Question #311403

Find the dimensions of a right circular cone of minimum volume V that can be

circumscribed about a sphere of radius 8 inches.


1
Expert's answer
2022-03-15T17:12:19-0400




The volume of a cone = 13πr2h​ \tfrac{1}{3}*π *r^2*h

Let the radius of the sphere = a

the radius of the cone =r

by similar triangles sin(θ)=ahasin (θ)=\tfrac{a}{h-a} = rh2+r2​​​\tfrac{r}{\sqrt{h^2+r^2}} ​ ​ ​


=a(h2+r2)=r(ha)=a*(\sqrt{h^2+r^2) }=r(h-a)

on squaring both sides

a2(h2+r2)=r2(ha)2a^2(h^2+r^2)=r^2(h-a)^2

=a2h2+a2r2=h2r22ahr2+a2r2=a^2h^2+a^2r^2= h^2r^2-2ahr^2+a^2r^2

making r2r^2 to be the subject of the formulae we have

r2=a2hh2ar^2= \frac{a^2h}{h-2a}

The volume V = 13πh2h2ah​ \tfrac{1}{3}*π *\frac{h^2}{h-2a}*h

at minimum volume ;


dVdh=πa23[(h2a)2hh2(1)(h2a)2]=0\frac{dV}{dh}= \frac{\pi a^2}{3}[\frac{(h-2a)2h-h^2(1)}{(h-2a)^2}]=0

2(h2a)hh2=02(h-2a)h-h^2=0


2h4ah=02h-4a-h=0

h4a=0h-4a=0

h=4ah=4a

but a =8 inches

therefore h=48=32h=4*8=32

The height of the cone is 32 inches.

r2=82323228r^2= \frac{8^2*32}{32-2*8}


r2=204816r^2=\frac{2048}{16}


r2=128r^2=128


r=128r= \sqrt{128}


r=11.3137r= 11.3137


The height of the cone is 32 inches and the radius is 11.3137inches.


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