Calculus 4 notes 2020, Study notes of Calculus

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LECTURE TWO
STATIONARY POINTS
They are also referred to extreme values, critical points. A necessary condition
that a function has critical point is
0
x
f
x
f
and
0
y
f
y
f
…………………………………..A
If point
00,yx
is a critical point satisfying equation A above and
2
))((xyyyxx FFF
……………………………………………..B
Then:
i. Point
00,yx
is maximum point if
0,0,0 yyxx ff
ii. Point
00,yx
is minimum point if
0,0,0 yyxx ff
iii. Point
00,yx
is saddle point (undefined) if
0
iv. If
0
no information is given and therefore further investigation is
required.
EXAMPLE
pf3
pf4
pf5
pf8
pf9
pfa

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LECTURE TWO

STATIONARY POINTS

They are also referred to extreme values, critical points. A necessary condition

that a function has critical point is

fx x

f

and   0

fy y

f

…………………………………..A

If point  x 0 , y 0 is a critical point satisfying equation A above and

 

2

( Fxx )( Fyy ) Fxy ……………………………………………..B

Then:

i. Point  x 0 , y 0 is maximum point if^ ^0 ,^ fxx ^0 , fyy ^0

ii. Point  x 0 , y 0 is minimum point if  0 , fxx  0 , fyy  0

iii. Point  x 0 , y 0 is saddle point (undefined) if 0

iv. If  0 no information is given and therefore further investigation is

required.

EXAMPLE

From equation D we can see that x  0 & x  1. Now use the fact that

2 yx

to get the critical points.

2 xy    0 , 0 ^ ………………………………………………..E

2 xy    1 , 1  …………………………………………………..F

So, we get two critical points. All we need to do now is classify them. To

do this we will need . Here is the general formula for

      

2  f (^) xx x , y. fyy x , yfxy x , y

   

36 9

2

xy

x y

……………………………………………………..G

To classify the critical points all that we need to do is plug in the critical

points and use the fact above to classify them.

For point  (^0) , 0  :   36 xy  9  36    0 0  9   9

So, for (^)  0 , 0  ,  is negative and so this must be a saddle point.

For point  1 , 1  :   36 xy  9  36    11  9  27  0 and

fxx  1 , 1   6  0

For (  1 , 1  ,  is positive and fxx is positive and so we must have a relative

minimum.

EXAMPLE 2.

Find and classify all the critical points of  ,  3 3 2

2 3 2 f x yx yyy

SOLUTION

As with the first example we will first need to get all the first and second

order derivatives.

f y f y f x

f xy x f x y y

xx yy xy

x y

6 6 , 6 6 , 6

2 2

We’ll first need the critical points. The equations that we’ll need to solve

this time are,

2 2   

x y y

xy x

These equations are a little trickier to solve than the first set, but once you

see what to do they really aren’t terribly bad.

First, let’s notice that we can factor out a 6 x from the first equation to get,

6 x ( y  1 ) 0

So, we can see that the first equation will be zero if x  0 or y  1. Be

careful to not just cancel the xx from both sides. If we had done that we

would have missed x  0.

b) Find the extreme values of the function

2 2

f x y ,  xy  x  y  2 x  2 y  4

f f  x , y   x y  1  x  y 

3 2

c) A rectangular box without a lid is to be made from

2

12 m of cardboard.

Find the maximum volume of such a box.

Part of assignment one 2,3,a and c

CAT 1 Q1&2 for T.I.E AND BSC. EEE

2. Construction Company has been contracted to build an auditorium rectangle

in shape and is to have a volume of 12000 square feet. It is estimated that the

annual heating and cooling cost will be sh 20 per square foot for the top and sh 40

per square foot for the front and back and sh 30 per foot for the sides. Find the

dimension of the building that will results in a minimum annual heating and

cooling cost.

CAT 1 Q1&2 for GEGIS, GIS AND BED. EEE

a) Classify extreme values of the function F  x , y  x y 6 x 6 y 9 xy

3 3 2 2     

b) A material for constructing a box costs per square meter for top

and per square meter for sides and bottom, if the volume of the

box is 12 cubic meters. Find the dimension of the box that would

minimize the cost of material for constructing.