BUSINESS MATHEMATICS (PART– 10)
Illustrative Examples
Example 1 : Find the maximum and minimum values of x 3  96 x  18 x 2 .
Solution : Let
(1)
y  x 3  18 x 2  96 x
dy
 3x 2  36 x  96
dx
Then
(2)
d2y
 6 x  36
dx 2
and
(3)
For stationary values,
i.e.
i.e.
i.e.
(4)
or
i.e.
(5)
Now
dy
0
dx
3x 2  36 x  96  0
3( x 2  12 x  32)  0
x 2  12 x  32  0 , ( 3  0)
( x  4)( x  8)  0
x  4, 8
d2y
 2   24  36  12  negative
 dx  x 4
(6)
Hence at x = 4, y is maximum ( x = 4 is called maxima).
Further
d2y
 2   48  36  12  positive
 dx  x 8
(7)
so that at x = 8 , y is minimum.
Example 2 : Examine maximum and minimum values of x 3  3x 2  6 x  7 .
Solution : Let
(8)
y  x 3  3x 2  6 x  7
.
dy
 3x 2  6 x  6
dx
Then
(9)
d2y
 6x  6
dx 2
and
(10)
For stationary values,
i.e.
i.e.
dy
0
dx
3x 2  6 x  6  0
x 2  2x  2  0
i.e.
x
2 48
2
2 4
x
2
(11)
This will give imaginary roots. There is neither maximum nor minimum.
Example 3 : Examine the maximum and minimum values of
8 x 5  15 x 4  10 x 2 .
Solution : Let
(12)
y  8 x 5  15x 4  10 x 2
dy
 40 x 4  60 x 3  20 x
dx
Then
(13)
and
d2y
 160 x 3  180 x 2  20
2
dx
(14)
For stationary values,
i.e.
(15)
i.e.
Roots are
i.e.
dy
0
dx
40 x 4  60 x 3  20 x  0
20 x(2 x 3  3x 2  1)  0
x  0 , 2 x 3  3x 2  1  0
( x  1)( x  1)( 2 x  1)  0
Thus stationary points are
Now
(16)
x  0, 1, 
1
2
d2y
 2   20 = positive (>0)
 dx  x 0
Hence at x = 0, y is minimum
d2y
 2   160  180  20  0
 dx  x 1
Further
(17)
so that at x = 8 , y is minimum.
Hence at x = 1, we have f ( x)  0
d2y
 1
1
 2 
 160    180   20  0
 8
4
 dx  x  12
 20  45  20  negative
(18)
Hence at x  
1
, y is maximum.
2
Now for x = 1, according to working rule (vi), we find
d3y
 f ( x)  480 x 2  360 x
3
dx
(19)
and
(20)
d3y
 3   480  360  0 .
 dx  x 1
Hence at x =1 there is neither maxima nor minima.
Maxima and Minima of Functions of Two Variables
We have
u  f ( x, y ) and
we want to find maximum or minimum at
x  a, y  b . Thus
f ( x, y ) becomes
f (a, b) . Theory of partial
differentiation is used.
The necessary conditions for f (a, b) to be an extreme value of f ( x, y )
are, that
f
0
x
and
f
0
y
at x  a, y  b .
If these conditions are satisfies, and if let
,
2 f
2 f
2 f
,
,
and

s

r
t
x y
x 2
y 2
(1)
evaluated at x  a, y  b , then the sufficient conditions for f (a, b) to be
an extreme value of f ( x, y ) are
(i) rt  s 2  0 (positive) and r  0 (negative), for f (a, b) to be maximum
value,
(ii) rt  s 2  0 (positive) and r  0 (positive), for f (a, b) to be minimum
value.
Example 1 : Examine the maximum and minimum of x 3 y 2 (1  x  y) .
Solution : Let
(2)
Then
or
u  x 3 y 2 (1  x  y)
u

 y 2 [ x 3 (1  x  y )]
x
x
2
3
 y [ x (1)  (1  x  y)3x 2 ]
 y 2 [  x 3  3x 2  3x 3  3x 2 y ]
u
 y 2 (4 x 3  3x 2  3x 2 y )
x
 y 2 x 2 (3  3 y  4 x)
(3)
and
u

 x 3 [ y 2 (1  x  y )]
y
y
 x 3 [ y 2 (1)  (1  x  y)2 y]
 x 3 ( y 2  2 y  2 xy  2 y 2 )
or
 x 3 (3 y 2  2 y  2 xy)
u
 x 3 y (2  2 x  3 y )
y
(4)
For maxima and minima (necessary condition)
or
(5)
u
 0  y 2 x 2 (3  3 y  4 x)
x
x  0, y  0, 4 x  3 y  3  0
u
 0  x 3 y (2  2 x  3 y )
y
or
x  0, y  0, 2 x  3 y  2  0
(6)
so the values are (stationary points)
x  0, y  0, x  1 / 2, y  1 / 3
Further (3) gives
 2u

 r  y 2 [ x 2 (3  3 y  4 x)]
2
x
x
r  6 xy  12 x 2 y 2  6 xy3
i.e.
(7)
Also (4) gives
  u   2 u

  
 s  [ x 3 y (2  2 x  3 y )]
x  y  x y
x
i.e.
(8)
and again (4) is
s  6 x 2 y  8x 3 y  9 x 2 y 2
  u   2 u

  
 t  [ x 3 y (2  2 x  3 y )]
2
y  y  y
y
or
(9)
Now
t  2x 3  2x 4  6x 3 y .
(rt  s 2 )1 / 2,1 / 3 
1

4
positive (i.e., >0)
1
( r )1 / 2,1 / 3   =negative (i.e., <0)
9
1 1
Hence by sufficient conditions, at
function,
 , 
 2 3
and
in
(2),
is
maximum.
The maximum value is obtained by putting
3
i.e.
(10)
x
1
1
,y 
2
3
in
(2),
i.e.
2
1
1 1 1  1 1

u  x  , y        1   
2
3  2  3  2 3

1 1 1
1
u max      
8 9  6  432
.