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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`. Also find maximum volume in terms of volume of the sphere
Show that the altitude of a right circular cone of maximum volume that can be inscribed in a sphere of radius r is `(4r)/3` . Also, show that the maximum volume of the cone is `8/27` of the volume of the sphere.
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Solution 1
`V=1/3piR^2H`
It is clear from the figure that
`R^2+(H−r)^2=r^2`
`⇒R^2+H^2+r^2−2Hr=r^2`
`⇒R^2=2Hr−H^2`
Substituting the value of R^{2} in the formula for the volume of the cone, we get
`V=1/3pi(2HrH^2)H`
` V=2/3πrH^2−π/3H^3`
Differentiating with respect to H both sides, we get:
`(dV)/(dH)=4/3πrH−πH^2`
At critical point,`(dV)/(dH)` is 0.
`⇒4/3πrH−πH^2=0`
`⇒H=4/3r`
Differentiating V w.r.t H again, we get:
`(d^2V)/(dH^2)=4/3πr−2πH`
`(d^2V)/(dH^2)_(H=4/3r)=−4/3πr <0`
Hence maxima.
Volume of cone =`1/3πR^2H`
` V=2/3πrH^2−π/3H^3`
Substituting the value of H, we get:
`V=2/3πr(4/3r)^2−π/3(4/3r)^3`
`=>V=8/27(4πr^3−8/3πr^3)`
`=>V=8/27(4/3πr^3)`
`=>V=8/27(volume of sphere)`
Solution 2
A sphere of fixed radius (r) is given.
Let R and h be the radius and the height of the cone, respectively.
The volume (V) of the cone is given by,
`V=1/3piR^2h`
Now, from the right triangle BCD, we have:
`BC=sqrt(r^2R^2)`
`:.h=r+sqrt(r^2R^2)`
`V=1/3piR^2(r+sqrt(r^2R^2))=1/3piR^2r+1/3piR^2sqrt(r^2R^2)`
`(dV)/(dR)=2/3piRr+2/3piRsqrt(r^2R^2)+(piR^2)/3 (2R)/(2sqrt(r^2R^2))`
`=2/3piRr+2/3piRsqrt(r^2R^2)(piR^3)/(3sqrt(r^2R^2))`
`=2/3piRr+(2piR(r^2R^2)piR^3)/(3sqrt(r^2R^2))`
`2/3piRr+(2piRr^23piR^3)/(3sqrt(r^2R^2))`
Now
`(dV)/(dR^2)=0`
`=>(2pirR)/3=(3piR^32piRr^2)/(3sqrt(r^2R^2))`
`=>2rsqrt(r^2R^2)=3R^22r^2`
`=>4r^2(r^2R^2)=(3R^22r^2)^2`
`=>4r^44r^2R^2=9R^4+4r^412R^2r^2`
`=>9R^48r^2R^2=0`
`=>9R^2=8r^2`
`=>R^2=(8r^2)/9`
Now,
`(d^2V)/(dR^2)=(2pir)/3+(3sqrt(r^2R^2)(2pir^29piR^2)(2piRr^23piR^3)(6R)1/(2sqrtr^2R^2))/(9(r^2R^2))`
`=(2pir)/3+(3sqrt(r^2R^2) (2pir^29piR^2)+(2piRr^23piR^3)(3R)1/(2sqrt(r^2R^2)))/(9(r^2R^2))`
Now when `R^2=(8r^2)/9`it can be shown that `(d^2V)/(dR^2)<0`
∴ The volume is the maximum when `R^2=(8r^2)/9`
When `R^2=(8r^2)/9`height of cone= `r+sqrt(r^2(8r^2)/9)=r+sqrt(r^2/9)=r+r/3=(4r)/3`
Hence, it can be seen that the altitude of a right circular cone of maximum volume that can be inscribed in a sphere of radius r is `(4r)/3`
Let volume of the sphere be `V_s=4/3pir^3`
`r=3sqrt((3V_s)/(4pi))`
∴ Volume of cone, V = `1/3piR^2h`
⇒R = `(2sqrt2)/3r`
V = `1/3pi((2sqrt2)/3r)xx(4r)/3`
⇒V = `1/3pi(8r^2)/9xx(4r)/3`
`V=(32pir^3)/81=32/81pi[(3V_s)/(4pi)]`
∴ Volume of cone in terms of sphere
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