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```Homework Assignment #4
13.11 Write the following linear functions in matrix form:
a) f (x1 , x2 , x3 ) = 2x1 − 3x2 + 5x3 ;
b) f (x1 , x2 ) = (2x1 − 3x2 , x1 − 4x2 , x1 );
c) f (x1 , x2 , x3 ) = (x1 − x3 , 2x1 + 3x2 − 6x3 , x3 + 2x2 ).

x1
2x1 − 3x2 + 5x3 = (2, −3, 5)  x2 
x3


2 −3 x1
(2x1 − 3x2 , x1 − 4x2 , x1 )T =  1 −4 
x2
1 0
 

x1
1 0 −1
T



x2 
(x1 − x3 , 2x1 + 3x2 − 6x3 , x3 + 2x2 ) = 2 3 −6
x3
0 2 1

(a)
(b)
(c)
13.12 Write the following quadratic forms in matrix form:
a) x21 − 2x1 x2 + x22 .
b) 5x21 − 10x1 x2 − x22 .
c) x21 + 2x22 + 3x23 + 4x1 x2 − 6x1 x3 + 8x2 x3 .
Answer: If we require the matrices to be symmetric, the solutions are:


1
2 −3
1
−1
5
−5
xT
x,
xT
x,
xT  2
2
4  x.
−1
1
−5 −1
−3 4
3
13.20 Prove carefully that h : Rk → R1 defined by h(x1 , x2 , . . . , xk ) = xi is continuous on Rk . Conclude that
any monomial g(x1 , x2 , . . . , xk ) = cxn1 1 xn2 2 · · · xnk k is continuous on Rk and that any polynomial from Rk
to Rm is continuous on Rk .
Answer: Suppose xn → x. Then |h(xn ) − h(x)| = |xni − xi | ≤ kx− xk. Since xn → x, for any > 0 we
can choose N such that kxn − xk < for all n ≥ N . Then |h(xn ) − h(x)| < for all n ≥ N , showing
h(xn ) → h(x), which establishes continuity of h.
We also known that multiplying a continuous function by a continuous function (including constant
functions) yields a continuous function.It follows that monomial g from Rk to R1 is continuous.
Since the addition of continuous functions yields a continuous function, any real-valued polynomial
on Rk is continuous. Finally, since each component is continuous, and polynomial from Rk to Rm is
continuous.
14.5 The demand function Q1 = K1 P1a11 P2a12 I b1 is called a constant elasticity demand function.
a) Compute the three elasticities (own price, cross price, and income) and show that they are all
constants.
The own price elasticity is (P1 /Q1 )∂Q1 /∂P1
= (P1 /Q1 )a11 Q1 /P1 = a11 .
=
(P1 /Q1 )a11 K1 P1a11 −1 P2a12 I b1
HOMEWORK ASSIGNMENT #4
Page 2
The cross price elasticity
= (P2 /Q1 )a12 Q1 /P2 = a12 .
is
(P2 /Q1 )∂Q1 /∂P2
=
(P2 /Q1 )a12 K1 P1a11 P2a12 −1 I b1
The income elasticity is (I/Q1 )∂Q1 /∂I = (I/Q1 )b1 K1 P1a11 P2a12 I b1 −1 = (I/Q1 )a12 Q1 /I = b1 .
b) What are reasonable ranges for the four parameters in Q1 ?
Answer: Demand should be positive, requiring K1 > 0. The own price elasticity should be negative,
requiring a11 < 0. Since goods can be either substitutes or complements, we can’t sign a12 , and
because good one can be either normal or inferior, we can’t sign b1 .
14.8 Use differentials to approximate each of the following:
a) f (x, y) = x4 + 2x2 y 2 + xy 4 + 10y at x = 10.36 and y = 1.04;
Answer: df = (4x3 + 4xy 2 + y 4 ) dx + (4x2 y + 4xy 3 + 10) dy. We evaluate this at the point
(10, 1), obtaining df = 4041 dx + 450 dy. Replacing dx by ∆x = .36 and dy by ∆y = .04, we find
∆f ≈ 4041(.36) + 450(.04) = 1454.76 + 18 = 1472.76. Since f (10, 1) = 10220, the approximate value
of the function is 11692.76 while the actual value is about 11744.3.
b) f (x, y) = 6x2/3 y 1/2 at x = 998 and y = 101.5;
Answer: Here we approximate around (x, y) = (1000, 100), where f (1000, 100) = 6000. Then
df = 4x−1/3 y 1/2 dx + 3x2/3 y −1/2 dy. Evaluating at (1000, 100), we find df = 4 dx + 30 dy. Replacing
dx by ∆x = −2 and dy by ∆y = 1.5, we find ∆f ≈ −8 + 45 = 37, so f (998, 101.5) ≈ 6037. In
comparison, the actual value is about 6036.77.
p
c) f (x, y, z) = x1/2 + y 1/3 + 5z 2 at x = 4.2, y = 7.95, and z = 1.02.
Answer: Here we evaluate around the point (x, y, z) = (4, 8, 1). Note that f (4, 8, 1) = 3. Then
1
df = p
2 x1/2 + y 1/3 + 5z 2
1 −2/3
1 −1/2
x
dx + y
dy + 10z dz .
2
3
1
dy + 10 dz]. We use ∆x = 0.2, ∆y = −0.05, and ∆z = 0.02.
At (4, 8, 1), this becomes 16 [ 14 dx + 12
1
This yields ∆f ≈ 6 [0.05 − 0.025 + 0.2] = 0.0375, so f (4.2, 7.95, 1.02) ≈ 3.0375, compared to the
actual value of about 3.041.
14.16 The university’s solar car is racing on Australia’s roads on the path (x, y) = (et + 5t2 , t4 − 4t). What
are the coordinates of the car when its velocity vector is pointing parallel to the x-axis?
Answer: The velocity vector is (et +10t, 4t3 −4). This will be parallel to the x-axis whenever 4t3 −4 = 0,
i.e., t = 1. At t = 1, the coordinates of the car are (x, y) = (5 + e, −3).
```