8/13/2019 MAT Ans

1/8

1. 2. Using first derivative test find the local extrema points

a) f(x) = x3/3+x2/2-6xf(x)= x

2+x-6

f(x)=(x+3)(x-2)

the extrema points are the values of x that make f(x)=0f(x)=0 if (x+3)=0 or (x-2)=0

so the extrema points are : -3 and 2b) f(x)= x4/4+x3/3-5x2/2+3x

f(x)= x3+x2-5x+3

f(x)=(x2+2x-3)(x-1)

f(x)=(x+3)(x-1)(x-1)=(x+3)(x-1) 2

so the extrema points are the values of x where: x+3=0 or x-1=0

x=-3 and x=1 are the extrema points

c) f(x)=2-|x|f(x)= -x\|x|

here for f(x)=0 x must be equal to zero (0) but the equation f(x) is not defined forthe value of x=0 so there will be no extrema points

3.

8/13/2019 MAT Ans

2/8

4. Find critical points if anya) f(x)=x2+4/x2

f(x)=2x-8x3

f(x)=(2x4-8)/x

3

here f(x)=0 where 2x4-8=0; x

4-4=0

x

4

=4x=21/2

b) f(x)=(x-1)/x2f(x)=[x2 (x-1)2x]/x4

f(x)= [-x2+2x]/x4

f(x) = [2x(x-1)]/x4

here for f(x)=0

-2x(x-1)=0

and x must not be equal to 0 hence the equation is not defined for x=0\

x-1= 0 the critical point for this equation is x=1c) f(x)=x 6/5-12x 1/5

f(x)= (6/5)x

6/5-1

-12(1/5)x

1/5-1

f(x)= (6/5)x1/5

- (12/5)x-4/5

to solve for the values of x that would make f(x)=0 let y=x 1/5

here f(y)=(6/5)y-(12/5y4)

f(y)= (6y5-12)/5y4

f(y)=0 where 6y5-12=0

y5=2

from the above replacement we have y= x1/5

so y5=x

and the critical point in this case will be at

x=2

d) f(x)=x/(x2-9)f(x)=(x

2-9)-2x

2/(x

2-9)

2

f(x)= -(x2+9)/(x2-9)

2

here f(x)=0 if x2+9=0

x2+9 =0

x2=-9 which is not defined for every real number so the function

f(x)=x/(x2-9) have no critical point5.

8/13/2019 MAT Ans

3/8

6. Find the absolute extremum of f on the given intervals (if any)a) f(x)=x2, (-3,2]

f(x)=2x , f(x)=0 when x=0

the absolute extremum in this case will be at x = 0, or x=2

f(0)=0

f(2)= 4f(-3)=9

so for the given range i.e. (-3,2] we have one extremum point at x=0that is an absolute minimum value on the defined range

b) f(x)=x3+2x2-4x+1, (-3,2)f(x)= 3x

2+4x-4

f(x)= 3(x+2)(x-2/3)

f(x)=0 when x=-2 or x=2/3

the absolute extremum in this case will be at x = -2, or x=2/3

f(-3)= 4f(-2)=9

f(2/3)= 1/9f(2)= 9here the given range doesnt include -3 on the left and 2 on the right and we have two

absolute extremum points on the range

Absolute minimum at x=2/3Absolute maximum at x=-2

7. 8. A farmer plans to fence off three rectangular grazing fields along a straight river by

erecting one fence parallel to the river and four fences perpendicular to the river. If thetotal area of the three plots is to be 6400sqrmet, what length of fence parallel to the river

to maximize the total length of fence required.

Let

Aarea of the total plot A=6400sqrmL1 length of the parallel fence

L2 length of the vertical fence

Ltot The total length of the fence

A=L1*L2 =6400L1=6400/L2

Ltot=L1+4L2

Ltot=L1+4(6400/L1)So here we have expressed the total length of the fence as a function of the parallel

length of the fence let L1 be represented by x and the total length Ltot by the function

f(x)

f(x)=x+(25600/x)

f(x)= 1 (25600/x2)

f(x) = (x2 25600)/x2

8/13/2019 MAT Ans

4/8

to find the optimal length of the horizontal/parallel length of the fence in order that

the total length of the fence will be as minimum as possible we have to find the value of xthat would make the equation f(x)=0

f(x)=0 if x2-25600=0 x2= 25600 =160 hence we are talking about length it must be positive

x = 160 metersthe horizontal/parallel length of the fence must be 160 meters

9. .10. the question is little confusing11.12.Refer to question 11 maximum sum area

Answer:

Let A1 be area of the equilateral triangle

A2 be area of the square

L1 length of the piece used for the triangleL2 length of the piece used for the triangle

Ltot the total length of the wire

Ltot =L1+L2=10mAsum the sum of the area of the triangle and the square Asum=A1+A2 Area of equilateral triangle

1 =34 Where sis the length of one side in this case s=L1/3

Area of square2 =

Where s is the length of one side in this case s=L2/4

The sum of the two Areas= 34 13 + 24

L1+L2=10

L2=10-L1

Let L1=x so that we can express the sum area as a function of x,let Asum =f(x)

() =34

3

+ 10

4

() = 34 3 +100 20 + 16

()= 318 +82016

8/13/2019 MAT Ans

5/8

8/13/2019 MAT Ans

6/8

14.Answer:Let P Perimeter of the triangle P=S1+S2+S3

For isosceles triangle P=2x+b=16inch b=16-2x

=12

= 2

=12 2

() =12 (162) 1 6 2 2

() =12 (162)(8 )

() =12 (16 2)16 64() = (8 )4 16

()= 4 16 +(8 )44 16()=(4 16) +(8 )4

4 16

()= 48 84 16

8/13/2019 MAT Ans

7/8

48 8 = 0

= 6

() = (8 )4 16(6) = (2)8(6) = 42sqrinch

15..16.Answer

a) Answer: =

= 15 (3 1) =2425 = 48.4b)

= +

= ( + ) (() +)= 0

8/13/2019 MAT Ans

8/8

17..18.Integration by sub

a)

2( 4)Answer: = 4 2 + 4() () = 2 + 4 = = 4

2(2 4)6 =17 (2 4)7b)

Answer: = =

(

)( )

( 1 )

( ) = 11 . ( 2 ) ( ) = 1 . (3) (2) (3) (1) 1 ()( )

=

= 6 = 17 7