Download Exercise 1 Calculate the partial derivatives of a) z = 5y 5 + 4x4y + 3x

BEE1020 — Basic Mathematical Economics
Dieter Balkenborg, Iannis Krassas
Class Exercises - Solution
Department of Economics
16-20/01/2006
Week 12
University of Exeter
Exercise 1 Calculate the partial derivatives of
a) z = 5y 5 + 4x4 y + 3x2 y 3 + 2xy 4 + 2x + 3y + 5
xy 2
b) z = 2 3
x y +1
Solution 1 a)
∂z
= 16x3 y + 6xy 3 + 2y 4 + 2
∂x
∂z
= 25y 4 + 4x4 + 9x2 y 2 + 8xy 3 + 3
∂y
b)
2 3
y 2 (x2 y 3 + 1) − 2xy 2 xy 3
∂z
2 x y −1
=
=
−y
∂x
(x2 y 3 + 1)2
(x2 y 3 + 1)2
2xy (x2 y 3 + 1) − 3xy 2 x2 y 2
x2 y 3 − 2
∂z
=
=
−xy
∂y
(x2 y 3 + 1)2
(x2 y 3 + 1)2
Exercise 2 Find all second derivatives
2
2
∂2z
, ∂z, ∂z
∂x2 ∂y∂x ∂x∂y
and
∂ 2z
∂y2
of
a) z = 5x2 y + 3x2 y 2 + 5y 3
5
b) z = x2 + y 3
Solution 2 a)
∂z
= 10xy + 6xy 2
∂x
2
2
∂ z
∂x2
∂2z
∂x∂y
∂ z
∂y∂x
∂2z
∂y2
=
∂z
= 5x2 + 6x2 y + 15y 2
∂y
10y + 6y 2 10x + 12xy
10x + 12xy 6x2 + 30y
b)
∂2z
∂x2
∂2z
∂x∂y
∂2z
∂y∂x
∂2z
∂y 2
=
4
∂z
= 10 x2 + y 3 x
∂x
3
4
∂z
= 15 x2 + y 3 y 2
∂y
80 (x2 + y 3 ) x2 + 10 (x2 + y 3 )
3
120 (x2 + y 3 ) y 2 x
4
3
120 (x2 + y 3 ) y 2 x
3
4
180 (x2 + y 3 ) y 4 + 30 (x2 + y 3 ) y
Exercise 3 Find the critical point of the function
z = 20x2 − 37xy + 31x + 15y 2 − 16y − 7
= (5x − 3y − 1) (4x − 5y + 7)
Solution 3 First order conditions determining a critical point
∂z
= 40x − 37y + 31 = 0
∂x
40x − 37y + 31 = 0
−37x + 30y − 16 = 0
∂z
= −37x + 30y − 16 = 0
∂y
A critical point is hence a solution to
40x − 37y + 31 = 0 | × 37
−37x + 30y − 16 = 0 | × 40
1480x − 1369y + 1147 = 0
−1480x + 1200y − 640 = 0 | +
−169y + 507 = 0
This yields y =
507
169
= 3 and so 40x = 111 − 31 = 80. The critical point is (x, y) = (2, 3).
The distance between two points in the plane with coordinates (a, b) and (x, y) can be
calculated using the formula
D = (x − a)2 + (y − b)2 .
This follows from the law of Pythagoras.
Exercise 4 Find the points (x∗ , y ∗ (x)) on the graph of the function y = f (x) = x2 − 9.5
which is closest to the origin (0, 0). (Hint: Minimize D2 .)
y
15
10
5
0
-5
-2.5
0
2.5
5
x
-5
y = f (x) = x2 − 9.5
Solution 4 The distance is
D (x) =
x2 + y 2 (x) =
x2 + (x2 − 9.5)2
The distance is minimized when the square of the distance is minimized.
2
D2 (x) = x2 + x2 − 9.5
dD2
= 2x + 2 x2 − 9.5 × 2x = 2x + 4x3 − 4 × 8.5x = 4x3 + 2x − 34x
dx
= 4x3 − 32x = 4x x2 − 9 = 4x (x + 3) (x − 3)
dD2
= 12x2 − 32
dx
2
y
250
200
150
100
50
-5
-2.5
0
2.5
5
x
The function D2 (x)
2
2
The critical points are x = −3, 0, +3. Since dD
< 0 and dD
> 0 the function
dx |x=0
dx |x=±3
2
D (x) has a peak at x = 0 and troughs at x = ±3, whereby D (3) = D (−3). The latter
are absolute minima, as can be seen from a sign diagram for the first derivative: For x ≥ 0
the function is decreasing when 0 ≤ x ≤ 3 and increasing when x ≥ 3. The absolute
minimum on the interval x ≥ 0 is hence at x = 3. Similarly the absolute minimum on the
interval x ≤ 0 is at x = −3. With respect to the whole number line the absolute minima
are therefore x = ±3. Consequently, the points (±3, y (±3)) = (±3, y (−0.5)) are closest
to the origin.
3