CHAPTER 62 MAXIMA, MINIMA AND SADDLE POINTS FOR
FUNCTIONS OF TWO VARIABLES
EXERCISE 254 Page 689
1. Find the stationary point of the surface f(x, y) = x 2 + y 2 and determine its nature. Sketch the
contour map represented by z.
Let f(x, y) = z = x 2 + y 2
(i)
∂z
= 2x
∂x
∂z
= 2y
∂y
and
(ii) For stationary points,
2x = 0 from which, x = 0
and
2y = 0 from which, y = 0
(iii) The coordinates of the stationary point are (0, 0)
∂2 z
=2
∂y 2
(iv)
∂2 z
=2
∂x 2
(v)
When x = 0, y = 0,
∂2 z
∂
= =
(2 y) 0
∂x∂y ∂x
∂2 z
=2
∂x 2
∂2 z
= 2 and
∂y 2
∂2 z
=0
∂x∂y
2
 ∂2 z 
(vi) 
 =0
 ∂x∂y 
2
 ∂2 z   ∂2 z   ∂2 z 
2
(vii) ∆=
− 4 which is negative
( 0 ) − ( 2 )( 2 ) =
(0,0)

 −  2  2  =
 ∂x∂y   ∂x   ∂y 
(viii) Since ∆ < 0 and
∂2 z
> 0 then (0, 0) is a minimum point.
∂x 2
The contour map for z is shown below.
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2. Find the maxima, minima and saddle points for the following functions:
(a) f(x, y) = x 2 + y 2 − 2 x + 4 y + 8
(b) f(x, y) = x 2 − y 2 − 2 x + 4 y + 8
(c) f(x, y) = 2 x + 2 y − 2 xy − 2 x 2 − y 2 + 4
(a) Let f(x, y) = z = x 2 + y 2 − 2 x + 4 y + 8
(i)
∂z
= 2x − 2
∂x
∂z
= 2y + 4
∂y
and
(ii) For stationary points,
2x – 2 = 0 from which, x = 1
and
2y + 4 = 0 from which, y = – 2
(iii) The coordinates of the stationary point are (1, –2)
∂2 z
=2
∂y 2
(iv)
∂2 z
=2
∂x 2
(v)
When x = 1, y = –2,
∂2 z
∂
=
4) 0
( 2 y +=
∂x∂y ∂x
∂2 z
= 2 and
∂y 2
∂2 z
=2
∂x 2
∂2 z
=0
∂x∂y
2
 ∂2 z 
(vi) 
 =0
 ∂x∂y 
2
 ∂2 z   ∂2 z   ∂2 z 
2
(vii) ∆ (1,=
−4 which is negative
( 0 ) − ( 2 )( 2 ) =
−2)

 −  2  2  =
∂
∂
∂
∂
x
y
x
y






(viii) Since ∆ < 0 and
∂2 z
> 0 then (1, –2) is a minimum point.
∂x 2
(b) Let f(x, y) = z = x 2 − y 2 − 2 x + 4 y + 8
(i)
∂z
= 2x − 2
∂x
and
∂z
=
−2 y + 4
∂y
(ii) For stationary points,
and
2x – 2 = 0 from which, x = 1
–2y + 4 = 0 from which, y = 2
(iii) The coordinates of the stationary point are (1, 2)
∂2 z
= −2
∂y 2
(iv)
∂2 z
=2
∂x 2
(v)
When x = 1, y = 2,
∂2 z
∂
=
( −2 y + 4 ) = 0
∂x∂y ∂x
∂2 z
=2
∂x 2
∂2 z
= −2 and
∂y 2
∂2 z
=0
∂x∂y
2
 ∂2 z 
(vi) 
 =0
 ∂x∂y 
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 ∂2 z   ∂2 z   ∂2 z 
(vii) ∆=
(1,2)

