Exercisa P, solutions

1) We want to find the min/max

off(x, y) x* =
+x y zy3 6y-
+

on the
rectangle -
1
[XC1 and
-2.53-2.
ay
2

&f (fX,fzz -
=

>
1
↓ 1
X
-

(3x2 2xy,X (y- 6) 0

=
+
+ =
/
2
/

R
-

x(3X
2y) X0 x zy
3
0
-

=> or
=

+ =

-
=

=>

x 2(y- 6 + 0
=

X 0
=
-
=
6y ( + 0
=

y
=
1 =
y 11 =
=
(0,11)

x =
-

y 6y- 6
yz +
0
=
=
y =

z y 133
=

=> x 1
=

) 35) =
=
=

2 z
(F23,135
Next, we use the Hessian to classify the critical points:

Exx 6X 2y,fxy 2x,

=
+ =

fyz = -

ky

det6X
1 1
2y +
2x
=> D =

24 -

12y

I
I2
for(0,I1) =1
det.
=

2 points
saka
253
For(75,135) P70
= Chech!

fxx( 2,3) 6( v) 2(35 -

-

+
-
=

=( - 253,353) loce man

fxx(23, 5)70 -

a
lumin

Next, we need to check boundaries.

2,ix 1, =
-

2fy7z = ay
2 22

23 ↑
f(1,y) =
-

2y3 7y
+
+
1 I
X
↳
-
1
/
/
1 >X
/
f(
- -
=

6yz +
7 0
=
=
y 1
=

6 -
2
/

R
24

=>
(1,F(s),(1, 1(6) -
C.
P.

the second derivative test:

using
(1,17(c) -T7/) load
"
7 012y loud (2,
= => max, min

4: y 2, = -
1(X11 =>

f(X,2) x3 = 2X - 4
+ =
f((X,2) 3X =
-
+
yX 0
=

=> X 0,X = -
=

A.

1 16X + 4 =
(0,2) is a look min
233x -
=
11 -

2.5y-2
f( 1,y)
-
=
-

2y3 7y + - 1 (
= -

1,f(),( -1, 56) -

Load may loal min

24:y =
-
2, -
(X4

f(x, -
2) x32X -
=

4 +
f =3X2
- -
4X 0
=

=> x 0, = X
Y,
=
N.A.

10,-2) loc max.

For globe min/max, we need to check endpoints, i.e.

the corners of the rectangle:1-2, -11, 12, 13, (9, -2),(1,2)
Globe min
71-1/), (1,((6)
 2) min/max of f(x,y)=5X-33 subject to x y
=
136
=

sol,
- >
-
we use the
Legrange multiplies g(x,y)
25x(K
=f 759
[
5
x(2X,2y)
=

(5,
= -
3) = =>

-
3 2xy()
=

=
we also have x+ 3 136 (ss

x x+CsX
z
3
(1), (2) give y
-

=>
= = =

=
z(1 2s)X
+ 136 =x
34

X 10
= =
y -
=
6,X -
=
10 =>
y6 =

f (10, -6) =- 68 min.

I ( 10,61 68
-
= max
3) min/maxof F(x,y) 2x = + y2 subjectto x +

y 3. =

g(x,y)
If x59 (4X,2y) x(1,1)
=
= =

->
4X x,2y x
=
= => 2X
y
=

substitute into the constraint:x + 2X = 3= x1 =

32
= =

So, we
only get one point (1,2)!
closed and bounded
x
y
+
=
y is
not a
region.
2x2 yz+ c
=
f(1,2)
=
x y3
+ =

min
Ifc6 => =

+5 1
=

2 -
(1,2)

- f(1,2) <4,4> =
"
at
The
tang atline (1,2) is;x
y
+
3
=

no nax!
4) If (2x y= -
y,XyX) -
f(X,y) =?

81. fx 2xz
=
-

y f(x,y) 5(2x3 y)dx+9(3)

=
= -

f(x,y) x-y
=> = -
yx+g(y)

fy g((y)
=

g((y)
-
x
= x + x x
=
= 0
=

g(y) c
=
=

f(X,y)
z x
=
y -
yx c
+
5)
(asSSxy-dxdy,k G(x,y)/1[X-35,0337,) =

-25x yz(X)(y (y2) 33re

=

5y2(25 -

1)dy 125y3y
=

12(53)? 4(8
=
= -
0) 32
=

Alternatively:

(2SSxyzdxdy ((x(x))))
=

[XY.(13) 1001(8
=
= -
8 32
(b)((
"

yny -

xby x

=
"22b
()) xa)
-

(22 11237121x
x
=

E)-z (e- 1)dX E) (1

+
x
x
= = - e

E(X x) =1(2 2 - 1)
-

=
e
+
+

1(1
=
+

E2)
(S), S
y dx
dy
f(3)
Sxx X tan0 ->dX=secodo
=

-
Stanio.Secio dO Ssecodo =

- >
-

ate S seco.seco do u sec0-du=secotano

=
do

Seco dO=dV- v=tanO

= SKOtanO
-
Stano seco so

Secotano.
=
5 (scc'0-1) Secode

=500 tanc-SSeiodo Ssecodo + ->

JSeiodo E S00tanp+El/sec0
= tanol
+

c
+
 tanc x =S0
=
VX
=
2
X
10
1

-> S(x2aX Ex, 1I(x 1

= x) x + + 1
+

=>

SVx32x =
E(x,
x +((X
yy))3 +

f(y) (πz
=
(n)1 1)
+ +
-
yy, ((y ry3)]
-
+

S'f(y)+y z)' (k
=
((1 ()
+ + -
3,5- k(3 15))d) +

=
Evz Eu(1 ()
+ + -

1)33idy
①
-

15((y 1y3)dy +

⑫
⑤53nz: dy u =
1
+
y2 du 2ydy
z =

y u1

5zu"
0 =
=

).
=

- au
1
=
( y =
4
17 2
=

f(2""1) =

5(2k - 1)

S'mn
u ((y vyz) =>
⑫ (3 193(d)
+
=

x
+

1y 1
du =

Sea
=
My
(3(n(3 13-)) - 2
= +

dv
dy
= + v y
=

- (n(1 12)
+
-

[(53)! (n(1 1)
=
+
-
(k -

(fy(t) 1vz Eu(()

=
+

E.5(2B -
1) -

zby)5) +

(1
= -

b)5 b + -
z =
flr-1)
S.(,
X1 =
xy =

dx
y
y =

1 -
. . .- .

S
**

(Y) (y)<x
,y
o
I

=S'(y.3); 4X 5'x,Xdx
=

u 1 X =du

1)usyhere
2xdX

=)z adu
= + =

=
X 0
=
n1
= =

X 1 =
42
- =

5(c" 1) 5
-
= =