Drill On August 23th-Maxima and Minima
Drill On August 23th-Maxima and Minima
Drill On August 23th-Maxima and Minima
1. Find all the critical points of the function 3. Determine whether the following statements are TRUE
or FALSE. Explain your reasoning.
f (x, y) = x3 + xy + y 2 .
3.1 After completing the square, we find that the graph
fx(x,y) 3x + 3x4y of function z = 13 − 6x + x2 + 4y + y 2 has no minimum
2
y 0
=
= -
i
fy(x,y) =
x 2y
+ x
2y
+
0
= -
2
value.
2:6x
+ 0
1
2y
=
1 x
0
2x,2xx(x,y) 2,2xy(x,y)
- =
2x(x,y)
-
= 6 +
1 -
2:6x2 -
x 0
=
2
2y(x,y) 2y,2yy(x,y)
=
4
+ =
0
x(6x -
1) =
0,46 02 I
x =
2
Hf(x,y) 4
=
y 0, 412
=
-
Hf(x,y):4>0
critical points are 10,01 and
- al 2xx(x,y):3)0
... it
has minimum value
&
fy(x,y) eXcos(y) =
excos(y) 0
=
and there's no
y satisfied the equation
2. Find Hf (x, y) of the function f (x, y) = x3 + y 2 + 3xy 2 then this statementi s false.
fx(x,y) 3x +
=
3y2,fxx(x,y) 0x,fxy =
by
=
fy(x,y) 2y Gxy,fyy(x,y)
=
+
z 6x
=
+
Hf(x,y)6x6y -
12x
=
38x2
+
-
36y2
4. Let k ∈ R be a non zero constant and consider the function
6y 2 +
6x
Hf(1,0)
+ -
2x
4ki
-
+
=
1
4 0;y
fy(x,y)
=
4y 4 4y
-
=
-
=
a 3,b 2,C 4k
=
=
=
-
the
for
<-Iz,
2
b -
4a <0 k
4
+
48k<0 no critical point
* - 1/12
5. In each part, compute fx (x, y) and fy (x, y) of the given 6.2 f (x, y) = x3 + y 2 − xy
function and use them to locate all critical points.
fx(x,y) 3x2 y,fxx1x,y) 6x,fxy(x,y) z
= -
= -
zy
-
=
2x - 2)
64 -
1
y2 + Hf(x,y) =
12x
= -
1
fy(x,y) (x 2x).)
+
1 2
zy)
-
=
-
(x y2 +
+
2x1
1
2)e 0
1 2x
1, 2
- - =
2
1- 0
=
2).(x2 y2
+
+
xx) Hf(x,y) - 1 1
n 2:1
+
-
2x -
zy
-
0
=
fxx(x,y) 8 1
2x 2 0; 2x zy 21
zy
- =
-
=
- - -
n 2: +
4x then
4,x 1 y 0
- =
=
- =
.
critical pointis 1-1,0)
5.2 f (x, y) = x3 + y 2 − xy
fx(x,y) 3x2 -
y 3x2 0 1
y
=
-
=
-
-
-
fy(x,y) zy x 2y -
x 0
=
-
2
6x 0 c
1.2: -
zy
+ =
2 1:
+
-
x 6x2
+ 0
=
However, this problem would help you develop your prob-
x(1 6x) 0
lem solving and understanding skills.
=
- -
x
0,46
=
2x - 2)
fy(x,y)
-
(4 y2 2x)
2)(x2 y2 2x) (
+
z)2
+
+
+
fxx(x,y) 1
=
-
+
-
2x -
4.(4 y2 2x), + +
= -
2 4x2
+
8x
+
+
2)(x2 y2 2x)
+
+
(4x2
=
8x
+
+
(4 ( z)e
-
8
=
-
-
+ =
(x y 2x)
z)(x2 y2 2x)
zy
+
fyy(x,y)
+
+ +
(
=
-
+
( -
y2
(4y2 z)j(x2
+ 2x)
+
Fyy) -
1,0) 1 =
27e
x+ y2
fxy(x,y) 2x!(
+
-
( 2x z).e zy)
= -
- -
fxy) -
1,0) 0
=
2e0
Hf) -
1,0)
-
4e2
=
ze
-
Hf( -
1,0)
>0
fxx) 1,0)(0 -