Application of Derivatives NCERT Solutions

NCERT Exercise 6.5 Q: 10-16

Question (10)Find the maximum value of `2x^3 - 24x +107 ` in the interval `[1,3]. Find the maximum value of the same function in `[-3, -1]`.

Solution:

`2x^3 - 24 x + 107`

` f'(x) =6x^2 - 24`

For maxima or minima, `f'(x) =0`

`:. 6x^2-24 =0`

`=>6x^2 = 24 `

`=> x^2 = 4 `

`=>x = +- 2`

Now `x=-2 notin [1, 3]`

At `x=1`

`f(1)= 2 xx (1)^3- 24 xx 1 + 107`

`=2 - 24 + 107 `

` 109-24 = 85`

At `x=2`

`f(2) = 2 xx (2)^3 - 24 xx 2 + 107`

`=16 - 48 + 107 `

`= 123 - 48 = 75`

At `x=3`

`f(3)= 2 xx (3)^3 - 24 xx 3 + 107`

` 54- 72 + 107 `

`=161 - 72 = 89`

Thus maximum value is ` 89` at `x =3`

Now `-2 in [-3, -1]` and ` 2 notin [-3, -1]`

`:.` At ` x= -1`

`f (-1) = 2x (-1)^3 -24 xx (-1)+ 107`

`= -2 + 24 + 107 `

`131 -2 = 129 `

At ` x=-2`

`f(-2) = 2 xx (-2)^3 - 24 xx (-2) + 107`

`=-16 + 48 + 107 `

`155 - 16 = 139`

At `x+ -3 `

`f(-3) = 2 xx(-3)^3 - 24 xx (-3)+ 107`

`=- 54+ 72 + 107 `

` =179 - 54 = 125`

Thus maximum value is ` 139` at `x=-2`

Question (11) It is given that `x-1 ` the function `x^4- 62x^2 + ax + 9 ` attains its maximum value , on the interval [0, 2], find the value of a.

Solution:

Let `f(x)= x^4 - 62x+ax + 9`

`f(x) = 4x^3 - 12 4x + a`

it is given that `f(x) ` have maximum value at `x =1`

`:. f'(1) = 0`

`f(1)= 4 xx (1)^3 - 124 xx 1+ a`

`=4 - 124+ a = a -120`

`:. a = 120 =0`

`=> a=120`

Question (12)Find the maximum and minimum values of `x+ sin 2x ` on `[0,2pi]`

Solution:

Let `f(x)= x+ sin 2x`

`f(x) = 1+ 2 cos 2x`

For maxima or minima, `f(x) =0`

`:. 1+ 2 cos 2x=0`

`=> cos 2x = 1/2`

`=>2x = (2pi)/3 , (4pi)/3 , (8pi)/3, (10pi)/3`

`=> x= pi/3, (2pi)/3, (4pi)/3, (5pi)/3`

At `x = 0`

`f(0)= 0+ sin (2 xx 0) `

`=sin 0 = 0`

At `x = pi/3 `

`f(pi/3)= pi/3 + sin (2 xx pi/3)`

`= pi/3 + sin (pi - pi/3)`

`=pi/3 + sin \ pi/3 `

` pi/3 + sqrt3/2`

At `x= (2pi)/3`

`f((2pi)/3)= (2pi)/3+ sin (2 xx (2pi)/3)`

`=(2pi)/3 + sin (pi + pi/3)`

`= (2pi)/3 - sin pi/3`

`=(2pi)/3 - sqrt3/2`

At`x = (4pi)/3`

`f((4pi)/3)= `

`(4pi)/3 + sin (2 xx (4pi)/3 )`

`(4pi)/3 + sin ( 3pi - pi/3)`

` =(4pi)/3 + sin pi/3 `

`= (4pi)/3 + sqrt3/2`

At `x= (5pi)/3`

` f((5pi)/3) `

`= (5pi)/3 + sin (2 xx 5pi/3)`

` (5pi)/3 + sin (3pi + pi/3)`

`(5pi)/3 - sin pi /3 `

` 5pi/ 3 - sqrt3/2`

At `x= 2x`

` f(2pi) `

`=2ppi + sin (2 xx 2pi)`

`=2pi + sin 4pi `

`= 2pi + 0 =2pi`

Thus maximum value in ` 2pi` at `x= 2pi and minimum value is `0` at `x=0`

Question (13)Find two numbers whole sum is `24` and whose product is as large as possible.

