Q1
a) False
This is so because, if f(x) = x, then y=x at every particular point and thus, the condition f(x) ≤ f(y) does not hold.
b)true
a function is said to have the convex nature if it is continuous over some certain interval say (a,b).
c)true
a function with a concave shape will most likely have a maxima. Only functions with a convex shape will have a minimizer instead.
d)false
this is so because, a convex function does not have argmax. It has a minimum set of values instead.
e)true
a monotonic function is either increasing or decreasing over some interval. Thus, for a monotone transformation of a function that has a maximizer, the transform will still have a maximizer too.
Q2
a)
Since the equation of x12 + x22 =1 is a curve running from 0-1, the it would be an intelligent guess that the solution we’re looking for lies at x1 = 0.5, and x2 =0.5
Maximize ln 0.5 + ln 0.5 = -1.3862
b)
Maximize x1 + x22 s.t x12 + x22 = 1
=f(x) – λ(g(x) –b)
= x1 + x22 –λ (x12 + x22-1)
= (x1 – x12λ) + (x22 – x22λ) + λ
dL/dx1 = 1-2x1λ = 0
thus, x1 = 1/2λ
DL/dx2 = x2(2-2λ) = 0
Thus, since x2 cannot be equal to zero, then
2-2λ = 0 implying that λ=1
Therefore x1 = 1/2 and x22 = 1-1/4 = ¾. So x2 = 0.866
We can therefore conclude that the maximized value is ½ + 0.8662 =1.25.
Q3
a)
L=u(x)-λ(px-I)
DL/dx=-λp
Dl/Di=-λI
Since both are equivalent to zero, the we can conclude that;
X(λp,λ I)=(P,X)
b)
from the information provided,,
u(x)=u(p,I)
y(x)=Nε(p,I)
therefore for all (p,I) and ε greater than zero, it can be clearly seen that,
px=p(p,I)=I
c)
since u(x) is strictly concave, then this implies that it has a maxima. Since x(p,I), then u(x)=u(p,I).
This in turn proves that the solution to the maximization problem, which must be concave, is a singleton; having only one single point for a solution.
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