 −
=

 ∂x∂y   ∂x 2   ∂y 2 
( 0 ) − ( 2 )( −2 )=
2
4 which is positive
(viii) Since ∆ > 0 then (1, 2) is a saddle point
(c) Let f(x, y) = z = 2 x + 2 y − 2 xy − 2 x 2 − y 2 + 4
(i)
∂z
=2 − 2 y − 4 x
∂x
∂z
=2 − 2 x − 2 y
∂y
and
(ii) For stationary points,
2 – 2y – 4x = 0
i.e.
1 – y – 2x = 0
and
2 – 2x – 2y = 0
i.e.
1–x–y=0
(1)
(2)
From (1), y = 1 – 2x
Substituting in (2) gives: 1 – x –(1 – 2x) = 0
i.e. 1 – x – 1 + 2x = 0
from which, x = 0
When x = 0 in equations (1) and (2), y = 1
(iii) The coordinates of the stationary point are (0, 1)
∂2 z
= −2
∂y 2
(iv)
∂2 z
= −4
∂x 2
(v)
When x = 0, y = 1,
∂2 z
∂
= ( 2 − 2x − 2 y ) =
−2
∂x∂y ∂x
∂2 z
= −4
∂x 2
∂2 z
= −2 and
∂y 2
∂2 z
= −2
∂x∂y
2
 ∂2 z 
(vi) 
(−2) 2 =
4
 =
∂
∂
x
y


2
 ∂2 z   ∂2 z   ∂2 z 
(vii) ∆=
(0,1)

 −
 =4 − ( −4 )( −2 ) =−4 which is negative

 ∂x∂y   ∂x 2   ∂y 2 
(viii) Since ∆ < 0 and
∂2 z
< 0 then (0, 1) is a maximum point
∂x 2
3. Determine the stationary values of the function f(x, y) = x3 – 6x2 – 8y 2 and distinguish between
them.
Let f(x, y) = z = x3 – 6x2 – 8y 2
(i)
∂z
= 3 x 2 − 12 x
∂x
and
(ii) For stationary points,
and
∂z
= −16 y
∂y
3 x 2 − 12 x = 0 i.e. 3x(x – 4) from which, x = 0 and x = 4
–16y = 0 from which, y = 0
(iii) The coordinates of the stationary points are (0, 0) and (4, 0)
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∂2 z
= −16
∂y 2
∂2 z
∂
= ( −16 y ) =0
∂x∂y ∂x
(iv)
∂2 z
= 6 x − 12
∂x 2
(v)
When x = 0, y = 0,
∂2 z
= −12
∂x 2
∂2 z
= −16 and
∂y 2
∂2 z
=0
∂x∂y
When x = 4, y = 0,
∂2 z
= 12
∂x 2
∂2 z
= −16 and
∂y 2
∂2 z
=0
∂x∂y
2
(vi) For (0, 0),
 ∂2 z 

 =0
 ∂x∂y 
For (4, 0),
 ∂2 z 

 =0
 ∂x∂y 
2
2
 ∂2 z   ∂2 z   ∂2 z 
2
(vii) ∆=
(0,0)

 −  2   2  =( 0 ) − ( −12 )( −16 ) =−192 which is negative
 ∂x∂y   ∂x   ∂y 
2
 ∂2 z   ∂2 z   ∂2 z 
2
+192 which is positive
∆=
( 0 ) − (12 )( −16 ) =
(4,0)

 −  2  2  =
∂
x
∂
y
∂
x
∂
y






(viii) Since ∆ (0,0) < 0 and
∂2 z
< 0 then (0, 0) is a maximum point
∂x 2
Since ∆ (4,0) > 0 then (4, 0) is a saddle point
4. Locate the stationary points of the function z = 12 x 2 + 6 xy + 15 y 2
(i)
∂z
= 24 x + 6 y
∂x
and
∂z
= 6 x + 30 y
∂y
(ii) For stationary points,
and
(iii) From (1),
6y = –24x
Substituting in (2) gives:
i.e.
24x + 6y = 0
(1)
6x + 30y = 0
(2)
i.e.
y = –4x
6x + 30(–4x) = 0
6x = 120x
i.e.
x=0
When x = 0, y = 0, hence the coordinates of the stationary point are (0, 0)
∂2 z
= 30
∂y 2
(iv)
∂2 z
= 24
∂x 2
(v)
When x = 0, y = 0,
∂2 z
∂
=
( 6 x + 30 y ) = 6
∂x∂y ∂x
∂2 z
= 24
∂x 2
∂2 z
= 30 and
∂y 2
∂2 z
=6
∂x∂y
2
 ∂2 z 
(vi) 
 = 36
 ∂x∂y 
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 ∂2 z   ∂2 z   ∂2 z 
(vii) ∆=
(0,0)