Solution:

Let first number be `x` .

then second number `= 24 -x`

Let `p` be the product of two numbers then.

`P= x ( 24 - x) = 24x -x^2`

`:. (dp)/(dx)= 24 - 2x`

` (d^2P)/(dx^2)=-2`

For maxima or minima, `(dP)/(dx)=0`

`:. 24 - 2x = 0`

`=> 2x = 24`

`=> x= 12`

At `x= 12`

`(d^P)/(dx)^2 = - 2 < 0`

`:. x= 12 ` is a point of maxima

Thus first number `=12`

Second number `= 24 - 12 = 12`

Question (14)Find two positive numbers `x` and `y` such that `x + y=60 ` and `xy^2` is maximum.

Solution:

Here `x+y = 60`

` => y = 60 -x`

Let `P` denote the product `xy^3`

`P = xy ^3`

From (i) and (ii), We have

`P = x (60 - x)^2`

`=4(60-x) [-60 + x - 30 + 2x]`

`4 ( 60 -x )(3x - 90)`

`= 12 (60- x)(x-30)`

For maxima or minima, `(dP)/(dx) =0`

`:. 4 (60 -x)^2 (15 - x)`

`0=> x = 60` or ` x=15`

`x=60 ` is not possible because `x+ y = 60` (given)

At `x= 15`

`(d^2P)/(dx^2)`

` = 12 ( 60 - 15)(15-30)`

`12 xx 45 xx - 15`

`=- 8100 < 0 `

`:. x= 15` is a point of maxima.

` :. y = 60 - 15 = 45`

Question (15)Find two positive numbers `x` and `y` such that their sum is `35` and the product `x^2y^2` is a maximum number.

Solution:

Here `x+y =35`

` => y=35-x`

Let `P` denote the product `x^2y^5`

`P = x^2y^5`

From (i)and (ii), we have

`P = x^2(35-x)^5`

`(dP)/(dx)`

`= x^2 xx 5(35-x)^4 xx(-1)` `+ (35-x)^5.2x`

`= x(35-x)^4[-5x+70-2x]`

`= x(35- x)^4(70-7x)`

`=7x (10-x)(35-x)^4`

`(d^2P)/(dx^2) `

`=7[x(10-x)xx 4 (35-x)^2(-1)` `+(35-x)^4(10-2x)] `

`=14(35-x)^3 [-20x+2x^2- (35-x)(5-x)]`

`=14(35-x)^3 [-20x+2x^2+ 175-40x +x^2]`

`=14(35-x)^3 [3x^2-60x + 175]`

For maxima or minima, (dP)/(dx)=0`

`:. 7x (10-x)(35-x)^4=0`

`=>x=0,x=10` or `x=35`

`x=35` is not possible because `x+y=35`

` At x = 0`

`(d^2P)/(dx^2)`

`=14 (35-0)^3(3 xx 0^2 - 60 xx 0 + 175)`

` =14 xx 35^3 xx 175>0`

`:. x= 0` is a point of minima.

A `x = 10`

`(d^2P)/(dx^2)`

`14(35-10)^3 [3 xx 10^2-60 xx 10 + 175]`

`14(25)^3[300-600+175]`

`14 xx 25^3 xx - 125 `

`=- 14 xx 25^3 xx125 <0`

`:. x= 10 ` is a point of maxima.

`x= 10` is a point of maxima.

`:. y = 35 - 10 =25`

Thus first number is `10` and second number is `25`

Question (16)Find two positive numbers whose sum is `16` and the sum of whose cubes is minimum.

Solution:

Let first number be `x` and second number be `y`

Then `x+y = 16 => y= 16 -x` ----(i)

Let `S` denote the sum of cubes of numbers

`:. S= x^3 +y^3 ` ----(ii)

From (i)and (ii), we have

`S= x^3+(16-x)^3`

` (dS)/(dx)`

`= 3x^2 + 3 ( 16-x)^2 xx(-1)`

`=3[x^2- (16- x)^2]`

`=3[x^2-256 - x^2+32x]`

`= 96(x-8)`

`(d^2S)/(dx^2)=96`

For maxima or minima, `(dS)/(dx) =0`

`:. 96(x-8)=0 `

`=> x=8`

At `x=8`

`(d^2S)/dx^2 = 96 > 0`

`:. x=8 ` is a point of minima.

:. ` y=16 - 8 = 8`

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