Application of Derivatives NCERT Solutions

NCERT Exercise 6.5:Q: 17-23

Question: (17) A square piece of tin of side `18` cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to from the box. What should be the side of the aquare to be cut off so that the volume of the box is the maximum possible?

Solution:

Let a square of side `x cm` to be cut from each corner of tin.

Then length, breath, and height of open box are `18-2x,18- 2x` and `x`,

Let `V` denote the volume of box, then

`V= (18- 2x)(18 - 2x)x `

`=4x(9-x)^2`

`(dv)/(dx) `

`=4[x xx 2(9-x) (-1)+(9-x)^2 xx 1]`

`=4(9-x) (-2x +9-x)`

`=4(9-x)(9-3x)`

`=12 (9-x)(3-x)`

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

`=12[(9-x)(-1) + (3-x) (-1)]`

`12 [-9+x-3+ x]`

`12(2x-12)`

`=24(x-6)`

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

`:. 12 (9-x)(3-x)=0`

`=> x = 9 or x = 3`

`x = 9` is not possible because at `x=9, V=0`

At `x = 3`

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

`24 (3-6) `

`=24 xx -3 `

`=-72 <0`

`:. x = 3 ` is a point of maxima.

Thus side of the square cut off is `3` cm.

Question: (18)Arectangular sheet of tin `45`cm by `24` is to be made into a box without top, by cutting off a square from each corner and folding up the flaps. What should be the side the square to be cut off so that the volume of the box is maximum? Fig - Page no 505

Solution:

Let a square of side `x` cm to be cut from each corner of the rectangular sheet.Then length, breadth and height of open box are `45-2x, 24- 2x` and ` x cm.

Let `V` denote the volume of box, them

`V = (45-2x) (24-2x) x`

`=2x (12-x) (45- 2x)`

`(dV)/(dx)`

`=2[x xx (12-x)(-2)` `+(45-2x) (12-2x)]`

` =2 [-24x+ 2x^2+540-114x+4x^2]`

`=2[6x^2 - 138x + 540]`

`=12[x^2-23x+90]`

`=12(x-5)(x-18)`

`(d^2V)/(dx^2)= 12(2x-23)`

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

`:. 12(x-5) (x-18)=0`

`=> x=5 or x=18`

`x= 18` is not possible because at `x= 18,V` is negative.

At `x=5`

`(d^2V)/(dx^2)= 12(2xx5 - 23) `

`=12 xx - 13 =- 156 <0`

Question: (19)Show that of all the rectangles inscribed in a given fixed circle the square has the maximum area`

Solution:

Let ABCD be a rectangle inscribed in a circle of radius a unita.

Let length and breadth of rectangle are `x` and `y` units respectively.

`:.x^2+y^2 = (2a)^2`

`=> x^2+ y^2 `

`= 4a^2 => y^2``=4a^2-x^2 => y`

`= sqrt 4a^2-x^2` ---(i)

Let `A` denote the area of rectangle then

`A = xy`

From (i) and (ii), we have

`A = x sqrt(4a^2-x^2)`

`:.(dA)/(dx)`

`=(2x[-8a^2+2x^2 + 2a^a -x^2])/ ((4a^2-x^2)^(3/2))`

`=(2x(x^2-6a^2))/((4a^2- x^2)^(3/2))`

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

`:. (2(2a^2-x^2))/(sqrt(4a^2-x^2))=0`

`=>2(2a^2-x^2)=0`

`=>x^2=2a^2`

` =>x= sqrt(2a)`

At `x= sqrt2a`

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

`=(2sqrt2a(2a^2 -6a^2))/((4a^2-2a^2)^(3/2))`

`=(2sqrt2a xx(-4a^2))/((2a^2)^(3/2))`

`= (-8sqrt(2a^3))/(2sqrt(2a^3))`

`=-4<0`

`:. x= sqrt2a` is a point of maxima .

Then `y= sqrt(4a^2-2a^2)`

`= sqrt2a^2 = sqrt 2a`

Thus the area is maximum when rectangle is a square.

Question: (20)Show that the right circular cylinder of given surface area and maximum volume is such that its height is equal to the diameter of the base.

Solution:

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

Let `S` denote the surface area and `V` denote the volume of cylinder.

Then `S=2pi r^2+2pi r h = K`

`:. h=(K-2pi r^2)/(2pi r)`

`V=pi r^2h`

From (i) and (ii), we have

`V = pi r^2(K-2pi r^2)/(2pi r)`

`= 1/2 r (K- 2pi r^2) `

`=1/2 (Kr - 2pi r^2)`

`(dV)/(dr) = 1/2 (K-6pi r^2`

`(d^2V)/ (dr)`

` = 1/2 (-12pir)=-6pir`

For maxima or minima`(dV)/(dr)=0`

`:. 1/2 (K-6pir^2)=0`

`=> K= 6pi r^2`

` => r^2= K/(6pi)`

`=> r = sqrt(K/(6pi))`

`=>` At `r = sqrt(K/(6 pi))`

`(d^2V)/(dx^2)= -6 pi xx sqrt(K/(6pi))`

`=-sqrt(6 pi K)<0`

`:. r= sqrt(K/(6 pi)) ` is a point of maxima.