 −

 =
 ∂x∂y   ∂x 2   ∂y 2 
(viii) Since ∆ < 0 and
( 6 ) − ( 24 )( 30 )
2
which is negative
∂2 z
> 0 then (0, 0) is a minimum point
∂x 2
5. Find the stationary points of the surface z = x3 – xy + y 3 and distinguish between them.
(i)
∂z
= 3x 2 − y
∂x
∂z
=− x + 3 y 2
∂y
and
(ii) For stationary points,
and
3x 2 – y = 0
(1)
−x + 3y2 = 0
(2)
(iii) From (1),
y = 3x 2
Substituting in (2) gives: – x + 3 ( 3x 2 ) = 0
2
– x + 27x 4 = 0
x ( 27 x 3 − 1) =
0
and
i.e.
x=0
27 x3 − 1 =0
or
Hence, x = 0 or x =
i.e.
x3 =
1
27
and
=
x
 1  1
=
 
 27  3
3
1
3
From (1), when x = 0, y = 0
2
1
1 1
and when x = , y = 3x 2 = 3   =
3
3 3
1 1
Hence, the stationary points occur at (0, 0) and  , 
3 3
(iv)
∂2 z
= 6x
∂x 2
∂2 z
= 6y
∂y 2
∂2 z
∂
= ( − x + 3 y 2 ) =−1
∂x∂y ∂x
(v)
For (0, 0),
∂2 z
=0
∂x 2
∂2 z
= 0 and
∂y 2
∂2 z
= −1
∂x∂y
1 1
For  ,  ,
3 3
∂2 z
=2
∂x 2
∂2 z
= 2 and
∂y 2
∂2 z
= −1
∂x∂y
2
 ∂2 z 
(vi) For (0, 0), 
 =1
 ∂x∂y 
2
 1 1   ∂2 z 
For  ,  , 
 =1
 3 3   ∂x∂y 
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(vii) ∆ (0,0)
 ∂2 z   ∂2 z   ∂2 z 
=
1 − ( 0 )( 0 ) =
1 which is positive

