Chapter 18 Maxima and Minima Ex 18.1

Question 1

Solution 1

Question 2

Find the maximum and minimum values, if any, without using derivatives of the following function given by f(x) = -(x-1)2 + 2 on R.Solution 2

Question 3

Solution 3

Question 4

Find the maximum and minimum values, if any, without using derivatives of the following function given by h(x) = sin(2x) + 5 on R.Solution 4

Question 5

Find the maximum and minimum values, if any,  usingwithout derivatives of the following function given by begin mathsize 12px style f left parenthesis x right parenthesis equals open vertical bar sin space 4 straight x space plus space 3 close vertical bar end style on R.Solution 5

Question 6

Solution 6

Question 7

Find the maximum and minimum values, if any, without using derivatives of the following function given by begin mathsize 12px style g italic left parenthesis x italic right parenthesis space equals negative 1 open vertical bar straight x plus 1 close vertical bar plus 3 end style on R.Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 18 Maxima and Minima Ex. 18.2

Question 1

Solution 1

Question 2

Find the local maxima and local minima, if any, of the following functions using first derivative test. Find also the local maximum and the local minimum values, as the case may be:

f(x) =x– 3xSolution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Find the local maxima and local minima, if any, of the following functions using first derivative test. Find also the local maximum and the local minimum values, as the case may be:

f(x) = sinx – cos x, 0 < x < 2πSolution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

f open parentheses x close parentheses equals 2 sin x space minus space x comma space minus straight pi over 2 space less or equal than x space less or equal than straight pi over 2

F o r space c h e c k i n g space t h e space m i n i m a space a n d space m a x i m a comma space w e space h a v e
f apostrophe open parentheses x close parentheses equals 2 cos x space minus 1 space equals space 0

rightwards double arrow cos x equals 1 half equals cos straight pi over 3
rightwards double arrow x equals minus straight pi over 3 comma space straight pi over 3
A t space x equals minus straight pi over 3 comma space f open parentheses x close parentheses space c h a n g e s space f r o m space minus v e space t o space plus space v e
rightwards double arrow x equals minus straight pi over 3 space i s space p o i n t space o f space l o c a l space m i n i m a space w i t h space v a l u e space equals space minus square root of 3 minus straight pi over 3

A t space x equals straight pi over 3 comma space f open parentheses x close parentheses space c h a n g e s space f r o m space plus v e space t o space plus space v e
rightwards double arrow x equals straight pi over 3 space i s space p o i n t space o f space l o c a l space m a x i m a space w i t h space v a l u e space equals space square root of 3 minus straight pi over 3

Question 12

Find the local maxima and local minima, if any, of the following functions using first derivative test. Find also the local maximum and the local minimum values, as the case may be:

begin mathsize 12px style f begin italic style left parenthesis x right parenthesis end style equals straight x square root of 1 minus x end root comma space x greater than 0 end style

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Chapter 18 Maxima and Minima Ex. 18.3

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

L o c a l space M a x i m u m space v a l u e space equals space f left parenthesis 4 right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 4 square root of 32 minus 4 squared end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 4 square root of 32 minus 16 end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 4 square root of 16
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 16
L o c a l space m i n i m u m space a t space x equals minus 4 ;
L o c a l space M i n i m u m space v a l u e space equals space f left parenthesis minus 4 right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals minus 4 square root of 32 minus open parentheses minus 4 close parentheses squared end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals minus 4 square root of 32 minus 16 end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals minus 4 square root of 16
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals minus 16

Question 1(ix)

Solution 1(ix)

Question 1(x)

Solution 1(x)

Question 1(xi)

Solution 1(xi)

Question 1(xii)

Solution 1(xii)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 3

Solution 3

Question 4

Show that the function given by f(x)=begin mathsize 12px style fraction numerator log x over denominator x end fraction end style has maximum value at x = e.Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

If f(x) = x3 + ax2 + bx + c has a maximum at x = -1 and minimum at x = 3. Determine ab and c.Solution 7

Question 8

Prove that   has maximum value at  Solution 8

Given: 

Differentiating w.r.t x, we get

Take f'(x) = 0

Differentiating f'(x) w.r.t x, we get

At 

Clearly, f”(x) < 0 at 

Thus,   is the maxima.

Hence, f(x) has maximum value at  .

