Question #280248

Find the dimensions of the right circular cone of the greatest lateral area that can be inscribed in a sphere of radius α


1
Expert's answer
2021-12-16T17:08:29-0500

Solution:


Given, a cone inscribed in a sphere of radius a.

Let the radius, height and slant height of cone be r,h,lr,h,l respectively.

OB=OA=a,BC=r,AC=h,OC=ha,AB=lOB=OA=a,BC=r,AC=h,OC=h-a,AB=l

Using pythagoras theorem in triangle BOC,

BC2+CO2=OB2r2+(ha)2=a2r2+h2+a22ha=a2r2+h2=2ha ...(i)BC^2+CO^2=OB^2 \\\Rightarrow r^2+(h-a)^2=a^2 \\\Rightarrow r^2+h^2+a^2-2ha=a^2 \\\Rightarrow r^2+h^2=2ha \ ...(i)

Using pythagoras theorem in triangle BAC,

BC2+CA2=AB2r2+h2=l2 ...(ii)BC^2+CA^2=AB^2 \\\Rightarrow r^2+h^2=l^2\ ...(ii)

From (i) and (ii)

2ha=l22ha=l^2

Now, lateral area of cone, L=πrlL=\pi rl

L2=π2r2l2=π2(l2h2)l2 [From (ii)]=π2(2hah2)(2ha)=π2(4h2a22h3a)L^2=\pi^2 r^2 l^2 \\=\pi^2 (l^2-h^2)l^2\ [From\ (ii)] \\=\pi^2(2ha-h^2)(2ha) \\=\pi^2(4h^2a^2-2h^3a)

On differentiating w.r.t. h

2LdLdh=π2(8ha26h2a)...(iii)dLdh=π2(2haπ2(4a3h))2L2L\dfrac{dL}{dh}=\pi^2(8ha^2-6h^2a) ...(iii) \\ \Rightarrow\dfrac{dL}{dh}=\dfrac{\pi^2 (2ha\pi^2 (4a-3h))}{2L}

Now, put dLdh=0\dfrac{dL}{dh}=0

π2(2haπ2(4a3h))2L=04a3h=0h=4a3 ...(iv)\\ \Rightarrow\dfrac{\pi^2 (2ha\pi^2 (4a-3h))}{2L}=0 \\ \Rightarrow 4a-3h=0 \\ \Rightarrow h=\dfrac{4a}3\ ...(iv)

Again, differentiating (iii) w.r.t h,

2Ld2Ldh2+2(dLdh)2=π2(8a212ha)=π2(8a212a(4a3)) [Using (iv)]=8π2a2<0 maxima2L\dfrac{d^2L}{dh^2}+2(\dfrac{dL}{dh})^2=\pi^2(8a^2-12ha) \\=\pi^2(8a^2-12a(\dfrac{4a}3))\ [Using\ (iv)] \\=-8\pi^2a^2<0 \\\Rightarrow\ maxima

Put (iv) in (i)

r2+(4a3)2=2(4a3)ar2=8a2316a29r2=8a29r=22a3\\r^2+(\dfrac{4a}3)^2=2(\dfrac{4a}3)a \\\Rightarrow r^2=\dfrac{8a^2}3-\dfrac{16a^2}9 \\\Rightarrow r^2=\dfrac{8a^2}9 \\\Rightarrow r=\dfrac{2\sqrt2a}3

Thus, the dimensions of cone are r=22a3;h=4a3r=\dfrac{2\sqrt2a}3;h=\dfrac{4a}3


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS