What is the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius R?

Let x be the radius of base and h be the height of the con which is inscribed in a sphere of radius r.

In the figure, AD = h and CD = x = BD

Since, ΔABD and ΔBDE are similar,

`"AD"/"BD" = "BD"/"DE"`

∴  BD2 = AD-DE = AD.(AE – AD)

∴ x2 = h(2r – h)                               ...(1)
Let V be the volume of the cone.

Then V = `(1)/(3)pix^2h`

= `pi/(3)h(2r - h)h`                        ...[By (1)]

∴ V = `pi/(3)(2rh^2 - h^3)`

∴ `"dV"/"dh" = pi/(3) d/"dh"(2rh^2 - h^3)`

= `pi/(3)(2r xx 2h - 3h^2)`

= `pi/(3)(4rh - 3h^2)`and

`(d^2V)/(dh^2) = pi/(3).d/"dh"(4rh - 3h^2)`

= `pi/(3)(4r xx 1 - 3 xx 2h)`

= `pi/(3)(4r - 6h)`

For maximm volume, `"dV"/"dh"` = 0

∴ `pi/(3)(4rh - 3h^2)` = 0

∴ `4rh = 3h^2` 

∴ h = `(4r)/(3)`                      ...[∵ h ≠ 0]and

`((d2V)/"dh"^2)_("at"  h = (4r)/(3)`

= `pi/(3)(4r - 6 xx (4r)/3)`

= `pi/(3)(4r - 8r)`

= `-(4pir)/(3) < 0`

∴ V is maximum when h = `(4r)/(3)`

Hence, the attitude (i.e. height) of the right circular cone of maximum volume = `(4r)/(3)`.

Page 2

Let R be the radius and h be the height of the cylinder which is inscribed in a sphere of radius r cm.

Then from the figure,

`"R"^2 + (h/2)^2` = r2

∴ R2  = `r^2 - h^2/(4)`   ...(1)

Let V be the volume of the cylinder.
Then V = πR2h

= `pi(r^2 - h^2/(4))h`   ...[By (1)]

= `pi(r^2 - h^3/(4))`

∴ `"dV"/"dh" = pid/"dh"(r^2h - h^3/(4))`

= `pi(r^2 xx 1 - 1/4 xx 3h^2)`

= `pi(r^2 - 3/4h^2)`and

`(d^2V)/("dh"^2) = pid/"dh"(r^2 - 3/4h^2)`

= `pi(0 - 3/4 xx 2h)`

= `-(3)/(2)pih`

Now, `"dV"/"dh" = 0  "gives", pi(r^2 - 3/4h^2)` = 0

∴ `r^2 - 3/4h^2` = 0

∴ `(3)/(4)h^2` = r2 

∴ h2 = `(4r^2)/(3)`

∴ h = `(2r)/sqrt(3)`          ...[∵ h > 0]and

`((d^2V)/(dh^2))_("at"  h = (2r)/sqrt(3)`

= `-(3)/(2)pi xx (2r)/sqrt(3) < 0`

∴ V is maximum at h = `(2r)/sqrt(3)`

If h = `(2r)/sqrt(3)`, then from (1)

R2 = `r^2 - (1)/(4) xx (4r^2)/(3) = (2r^2)/(3)`

∴ volumeof the largest cylinder

= `pi xx (2r^2)/(3) xx (2r)/sqrt(3) = (4pir^3)/(3sqrt(3)`cu cm.

Hence, the volume of the largest cylinder inscribed in a sphere of radius 'r' cm = `(4R^3)/(3sqrt(3)`cu cm.

A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 Cm, how fast is the enclosed area increasing?

Let r be the radius of the circular wave and A the enclosed area at time t.
 

Last updated at April 19, 2021 by Teachoo

Misc 15 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4𝑟/3 . Given, Radius of sphere = r Let R be the radius of the cone and H be its height. Let ∠ BOC = θ Now, AC = AO + OC H = r + r cos θ H = r (1 + cos θ) And, R = r sin θ We need to maximize volume of cone. Volume of the cone is V = 1/3 𝜋𝑅^2 𝐻 V = 1/3 𝜋𝑟^2 sin^2⁡"θ" r (1 + cos θ) V = 𝟏/𝟑 𝝅𝒓^𝟑 〖𝒔𝒊𝒏〗^𝟐⁡"θ" (1 + cos θ) Finding 𝒅𝑽/𝒅𝜽 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 (𝑑[sin^2⁡〖θ 〗 (1 + cos⁡θ))/𝑑𝜃 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 [2 sin⁡〖"θ" cos⁡〖"θ" (1+cos⁡〖"θ" )+sin^2⁡〖"θ" (−sin⁡〖"θ" )〗 〗 〗 〗 〗 ] 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 (2 sin⁡〖"θ" cos⁡〖"θ" (1+cos⁡〖"θ" )−sin^3⁡"θ" 〗 〗 〗 ) 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 (2 sin⁡〖"θ" cos⁡〖"θ" (1+cos⁡〖"θ" )−sin^3⁡"θ" 〗 〗 〗 ) 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (2 cos⁡〖"θ" +〖2 cos^2〗⁡〖"θ" −〖𝒔𝒊𝒏〗^𝟐⁡"θ" 〗 〗 ) 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (2 cos⁡〖"θ" +〖2 cos^2〗⁡〖"θ" −(𝟏−〖𝒄𝒐𝒔〗^𝟐⁡"θ" 〗 〗)) 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (𝟑 〖𝒄𝒐𝒔〗^𝟐⁡〖"θ" +〖𝟐 𝒄𝒐𝒔〗⁡〖"θ" −𝟏〗 〗 ) 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (𝟑 〖𝒄𝒐𝒔〗^𝟐⁡〖"θ" +〖𝟑 𝒄𝒐𝒔〗⁡𝜽 〗 −𝒄𝒐𝒔⁡𝜽 −𝟏) 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (3 〖𝑐𝑜𝑠 〗⁡"θ" (cos⁡𝜃+1)−1(cos⁡𝜃+1)) 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (3 𝑐𝑜𝑠⁡〖"θ" −1〗 )(𝑐𝑜𝑠⁡〖"θ" +1〗 ) Putting 𝒅𝑽/𝒅𝜽 = 0 1/3 𝜋𝑟^3 sin θ (𝑐𝑜𝑠⁡〖"θ" +1〗 ) (3 𝑐𝑜𝑠⁡〖"θ" −1〗 ) = 0 𝐬𝐢𝐧 θ (cos θ + 1) (3 cos "θ" − 1) = 0 𝐬𝐢𝐧 θ = 0 θ = 0° θ cannot be 0° for cone 𝐜𝐨𝐬 θ + 1 = 0 cos θ = −1 For cone, 0° < θ < 90° & cos θ is negative in 2nd & 3rd quadrant. So cos θ = −1 is not possible 𝟑 𝐜𝐨𝐬 θ − 1 = 0 cos θ = 1/3 θ = cos−1 1/3 cos θ = 𝟏/𝟑 is possible So, 𝒄𝒐𝒔⁡"θ" = 𝟏/𝟑 Thus, H = r (1 + cos θ) H = r ("1 + " 1/3) H = 𝟒𝒓/𝟑 Finding (𝒅^𝟐 𝑽)/(𝒅𝜽^𝟐 ) 𝑑𝑉/𝑑𝜃 = (1/3 𝜋𝑟^3 sin⁡𝜃 (3 cos2 𝜃 + 2 cos⁡𝜃− 1)) = 1/3 𝜋𝑟^3 [cos⁡𝜃 (3 cos^2⁡𝜃+2 cos⁡𝜃−1)+sin⁡𝜃 (6 cos⁡𝜃 (−𝑠𝑖𝑛)−2 sin⁡𝜃 )] = 1/3 𝜋𝑟^3 [3 cos^3⁡𝜃+2 cos^2⁡𝜃−cos⁡𝜃+sin⁡𝜃 (−6 cos⁡𝜃 sin⁡𝜃−2 sin⁡𝜃 )] = 1/3 𝜋𝑟^3 [3 cos^2⁡𝜃+2 cos^2⁡𝜃−cos⁡𝜃−6 sin^2⁡𝜃 cos⁡𝜃−2 sin^2⁡𝜃 ] Now, 𝐜𝐨𝐬⁡𝜽=𝟏/𝟑 And sin2 "θ" = 1 − cos2 "θ" = 1 − (1/3)^2= 1 − 1/9 = 𝟖/𝟗 Putting values in (𝒅^𝟐 𝑽)/(𝒅𝜽^𝟐 ) (𝑑^2 𝑉)/(𝑑𝜃^2 ) = 1/3 𝜋𝑟^3 [3(1/3)^3+2(1/3)^2−1/3−6(1/3)(8/9)−2(8/9)] = 1/3 𝜋𝑟^3 [3(1/27)+2(1/9)−1/3−6(8/27)−2(8/9)] = 1/3 𝜋𝑟^3 [1/9+ 2/9−1/3−16/9−16/9] = 1/3 𝜋𝑟^3 [(−32)/9] = (−𝟑𝟐𝝅𝒓^𝟑)/𝟐𝟕 Thus, (𝑑^2 𝑉)/(𝑑𝜃^2 ) < 0 for 𝐜𝐨𝐬⁡𝜽=𝟏/𝟑 So, V is maximum for 𝐜𝐨𝐬⁡𝜽=𝟏/𝟑 Hence, altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 𝟒𝒓/𝟑 .