Intermediate Mathematics for Economics

EC206 Intermediate Mathematics for Economics
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Question 1
The given function f(x) is strictly convex. The definition of a convex function is highlighted
below.
When its domain is also a convex set and is true for all x, y and for
The condition for strict convexity of the given function is shown below.
It is an imperative use of the strictly convexity that the obtained optimal solution is itself
unique.
In other words, the strict convexity is to ensure the uniqueness of the optimal solution.
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Question 2
(a) Partial and total derivative of z with respective to x and y
z=ln ( x2 y )
x + y=1
Partial derivative of z with respect to x
∂ z
∂ x = ∂ ln ( x2 y )
∂ x
Let u=x2 y
∂ z
∂ x = ∂ ln ( u )
∂ x . ∂ ( x2 y )
∂ x
∂ z
∂ x = 1
u∗2 y∗x2 y−1= 1
x2 y∗2 y x2 y−1=2 y x2 y−1 +2 y=2 y x−1= 2 y
x
Now ,
With respect to y
∂ z
∂ y =∂ ln ( x2 y )
∂ y
Let u=x2 y
∂ z
∂ y =∂ ln ( u )
∂ y . ∂ ( x2 y )
∂ y
From exponent rule: x2 y=e2 yln(x)
∂ z
∂ y =∂ ln ( u )
∂ y . ∂ ( e2 yln ( x ) )
∂ y =1
u . 2 x2 y ln ( x )
∂ z
∂ y = 1
x2 y . 2 x2 y ln ( x ) =1.2 ln ( x ) =2 ln(x)
Hence, the partial derivatives would be 2 y
x ,2 ln(x)
Total derivative of z
dz= ∂ z
∂ x dx + ∂ z
∂ y dy
dz= 2 y
x dx +2 ln(x )dy
Here,
x + y=1
2

dx +dy =0
dy =−dx
Thus,
dz= 2 y
x dx +2 ln ( x ) ( −dx )
dz= ( 2 y
x −2 ln ( x ) ) dx
dz= 2
x ( y −x ln ( x ) ) dx
(b) Partial and total derivative of z with respective to x and y
z=x + y
A xa yb=k
Where, a , b∧k are constants
Partial derivative of z with respect to x
∂ z
∂ x = ∂(x + y )
∂ x =1
With respect to y
∂ z
∂ y =∂ (x+ y)
∂ y =1
Total derivative of z
dz= ∂ z
∂ x dx + ∂ z
∂ y dy
Divide by dx
dz
dx = ∂ z
∂ x + ∂ z
∂ y
dy
dx
dz
dx =1+ 1. dy
dx
Now,
A xa yb=k
x= ( k
yb A )1
a
Similarly,
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