Chapter 18 Maxima and Minima Ex. 18.4

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Find both the absolute maximum and absolute minimum of 3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3] Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 18 Maxima and Minima Ex. 18.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Divide 15 into two parts such that the square of one multiplied with the cube of the other is maximum.Solution 4

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

W e space h a v e comma space 4 l plus 3 a equals 20
rightwards double arrow 4 l equals 20 minus 3 a
rightwards double arrow l equals fraction numerator 20 minus 3 a over denominator 4 end fraction
F r o m space left parenthesis i right parenthesis comma space w e space h a v e comma
s equals open parentheses fraction numerator 20 minus 3 a over denominator 4 end fraction close parentheses squared plus fraction numerator square root of 3 over denominator 4 end fraction a squared
fraction numerator d s over denominator d a end fraction equals 2 open parentheses fraction numerator 20 minus 3 a over denominator 4 end fraction close parentheses open parentheses fraction numerator minus 3 over denominator 4 end fraction close parentheses plus 2 a cross times fraction numerator square root of 3 over denominator 4 end fraction
T o space f i n d space t h e space m a x i m u m space o r space m i n i m u m comma space fraction numerator d s over denominator d a end fraction equals 0
rightwards double arrow 2 open parentheses fraction numerator 20 minus 3 a over denominator 4 end fraction close parentheses open parentheses fraction numerator minus 3 over denominator 4 end fraction close parentheses plus 2 a cross times fraction numerator square root of 3 over denominator 4 end fraction equals 0
rightwards double arrow minus 3 open parentheses 20 minus 3 a close parentheses plus 4 a square root of 3 equals 0
rightwards double arrow minus 60 plus 9 a plus 4 a square root of 3 equals 0
rightwards double arrow 9 a plus 4 a square root of 3 equals 60
rightwards double arrow a open parentheses 9 plus 4 square root of 3 close parentheses equals 60
rightwards double arrow a equals fraction numerator 60 over denominator 9 plus 4 square root of 3 end fraction
D i f f e r e n t i a t i n g space o n c e space a g a i n comma space w e space h a v e comma
fraction numerator d squared s over denominator d a squared end fraction equals fraction numerator 9 plus 4 square root of 3 over denominator 8 end fraction greater than 0
T h u s comma space t h e space s u m space o f space t h e space a r e a s space o f space t h e space s q u a r e space a n d space t r i a n g l e space i s space m i n i m u m space w h e n space a equals fraction numerator 60 over denominator 9 plus 4 square root of 3 end fraction
W e space k n o w space t h a t comma space l equals fraction numerator 20 minus 3 a over denominator 4 end fraction
rightwards double arrow l equals fraction numerator 20 minus 3 open parentheses fraction numerator 60 over denominator 9 plus 4 square root of 3 end fraction close parentheses over denominator 4 end fraction
rightwards double arrow l equals fraction numerator 180 plus 80 square root of 3 minus 180 over denominator 4 open parentheses 9 plus 4 square root of 3 close parentheses end fraction
rightwards double arrow l equals fraction numerator 20 square root of 3 over denominator 9 plus 4 square root of 3 end fraction

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

A square piece of tin of side 18 cm is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible? Also, find this maximum volume.Solution 12

M a x i m u m space v o l u m e space i s space V subscript x equals 3 end subscript equals 3 cross times open parentheses 18 minus 2 cross times 3 close parentheses squared
rightwards double arrow V equals 3 cross times 12 squared
rightwards double arrow V equals 3 cross times 144
rightwards double arrow V equals 432 space c m cubed

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

An isosceles triangle of vertical angle 2θ is inscribed in a circle radius a. show that the area of the triangle is maximum when  Solution 22

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

S i n c e space fraction numerator d squared S over denominator d x squared end fraction less than 0 comma space t h e space s u m space i s space l a r g e s t space w h e n space x equals y equals fraction numerator r over denominator square root of 2 end fraction

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

A given quantity of metal is to be cast into a half cylinder with a rectangular base and semicircular ends. Show that in order that the total surface area may be minimum, the ratio of the length of the cylinder to the diameter of its semi – circular ends is π : ( π + 2) Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 45

Solution 45

Question 5

Amongst all open (from the top) right circular cylindrical boxes of volume 125π cm3, find the dimensions of the box which has the least surface area.Solution 5

Let r and h be the radius and height of the cylinder.

Volume of cylinder   

  … (i)

Surface area of cylinder   

From (i), we get

Question 23

Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is   Solution 23

Let   be an isosceles triangle with AB = AC.

Let   

Here, AO bisects 

Taking O as the centre of the circle, join OE, OF and OD such that

OE = OF = OD = r (radius)

Now, 

In 

Similarly, AF = r cot x

In 

As OB bisect   we have

In 

Similarly, BD = DC = CE = 

We have, perimeter of 

P = AB + BC + CA

 = AE + EC + BD + DC + AF + BF

Differentiating w.r.t x, we get

Taking 

As 

Therefore,   is an equilateral triangle.

Taking second derivative of P, we get

At 

Therefore, the perimeter is minimum when 

Least value of P 


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