 −
=

 ∂x∂y   ∂x 2   ∂y 2 
1 − ( 2 )( 2 ) =
∆ 1 1  =
−3 which is negative
 , 
 3 3
(viii) Since ∆ ( 0,0) > 0 then (0, 0) is a saddle point
∆  1 1  < 0 and
 , 
 3 3
∂2 z
1 1
> 0 then  ,  is a minimum point
2
∂x
3 3
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EXERCISE 255 Page 693
1. The function z = x 2 + y 2 + xy + 4 x − 4 y + 3 has one stationary value. Determine its coordinates
and its nature.
∂z
= 2x + y + 4
∂x
(i)
and
∂z
= 2y + x − 4
∂y
(ii) For stationary points,
and
(iii) (1) + (2) gives:
2x + y + 4 = 0
(1)
2y + x – 4 = 0
(2)
3x + 3y = 0
Substituting in (1),
2x – x + 4 = 0
from which, y = –x
i.e. x = –4, thus y = +4
Hence, the stationary point occurs at (–4, 4)
∂2 z
=2
∂y 2
(iv)
∂2 z
=2
∂x 2
(v)
When x = –4, y = 4,
∂2 z
∂
=
4) 1
( 2 y + x −=
∂x∂y ∂x
∂2 z
=2
∂x 2
∂2 z
= 2 and
∂y 2
∂2 z
=1
∂x∂y
2
 ∂2 z 
(vi) 
 =1
 ∂x∂y 
2
 ∂2 z   ∂2 z   ∂2 z 
2
(vii) ∆ ( −4,4)
−3 which is negative
= 
(1) − ( 2 )( 2 ) =
 −  2  2  =
 ∂x∂y   ∂x   ∂y 
(viii) Since ∆ < 0 and
∂2 z
> 0 then (–4, 4) is a minimum point
∂x 2
2. An open rectangular container is to have a volume of 32 m3. Determine the dimensions and the
total surface area such that the total surface area is a minimum.
Let the dimensions of the container be x, y and z as shown below.
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Volume V = xyz = 32
(1)
Surface area, S = xy + 2yz + 2xz
(2)
From equation (1), z =
32
xy
Substituting in equation (2) gives:
 32 
 32 
S = xy + 2y   + 2x  
 xy 
 xy 
i.e. S = xy +
64
64
+
y
x
which is a function of two variables
∂S
64
=y–
= 0 for a stationary point, hence x2y = 64
2
∂x
x
64
∂S
=x–
= 0 for a stationary point, hence xy2 = 64
2
∂y
y
(3)
(4)
Dividing equation (3) by (4) gives:
x2 y
= 1 i.e.
xy 2
x
= 1 i.e.
y
x=y
Substituting y = x in equation (3) gives x3 = 64, from which, x = 4 m
Hence y = 4 m also
From equation (1), (4)(4)z = 32 from which, z =
32
=2m
16
∂ 2S
∂ 2 S 128 ∂ 2 S 128
=
,
=
and
=1
∂ y2
y3
∂ x∂ y
∂ x2
x3
When x = y = 4,
∂ 2S
∂ 2S
∂ 2S
= 2,
= 2 and
=1
∂ y2
∂ x∂ y
∂ x2
∆ = (1)2 – (2)(2) = –3
Since ∆ < 0 and
∂ 2S
> 0, then the surface area S is a minimum
∂ x2
Hence the minimum dimensions of the container to have a volume of 32 m3 are 4 m by 4 m by 2 m
From equation (2), minimum surface area, S = (4)(4) + 2(4)(2) + 2(4)(2)
= 16 + 16 + 16 = 48 m2
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3. Determine the stationary values of the function f(x, y) = x 4 + 4 x 2 y 2 − 2 x 2 + 2 y 2 − 1 and
distinguish between them.
Let f(x, y) = z = x 4 + 4 x 2 y 2 − 2 x 2 + 2 y 2 − 1
(i)
∂z
=4 x3 + 8 xy 2 − 4 x
∂x
(ii) For stationary points,
∂z
= 8x2 y + 4 y
∂y
and
4 x3 + 8 xy 2 − 4 x =
0
(1)
8x2 y + 4 y =
0
and
(2)
4 y ( 2 x 2 − 1) =
0 from which, y = 0
(iii) From (2),
0
4 x3 − 4 x =
0 i.e. 4 x ( x 2 − 1) =
From (1), if y = 0,
x = 0 or x = ± 1
from which,
Hence, the stationary points occur at (0, 0) and (1, 0) and (–1, 0)
∂2 z
∂
y ) 16 xy
=
(8 x 2 y + 4=
∂x∂y ∂x
∂2 z
= 8x2 + 4
∂y 2
(iv)
∂2 z
= 12 x 2 + 8 y 2 − 4
∂x 2
(v)
For (0, 0),
∂2 z
= −4
∂x 2
∂2 z
= 4 and
∂y 2
For (1, 0),
∂2 z
=8
∂x 2
∂2 z
= 12
∂y 2
For (–1, 0),
∂2 z
=8
∂x 2
∂2 z
=0
∂x∂y
and
∂2 z
= 12
∂y 2
∂2 z
=0
∂x∂y
and
∂2 z
=0
∂x∂y
2
 ∂2 z 
(vi) For all three stationary points, 
 =0
 ∂x∂y 
2
 ∂2 z   ∂2 z   ∂2 z 
(vii) ∆=
=
(0,0)

