UNIVERSITY OF TORONTO SCARBOROUGH
DEPARTMENT OF MANAGEMENT
MGEB02: Price Theory: A Mathematical Approach
(Intermediate Microeconomics I)
Problem Set-0
The following “calculus” problems are assigned to help you condition yourself for the
course. The main component is Part-4, which will be used throughout the course. Make
sure you clearly understand the process, if you don’t, come to see me ASAP.
Part-1: Graph the following functions. What are their maximum and minimum? If the
function is linear, what is its slope?
a) y = 3x − 2
b) y = x2 + 1
Part-2: Find the derivatives of the following functions with respect to x.
a) y = −7x3
b) y = 5x-5/2
1
c) y 
x
2
d) y = (3x + 1)2
Part-3: Find the partial derivatives for the following functions with respect to all
variables. You may assume that px, py, and I are constants, so you will not need to find
the partial derivative with respect to these variables:
f ( x, y )  x  x( x  y 2 )
f ( p, q )  ( q  p ) q  q
1
c) f ( x, y )  2
x  y2
d) U ( X ,Y )  p x X  p yY  I
a)
b)
Part-4: Maximization (Minimization). Find the maximum or minimum of the following:
a)
f ( x, y )  3 x 2  4 y 2  2 x  8 y
b)
c)
d)
f ( x, y )  x 2  2 xy  2 y 2  4 y
f ( x, y )  4 x 2  2 y 2  3
Subject to : x  2 y  9
U ( x, y )  x 2 y
Subject to : 2 x  3 y  18
e)
U ( x, y )  x 0.5 y 0.5
Subject to : x  2 y  18
f)
Q ( L, K )  100 L  50 K  L2  K 2
Subject to : 6 L  3K  30
g)
Q ( L, K )  2 L0.5  2 K 0.5
Subject to : 6 L  K  60
Problem Set-0 (Solutions)
Part-1: Graph the following functions. What are their maximum and minimum? If the
function is linear, what is its slope?
c) y = 3x − 2
-2
2/3
d) y = x2 + 1
y
 2 x  0  x  0
x
The 2nd derivative (2) is >0 => this is a minimum.
2
1
-1
1
Part-2: Find the derivatives of the following functions with respect to x.
a) −21x2
b) -12.5x-7/2
1
c)
4 x
d) 6(3x + 1)
Part-3: Find the partial derivatives for the following functions with respect to all
variables. You may assume that px, py, and I are constants, so you will not need to find
the partial derivative with respect to these variables:
e)
f)
g)
f x  1  2 x  y 2 , f y  2 xy
f p  q, f q  2 q  p  1
fx 
x
 2x
2
y

2 2
, fx 
x
 2y
2
 y2

2
h) U X  p x ,U y  p y
Part-4: Maximization (Minimization). Find the maximum or minimum of the following:
a)
b)
f x  6 x  2  0  x  1 / 3
f y  8 y  8  0  y  1
f x  2x  2 y  0
f y  2x  4 y  4  0
 x  2, y  2
L  4 x 2  2 y 2  3   ( x  2 y  9)
L
 8x    0
x
L
 4 y  2  0
c) y
L
 ( x  2 y  9 )  0

(1), (2 ) : y  4 x  x  2( 4 x )  9
 x  1, y  4
L  x 2 y   (2 x  3 y  18)
L
 2 xy  2  0
x
L
 x 2  3  0
d) y
L
 ( 2 x  3 y  18)  0

(1), (2 ) : x  3 y  2(3 y )  3 y  18
 y  2, x  6
L  x 0.5 y 0.5   ( x  2 y  18)
L
 0.5 x 0.5 y 0.5    0
x
L
 0.5 x 0.5 y 0.5  2  0
e) y
L
 ( x  2 y  18)  0

(1), ( 2) : x  2 y  2 y  2 y  18
 y  4.5, x  9
L  100 L  50 K  L2  K 2   (6 L  3K  30 )
L
 100  2 L  6  0
L
L
 50  2 K  3  0
f) K
L
 (6 L  3K  30 )  0

(1), ( 2 ) : L  2 K  6( 2 K )  3K  30
 K  2, L  4
L  2 L0.5  2 K 0.5   (6 L  K  60 )
L
 L0.5  6  0
L
L
 K 0.5    0
g) K
L
 (6 L  K  60 )  0

(1), (2 ) : K  36 L  6 L  (36 L)  60
 L  60 / 42, K  2160 / 42