WS 21/22
Assignment 01
Microeconomic Analysis
Mathematical Basics I
Solutions
1. Determine the slope of the following functions at the given points.
a) 𝒇(𝒙) = 𝟒√𝒙 − 𝒙² at 𝒙 = 𝟒
𝑓′(𝑥) = 2𝑥 −0.5 − 2𝑥
𝑓 ′(𝑥) = −7
b) 𝒇(𝒙) = 𝐥𝐧(𝟓𝒙𝟐 ) at 𝒙 = 𝟐
1
2
𝑓 ′(𝑥) = 5𝑥 2 ∙ 10𝑥 = 𝑥
𝑓 ′(𝑥) = 1
c) 𝒇(𝒙) =
𝑓(′𝑥) =
𝒙³−𝒙²
𝒙+𝟓
at 𝒙 = 𝟏
(3𝑥 2 −2𝑥)(𝑥+5)−(𝑥 3 −𝑥²)
(𝑥+5)²
6
=
3𝑥 3 +15𝑥²−2𝑥2 −10𝑥−𝑥 3 +𝑥²
(𝑥+5)²
=
2𝑥 3 +14𝑥 2 −10𝑥
(𝑥+5)²
1
𝑓(′𝑥) = 36 = 6
2. Decide whether the function 𝒇(𝒙) depicted in the graph below is concave or not,
using the following definition of a concave function.
𝒇(𝝀𝒙𝟏 + (𝟏 − 𝝀)𝒙𝟐 ) ≥ 𝝀𝒇(𝒙𝟏 ) + (𝟏 − 𝝀)𝒇(𝒙𝟐 ) 𝐟𝐨𝐫 𝐚𝐥𝐥 𝝀 ∈ [𝟎, 𝟏] 𝐚𝐧𝐝 𝒙𝟏 , 𝒙𝟐 𝝐𝑫
Make sure to include the points A-D in your verbal explanation.
WS 21/22
Assignment 01
•
Microeconomic Analysis
𝑓(𝜆𝑥1 + (1 − 𝜆)𝑥2 ) is the function’s value at a convex combination of any
two points (D)
(defines the value of 𝑓 when 𝑥 is one point on the interval [𝑥1 , 𝑥2 ] where its
location on that interval is determined by 𝜆; is a point on the graph of the
function)
•
this is larger than
•
the convex combination of the function values at these two points (𝜆𝑓(𝑥1 ) +
(1 − 𝜆)𝑓(𝑥2 ))(C) (a point on the straight-line chord connecting (𝑥1 , 𝑓(𝑥1 ) A
and (𝑥2 , 𝑓(𝑥2 ) B where its location is again determined by 𝜆)
➔ Because this is true for all 𝜆 ∈ [0,1], every straight line connecting any two
points of the graph of the functions is below the graph
3. The following function 𝒇(𝒙, 𝒚, 𝒛) has one maximum point. Determine this point
and propose two transformations that are characterized by the same maximum
point but different maximum values.
𝒇(𝒙, 𝒚, 𝒛) = 𝟒𝒙 − 𝒙𝟐 + 𝟖𝒚 − 𝒚𝟐 − 𝟐𝒛³ + 𝟒
𝑑𝑓
= 4 − 2𝑥 = 0 → 2𝑥 = 4 → 𝑥 = 2
𝑑𝑥
𝑑𝑓
= 8 − 2𝑦 = 0 → 2𝑦 = 8 → 𝑦 = 4
𝑑𝑦
𝑑𝑓
= −6𝑧² = 0 → 𝑧 = 0
𝑑𝑧
Positive monotonic transform, like:
(i) 𝐹 = 𝑎 ∙ 𝑓 + 𝑏 (with 𝑎 > 0)
(ii) 𝐹 = 𝑒 𝑓
(iii) F = 𝑙𝑛𝑓
4. Solve the following problem using the Lagrange multiplier method.
𝒇(𝒙, 𝒚) = 𝒙𝟎.𝟐 ∙ 𝒚𝟎.𝟖 → 𝐦𝐚𝐱!
s. t. 𝟐𝒙 + 𝟒𝒚 = 𝟏𝟎𝟎
𝐿 = 𝑥 0.2 ∙ 𝑦 0.8 + 𝜆(100 − 2𝑥 − 4𝑦)
𝑑𝐿
= 0.2𝑥 −0.8 𝑦 0.8 − 2𝜆 = 0
𝑑𝑥
WS 21/22
Assignment 01
𝑑𝐿
= 0.8𝑥 0.2 𝑦 −0.2 − 4𝜆 = 0
𝑑𝑦
𝑑𝐿
= 100 − 2𝑥 − 4𝑦 = 0
𝑑𝜆
I/II
1𝑦 2
=
4𝑥 4
𝑦
=2
𝑥
𝑦 = 2𝑥
in III
100 − 2𝑥 − 8𝑥 = 0
𝑥 = 10
𝑦 = 20
Microeconomic Analysis