 −


 ∂x∂y   ∂x 2   ∂y 2 
( 0 ) − ( −4 )( 4=)
2
32 which is positive
2
 ∂2 z   ∂2 z   ∂2 z 
2
−96 which is negative
∆=
( 0 ) − (8)(12 ) =
(1,0)

 −  2  2  =
 ∂x∂y   ∂x   ∂y 
2
 ∂2 z   ∂2 z   ∂2 z 
2
−96 which is negative
∆ ( −=
( 0 ) − (8)(12 ) =
1,0)

 −  2  2  =
 ∂x∂y   ∂x   ∂y 
(viii) Since ∆ (0,0) > 0, then (0, 0) is a saddle point
Since ∆ (1,0) < 0 and
∂2 z
> 0 then (1, 0) is a minimum point
∂x 2
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∂2 z
> 0 then (– 1, 0) is a minimum point
∂x 2
Since ∆ ( −1,0) < 0 and
4. Determine the stationary points of the surface f(x, y) = x3 – 6x2 – y2
Let f(x, y) = z = x 3 − 6 x 2 − y 2
∂z
= 3 x 2 − 12 x
∂x
(i)
∂z
= −2 y
∂y
and
(ii) For stationary points,
3 x 2 − 12 x = 0
(1)
–2y = 0
(2)
and
(iii) From (1)
3 x( x − 4) = 0
From (2),
y=0
from which, x = 0 or x = 4
Hence, the stationary points occur at (0, 0) and (4, 0)
∂2 z
∂
=
( −2 y ) = 0
∂x∂y ∂x
∂2 z
= −2
∂y 2
(iv)
∂2 z
= 6 x − 12
∂x 2
(v)
At (0, 0),
∂2 z
= −12
∂x 2
∂2 z
∂2 z
= −2 and
=0
∂y 2
∂x∂y
At (4, 0),
∂2 z
= 12
∂x 2
∂2 z
= −2 and
∂y 2
∂2 z
=0
∂x∂y
2
 ∂2 z 
(vi) For both points, 
 =0
 ∂x∂y 
2
 ∂2 z   ∂2 z   ∂2 z 
2
(vii) ∆=
(0,0)

 −  2   2  =( 0 ) − ( −12 )( −2 ) =−24 which is negative
 ∂x∂y   ∂x   ∂y 
∆ (4,0) =
( 0 ) − (12 )( −2 )=
2
(viii) Since ∆ (0,0) < 0 and
24 which is positive
∂2 z
< 0 then (0, 0) is a maximum point
∂x 2
Since ∆ (4,0) >0 then (4, 0) is a saddle point
5. Locate the stationary points on the surface
f(x, y) = 2x 3 + 2y 3 – 6x – 24y + 16
and determine their nature.
Let f(x, y) = z = 2x 3 + 2y 3 – 6x – 24y + 16
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(i)
∂z
= 6x2 − 6
∂x
∂z
= 6 y 2 − 24
∂y
and
(ii) For stationary points,
and
6x2 − 6 =
0
(1)
6 y 2 − 24 =
0
(2)
6 x2 = 6
(iii) From (1),
6 y 2 = 24
From (2),
from which, x = ± 1
i.e.
y 2 = 4 and y = ± 2
Hence, the stationary points occur at (1, 2) and (–1, –2) and (–1, 2) and (1, –2)
(iv)
∂2 z
= 12 x
∂x 2
(v)
For (1, 2),
∂2 z
= 12 y
∂y 2
∂2 z
= 24 and
∂y 2
∂2 z
= 12
∂x 2
For (–1, –2),
∂2 z
∂
=
( 6 y 2 − 24=) 0
∂x∂y ∂x
∂2 z
= −12
∂x 2
For (–1, 2),
∂2 z
= −12
∂x 2
For (1, –2),
∂2 z
= 12
∂x 2
∂2 z
=0
∂x∂y
∂2 z
= −24
∂y 2
and
∂2 z
=0
∂x∂y
∂2 z
= 24
∂y 2
and
∂2 z
=0
∂x∂y
∂2 z
= −24
∂y 2
and
∂2 z
=0
∂x∂y
2
 ∂2 z 
(vi) For all four stationary points, 
 =0
 ∂x∂y 
2
 ∂2 z   ∂2 z   ∂2 z 
2
(vii) ∆=
−288 which is negative
( 0 ) − (12 )( 24 ) =
(1,2)