` h= (6 pi r^2 - 2pir^2)/(2pi r)`

` (4pi r^2)/(2pi r)`

`2r`

`h= 2r`

Thusvolume of cylinder is maximum when height of cylinder is equal to diameter of base.

Question: (21)Of all closed cylindrical cans (right circular) of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?

Let `r` be the radius and `h` be the height of cylinderical can.

Then `V= pi r^2h = 100 `

`=> h= 100/(pi r^2)`

Let `S` denote the surface area of cylindrical can, then

`S= 2pi r h + 2pi r^2`

From (i) and (ii), we have

`S = 2pi r xx 100/(pi r^2)+ 2pi r^2`

`= (200)/r + 2pi r^2`

`(dS)/(dr) =- (200)/r^2 + 4pi r `

`(-200+4 pi r^3)/r^2`

`(d^S)/dr^2`

=`(400)/r^3 + 4 pi`

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

` :. (-200 + 4 pi r^3)/r^2 =0`

`=>-200+ 4pi r^3= 0`

` => 4pi r^3 = 200 => `

`=>r^3=(200)/(4pi)`

`=> r= (50/pi)^(1/3)`

At` r = ((50)/pi)^(1/3)`

`(d^2S)/(dr^2)= (400)/[(50/pi)^(1/3)]^3 + 4pi`

`=(400)/(50) xx pi + 4pi`

`12 pi > 0`

`:. r= (50/pi)^1/3 ` is a point of minima.

`:.h= (100)/(pi(50/pi)^(2/3))`

`(2 xx 50)/(pi^(1/3 (50)^(2/3))`

`=2(50/pi)^(1/3)`

Question: (22) A wire of length 28 `m` is to be cut into two pices. One of the pices is to be made into squire and the other into a circle. What should be the length of the two pices so that the combined area of the square and the circle is minimum?

Solution:

Let `x` be the length of the side of a square and `y` be the radius of circle the

`4x + 2pi y = 28 `=>

`y = (28- 4x)/(2pi)`

`(14-2x)/pi`

Let `A` be the combined area of square and circle then

A `= x^2+piy^2`

A `= x^2 + pi ((14-2x)pi)^2`

`(dA)/(dx)= 2x +1/pi . 2(14 - 2x)(-2)`

`= 2x - 4/pi(14- 2x)`

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

`= 2- 4/pi(-2)`

` =2+ 8/pi`

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

`:. 2x - 4/pi ( 14- 2) = 0`

=> `2x-(56)/pi+ (8x)/pi =0`

`=>2x(1+ 4/pi)=(56)/pi`

`=> 2x ((pi+4) /pi)= 56/pi`

`x= 56/pi xx pi/(2(pi+4)`

`= 28/ (pi+4)`

At `x = 28/(pi+ 4)`

`(d^2A)/(dx^2) = 2+ 8/pi >0`

`:. x= 28/(pi+4)` is a point of minima.

`y= (14-2 xx 28/(pi+4))/pi`

` =(14 pi + 56 - 56)/(pi(pi+4))`

` =14/(pi+4)`

Length of one piece `= 4x= 4 xx (28)/(pi+4)`

`(112)/(pi+4)m`

Length of second piece `= 2piy =2pi xx 14/(pi+4)`

`(28pi)/pi+4 m`

Question: (23)Prove that the volume of the largest cone that can be inscribed in a sphere of radius `R` is 8/ (27) of the volume of the sphere.

Solution:

Let `x` and `y` be the radius of the base and altitude of the cone respectively inscribed in a sphere of radius `R`

Then `OA = OB = OC = R`

`PD = y,BD = x`

` :.OD +CD - OC +Y-R`

Now `OB^2= OD^2 + BD^2`

`:. R^2 = (y-R)^2 + x^2 `

`=> x^2=R^2 - (Y-R)^2`

`x^2 = R^2- y^2 - R^2+ 2yR`

`=>x^2 =- y^2 + 2YR `

` Y(2R-y)`

Let `V` denote the volume of cone then

`V = 1/3 pix^2y`

From(i)and (ii), we have

`V = 1/3 pi [y(2R-y)]`

` = pi/3[2Ry^2 -y^3]`

`(dV)/(dy) = pi/3[4Ry-3y^2]`

`(d^2V)/(dy^2)`

`= pi/3 [4R-6y]`

For maxima or minima,`(dV)/(dy)=0`

`:. pi/3 [4Ry-3y^2]=0`

`=> 4Ry- 3y^2 =0 `

` =>3y^2`

`= 4Ry`

`=> y= 4/3 R`

` At `y= 4/3 R`

`(d^2V)/(dy^2)`

`= pi/3[4R-6 xx 4/3 R]`

`= pi/3[4R- 8R]`

`4/3 pi R<0`

`:. y= 4/ 3 R ` isv a point of maxima

`:. x^2 = 4/3 R [2R- 4/3 R]`

`=(4R)/3 xx (2R)-3`

`= 8/9 R^2`

So volume of largest cone `= 1/3 pi x^2 y`

`=1/3 pi xx 8/9 R^2 xx 4/3 R`

`(32piR^3)/81`

`= 8/27(4/3 piR^3)`

`8/27` [ Volume of sphere of radius `R`]

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