Business Analysis II Mat 211
Test 1 Review: Do not forget to do all homework problems before
you try the following problems:
Chapter: Function of several variables
Function notation:
1. Given f ( x, y )  x  2 y , find
f (0,1), f (2,1), f (2, 1), f (a, b), f (a  h, b)  f (a, b)
2. Given f ( x, y )  2 x 2 y , find f (0,1), f (2,1), f (2, 1), f (a, b), f (a  h, b)  f (a, b)
3. Given f ( x, y )  x  2 y , find
f (1/ x,1/ y ), F ( x, y )  f ( x  h, y  h)  f ( x. y ), F ( x, y ) / h
4. Given f ( x, y)  10 x 3 y , x  0, y  0. Find f (9, 27), f (2 x, 2 y )
Answer: 1. 2, 4, 0, a  2b, h
3.
1 2
 , 3h, 3
x y
4. 90, 25/ 6 f ( x, y)
Domain: Plot the domain:
5. Determine the domain of z  x 2  y 2
Answer: x 2  y 2  0
6. Determine the domain of z  x 2  y 2  36
x2  y 2
x2  y 2
7. Determine the domain of z 
Answer: x 2  y 2  0
8. Determine the domain of z  x 2  y 2  25  36  x 2  y 2
1
9. Determine the domain of z  x 2  y 2  25 
36  x 2  y 2
Answer: 25  x 2  y 2  36
10. Determine the domain of z 
x 2  y 2  36
x y
x 2  y 2  36
x y
11. Determine the domain of z 
12. Determine the domain of z 
13. Determine the domain of z 
3
Answer: x 2  y 2  36, x  y
x 2  y 2  36
x y
x 2  y 2  36
x y
Answer: 36  x 2  y 2 , x  y
14. Determine the domain of z  7
x 2  y 2  36
x y
15. Determine the domain of z  7
x 2  y 2  36
 x  2y
x y
Answer: x  y
Answer: x  y
Level curves:
Definition: If g ( x, y )  c is a level curve of z  f ( x, y ) then g ( x, y )  c will make
z  f ( x, y ) a constant.
16. Show that x 2  y 2  c is a level curve of f ( x, y)  e x
2
 y2
 x4  2x2 y 2  y 4 .
Solution: Given x 2  y 2  c , we have
f ( x, y)  e x  y  x 4  2 x 2 y 2  y 4  e c  ( x 2  y 2 ) 2  e c  c 2 which is a constant.
Therefore x 2  y 2  c is a level curve of f ( x, y ) at a distance e c  c 2 .
2
2
6. Draw the graphs of the following functions in 3D space and draw atleast 3 level
curves in each.
a) z  f ( x, y )  12  3x  4 y
b) z  f ( x, y )   9  x 2  y 2
c) z  f ( x, y )  9  x 2  y 2
Answer: a) a plane b) Half of a sphere below the xy plane, z is negative
c) Half of a sphere above the xy plane, z is positive
7. Draw the hyperbolic paraboloid z  x 2  y 2 (check the graph in your text book,
chapter 15, page# 1089)
Application:
8. Show that the function z  f ( x, y)  2 x 2  y 2  4 x  4 y  3 has a maximum. Find
the maximum.
Solution: f x  4 x  4  0  x  1 and f y  2 y  4  0  y  2 , the stationary point is
(1, 2).
Moreover f xx  4  0 and f yy  2  0 with f xx f yy  ( f xy ) 2  8  0 , we have
maximum at (1, 2).
9. For the given function f ( x, y)  x 2  2 xy2  2 y 2 , find all stationary points and
classify them.
Solution:
f
f
 f x  f1  2 x  2 y 2 ,
 f y  f 2  4 xy  4 y ,
x
y
2 f
2 f
2 f
2 f
f xy  f12  4 y,
 f yx  f 21  4 y ,

f

f

2,
 f 22  4 x  4
xx
11
yx
xy
x 2
y 2
Now, solve f1  2 x  2 y 2  0 , f 2  4 xy  4 y  0 and find that the solution gives you the
points (0, 0), (-1, 1), (-1, -1) and classify them as in 1.
10. A firm produces two different kinds A and B of a commodity. The daily cost of
producing x units of A and y units of B is
C ( x)  0.04 x 2  0.01xy  0.01y 2  4 x  2 y  500 . Suppose that the firm sells its
output at aprice per unit is 15 for A and 9 for B. Find the daily production levels x
and y that maximizes profit per day.
Answer: x = 100, y = 300, profit = 1100
The Lagrange Multiplier Method
To find max (min) of z  f ( x, y ) subject to g ( x, y )  c proceed as follows
1) Write down the Lagrangian function
L( x, y )  f ( x, y )   ( g ( x, y )  c) where  is a constant.
2) Find Lx  0 , L y  0
3) Solve the system Lx  0 , L y  0 and g ( x, y )  c .
4) The max/min of the given function will be found at the solution(s).
11. maximize f ( x, y )  xy subject to g ( x, y )  2 x  y  100
The Langrangian for this problem is
L( x, y )  f ( x, y )   ( g ( x, y )  c)  xy   (2 x  y  100)
Now we have Lx  y  2  0  y  2 , L y  x    0  x   .
We have y  2 x and from g ( x, y )  2 x  y  100 we find x = 25 and y = 50.
The solution for this maximization problem is x = 25 and y = 50 and z = 1250
Go through the example problem 2 and 3 at page # 270-271.
12. max xy subject to x  3 y  24
Notice that f ( x, y )  xy and g ( x, y )  x  3 y and c = 24
The Langrangian for this problem is
L( x, y )  f ( x, y )   ( g ( x, y )  c)  xy   ( x  3 y  24)
Lx  y    0  y   , L y  x  3  0  x  3 . So that x = 3y
Form x  3 y  24 we find x = 12, y = 4
The solution for the given problem is (12, 4)
13. (Compare with example 21) max Ax a y b subject to px  qy  m has the solution
x
a m
b m
 and y 

ab p
ab q
In our case max xy subject to x  3 y  24
A = 1, a = 1, b = 1, m = 24, p = 1, q = 3
a m
1 24
b m
1 24
Thus x 
 

 12 and y 
 
 4
a  b p 11 1
a  b q 11 3
Derivatives
14. Find f y , f x , f xy , f yy : for f ( x, y)  (1  xy)2
f x  2(1  xy) y, f xy  2(1  2 xy), f xx  2 y 2
f y  2(1  xy) x, f yy  2 x2
15. Find f y , f x , f xy , f yy : for f ( x, y)  (1  3xy 2 )2
16. Find f y , f x , f xy , f yy : for f ( x, y )  1 
fx 
3x
y
3
3
, f xy   2 , f xx  0
y
y
3x
6x
, f yy  3
2
y
y
1
17. Prove that if z  ln( x 2  y 2 ) then z xx  z yy  0
2
18. When f ( x, y)  x 2  2 xy  y 2 then f (2 x, 2 y )  4 f ( x, y)
19. If f ( x, y )  eax by find f xx f yy  f xy f yx
fy  
20. Show that x 2  y 2  6 is not a level curve of f ( x, y )  x 2  y 2  x 2  y 2  3 .
x2 y 2
21. Given f ( x, y)  3
. Compute and simplify xf x  yf y
ay  bx3