 −  2  2  =
 ∂x∂y   ∂x   ∂y 
2
 ∂2 z   ∂2 z   ∂2 z 
2
∆ ( −1,−=
2)

 −  2   2  =( 0 ) − ( −12 )( −24 ) =−288 which is negative
 ∂x∂y   ∂x   ∂y 
2
 ∂2 z   ∂2 z   ∂2 z 
2
∆ ( −1,2)
= 
 −  2   2  =( 0 ) − ( −12 )( 24 ) =+288 which is positive
 ∂x∂y   ∂x   ∂y 
2
 ∂2 z   ∂2 z   ∂2 z 
2
+288 which is positive
∆ (1,=
( 0 ) − (12 )( −24 ) =
−2)

 −  2  2  =
 ∂x∂y   ∂x   ∂y 
(viii) Since ∆ ( −1,2) > 0, then (–1, 2) is a saddle point
Since ∆ (1,−2) > 0, then (1, –2) is a saddle point
Since ∆ (1,2) < 0 and
Since ∆ ( −1,−2) < 0 and
∂2 z
> 0 then (1, 2) is a minimum point
∂x 2
∂2 z
< 0 then (–1, –2) is a maximum point
∂x 2
1031
© 2014, John Bird
6. A large marquee is to be made in the form of a rectangular box-like shape with canvas covering
on the top, back and sides. Determine the minimum surface area of canvas necessary if the
volume of the marquee is to be 250 m3.
A sketch of the marquee is shown above
Volume of marquee,
V = xyz = 250
Surface area,
S = xy + yz + 2xz
From (1),
z=
Substituting in (2) gives:
 250 
 250 
250 500
S = xy + y 
+
=xy + 250 x −1 + 500 y −1
 + 2x 
 =xy +
x
y
 xy 
 xy 
∂S
250
= y−
∂x
x2
from which,
y=
250
x2
∂S
500
= x−
∂y
y2
∂S
250
=−
y
=
0
∂x
x2
yx 2 = 250
or
(3)
500
∂S
x−
0
=
=
y2
∂y
and
from which,
(2)
250
xy
and
For a stationary point,
(1)
x=
500
y2
xy 2 = 500
or
(4)
Dividing equation (3) by equation (4) gives:
yx 2 250
=
xy 2 500
i.e.
x 1
=
y 2
Substituting y = 2x in equation (3) gives:
and
From equation (1),
and
y = 2x
2 x3 = 250
and
x = 3 125 = 5 m
y = 2x = 10 m
xyz = 250
i.e.
(5)(10)z = 250
1032
from which, z = 5 m
© 2014, John Bird
∂ 2 S 750
=
∂x 2
x3
When x = 5 and y = 10,
∂ 2 S 1000
=
∂y 2
y3
∂2S
=6
∂x 2
∂2S
=1
∂y 2
and
and
∂2S
=1
∂x∂y
∂2S
=1
∂x∂y
2
 ∂2S   ∂2S   ∂2S 
2
−6 which is negative
=
∆ 
(1) − ( 6 )(1) =
 −  2  2  =
 ∂x∂y   ∂x   ∂y 
Since ∆ < 0 and
∂2 z
> 0 then the surface area is a minimum.
∂x 2
Minimum surface area, S = xy + yz + 2xz
= (5)(10) + (10)(5) + (2)(5)(5)
= 50 + 50 + 50 = 150 m 2
1033
© 2014, John Bird