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example questions calc, Exercises of Mathematics

example questions for calc three

Typology: Exercises

2024/2025

Uploaded on 05/03/2025

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bg1
Exercises
14
.
7
:
2)
+(x
,
y)
=
(x
-
3
+
ye
,
Be s
X
=
3
See
y
=
0
2)
minimize
xy2
on
x2
+
y
+
z
=
1
boundary
points
:
com
a
xy
=
122
(1
,
1)
using
y
=
x2
-
f(x
,
xz)
=
(x
-
3)
+
x 4x3
+
2x
-
3
=
x
=
E
= =
f
(1
,
1)
=
(1
-
3)2
+
1
=
S
cross
multiply
-
>
222
=
2x22
results
in
-
>
y
=
X
absolute
max
on
a
cross
mulyx
x
,
y2243
10)
f(x
,
y)
=
xz
-
2xy
+
y2
,
D
=
E(x
,
4)
:
02x26
,
0
=
y
=
12
-
2x3
5
X
S
boundary
PS
(0
,
0
,
16
,
6)
,
10
,
12)
3)
maximize
2x
+
31
+
S2
on
xty
+
z
=
19
!
f10
,
07 0
,
f16
,
b)
=
0
,
flort
=
144
8
f (x
,
y
,
z)
=
(2
,
3
,
5)
=
x8g(x
,
y
,
2)
=
(2x
,
24
,
22)
absolute
min
of
2
at
1010)
=
absolute
minoto
where
absolute
max
of
6
at
boundary
points
.
3)
f(x
,
y)
=
y(x
-
3)
,
D
=
E(x
,
y)
:
x
+
y2293
2y
=
3x
+
y
= 2
x
&
Sx
=
2 -
2
=
Ex
Xy
(xzo
a
X
=
y
=
z
=
t
e
x2
+
x
+
Ex2
=
19
-
x
=
19
=
3
x
=
2
-
>
3sin(G)(3cos(0)
-
3)
=
volume
=
x yz
=
(t)3
=
1/2
al
positive b
maintaineriz
=
192
9
SinEcosO-9
sinG
=
-AsinG
+
9 cOPO-acost
=
O
X
=
0
So
stationary
point
:
10
,
0
coso
=
/2
,
o
=
13
,
5π/3
b)xz
+
y2
+
z2
=
1
,
P(z)
,
2)
1)
3
faces
on
planes
&
Vertex
on
2
=
4-x2y2
.
Maximize
Volume
:
a
=
M
x
(2
+
(y
-
12
+
(2
=
2)2
=
x
=
y
-
x
=
y
&
Of
=
((2x
-
u)
,
(zy
-
2)
,
(22
-
4))
-g
=
(2x
,
zy
,
zz)
en
a
a
E
So
S
coso
=
Sinocoso
~
absolute
max
a
a
2
-
2
=
xz
2
=
2/1
-
X
x
=
z
&
y
=
2/53
Sincecos(o)
=
0
at
0
=
0
,
12
,
i
=
first
actant
-
12
=
2
0
=
0
-
,
o
=
-
, =
=
-
t
absolute
max
of
ats
a
10-1
v
=
(F
,
3)(y)(2)
=
8/3
we

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Exercises 14. 7

:

  1. +(x , y)

= (x

+ ye

,

Be s

X = 3

See

y = 0

  1. minimize xy

on x2+

y

+z= 1

boundary points

:

coma

(1 , 1) xy= 122

using y

=x2 - f(x , xz) = (x

x 4x

2x

  • 3

=

x = E = =

f

,

1 = S

cross

multiply

= 2x

results in

y

= X

absolute maxona

cross

mulyx

x , y2243 10) f(x ,

y)

= xz

2xy

y2 ,

D =

E(x , 4)

: 02x ,

0 = y

= 12 - 2x

5 X

S

boundary

PS (0 , 0 ,

,

,

10 , 12) 3) maximize 2x +

+ S

on xty

  • z = 19

f ,

,

f16 , b)

= 0 , flort

8 f (x , y ,

z) = ( ,

=

x8g(x

, y ,

  1. = (2x , 24 ,

absolute

min of 2

at

absoluteminoto =

where

absolute max of 6 at boundary points

.

  1. f(x ,

y)

=

y(x

    1. ,

D = E(x , y)

: x+ y

2y

= 3x + y

= 2

x & Sx

= 2 - 2

= Ex

Xy

(xzoa

X = y

= z =

t

e

x2 +x + Ex2 = 19 - x= 19

= 3

x

= 2

3sin(G)(3cos(0) -

volume

=

x yz

=

(t)3 = 1/

alpositive b

maintaineriz

=

9

SinEcosO-9 sinG

=

-AsinG

9 cOPO-acost = O

X = 0

So

stationary point

: 10 , 0

coso

=

,

o = 13 ,

5 π/3 b)xz +

y

  • z2 = 1 ,

P(z) ,

1) 3 faces on

planes

& Vertex on

2

=

4-x2y.

Maximize Volume

:

a

= M x(2 + (y

=

= x = y

  • x

=

y

&

Of

=

((2x

  • u) , (zy - 2) , ( - 4))

-g

= (2x

,

zy

,

zz)

en

a

a

So E

S

coso = Sinocoso

absolutemax ~

a a 2 - 2 = xz 2 = 2/

  • X

x = z

y

=

Sincecos(o) 2/

= 0 at 0 = 0 , 12 ,

i

= first

actant

  • 12

= 0

,

o

= - , = =

  • t

absolute

maxofatsa 10-1 v

= (F,3)(y)(2)

=

8/

we