 before 9:30 AM in class

1 Unconstrained optimization in one variableConsider the function

f (x) = x3 + x2 where x ∈ [−2,2]

a) Find the critical values of x

(a) Sol: FOCs yieldf ′(x) = 3x2 +2x = 0

The solutions are x = 0, x =− 23

b) For each critical value, is it a maximum or a minimum?

(a) Sol: The second derivative is given by

f ′′(x) = 6x+2

Evaluating at our critical points, we have f ′′(0) = 2 and f ′′(− 23 ) =−2 . This establishes that x = 0

is a local minimum, and that x =− 23 is a local maximum. To check that these are a global minimum

and maximum, respectively, we evaluate the function at the endpoints x=−2 and x= 2 and comparewith the values of the function evaluated at its critical values.

f (−2) =−8+4 =−4

f (2) = 8+4 = 12

f (0) = 0

f (−23) = 4

27

Hence, the critical points are local, but not global maxima or minima. The function obtains its globalextrema at the end points x =−2 and x = 2.

c) Graph the function and label the critical values

1

(a) Sol:

2 Unconstrained optimization in two variablesConsider the function

f (x1,x2) =−2x1x2 + x31 + x3

2

a) Find a maximum in the region where x1 ≤ 1 and x2 ≤ 1

(a) Sol: FOCs yield

f1(x1,x2) =−2x2 +3x21 = 0, f2(x1,x2) =−2x1 +3x2

2 = 0

These give us two equations in two unknows. The solution is x1 = x2 = 0. and x1 = x2 = 23 . On

tests, you are not going to be asked how to verify local/global extrema for multivariable functions.However, we will review this here. The SOC are:

i. Local minimum: f11 > 0, f22 > 0, and f11 f22− f 212 > 0

ii. Local maximum: f11 < 0, f22 < 0, and f11 f22− f 212 > 0

iii. Inflection point: f11 f22− f 212 < 0

iv. The SOC is indeterminate if f11 f22− f 212 = 0. The second derivatives yield:

f11 = 6x1, f22 = 6x2, f12 =−2

Evaluating the SOCs shows an inflection point at (0,0), and a local minimum at ( 23 ,

23 ). So

where is this maximum? One possibility is that there is no maximum (i.e., the function ex-plodes to infinity in some direction). One way to easily rule this out is to use the followingthought experiment: if we choose some arbitrarily large value for f (x1,x2), such as 1,000,000,is there some (x1,x2) within the domain that achieves this? There isn’t. The remaining possi-bility is that the function is maximized along the boundary, x1 = 1 or x2 = 1 (or both).

To find where this occurs, we have to analyze the function along the boundaries. You willsee how much time the Lagrange method, practiced below, can save. We fix x1 = 1 and x2 = 1,

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one at a time, and maximize the resulting single-variable function with respect to the unfixedvariable. That is, we maximize the following functions:

g1(x2) =−2(1)x2 +(1)3 + x32 subject to x2 ≤ 1

andg2(x1) =−2(1)x1 +(1)3 + x3

1 subject to x1 ≤ 1

This is the same exact type of problem as problem 1. You need to take FOCs and comparethe function evaluated at your solution to the function evaluated at the boundary point, 1. The

solutions to the FOCs of these problems are (x1 = ±√

23 , x2 = 1) (x1 = 1, x2 = ±

√23 ), re-

spectively. Evaluating SOCs of g1 and g2 at the values for x2 and x1, respectively, shows that

we have two local minima and two local maxima. The local maxima are at (−√

23 ,1) and

(1,−√

23 ). Finally, we have to evaluate the function at these points and compare to the value

of the function at the boundary point (1,1).

f (1,1) = 0 , f (−√

23,1) =

16√

227

+1 = 1.838 = f (1,−√

23)

Hence, we have two global maxima: (−√

23 ,1) and (1,−

√23 ). The function is shown below.

b) Now consider (x1,x2)∈R2, that is, the entire two-dimensional space where x1 and x2 are in[−∞,+∞]. Arethere any global extrema?

(a) Sol: There is no global extremum. To see this, consider fixing one variable and moving the other topositive or negative infinity.

3 Constrained optimizationRicardo seeks to maximize the following utility function

u(x1,x2) = x1/41 x3/4

2

3

subject to the budget constraintp1x1 + p2x2 = I

where p1, p2,x1,x2, I > 0

a) Find Ricardo’s utility-maximizing bundle (x?1,x?2) as a function of p1, p2, and I

(a) Sol:

i. Identify the type of problem. This is a CONSTRAINED two-variable optimization problem.Therefore it is appropriate to apply the Lagrangean method.

ii. Set up the Lagrangean:

L(x1,x2,λ ) = x1/41 x3/4

2 +λ (I− p1x1− p2x2)

iii. Find critical points using FOCs:

L1 =14

x−3/41 x3/4

2 −λ p1 = 0 =⇒ 14

x−3/41 x3/4

2 = λ p1 (1)

L2 =34

x1/41 x−1/4

2 −λ p2 = 0 =⇒ 34

x1/41 x−1/4

2 = λ p2 (2)

L3 = I− p1x1− p2x2 = 0 (3)

Dividing (1) by (2) yields:x1

x2=

3p1

p2

Substituting into (3) yields:

x?1 =3I

4p1, x?2 =

I4p2

We will not check SOCs here since this was demonstrated above. But they do in fact hold for alocal maximum. Moreover, our solution is a global maximum because we turned the probleminto an UNCONSTRAINED one by writing down the lagrangean (hence, no need to checkboundaries).

b) Show that x?2 is decreasing in both p2 and I (Hint: use partial derivatives)

(a) Sol:∂x?2∂ p2

=−I4p2

2≤ 0 for I, p2 ≥ 0

∂x?2∂ I

=1

4p2> 0 for I, p2 ≥ 0

c) Does x?2 change if p1 changes?

(a) Sol: Clearly x?2 does not depend on p1.

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