A sheet of paper with area 25 cm^2 is used to make a cone that is open in the bottom. What is the maximum volume of the cone.



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justaguide's profile pic

Posted on (Answer #1)

The volume of a cone with height h and base radius r is `(1/3)*pi*r^2*h` and the curved surface area is `pi*r*sqrt(r^2+h^2)`

A cone with an open bottom is made using a sheet of paper with area 25 cm^2; the maximum volume of the cone has to be determined. If the radius of the base is r and the height is h, the surface area of the cone is `pi*r*sqrt(r^2+h^2) = 25`

=> `(pi)^2*r^2*(r^2+h^2) = 625`

=> `r^2 + h^2 = 625/((pi)^2*r^2)`

=> `h = sqrt(625/((pi)^2*r^2) - r^2)`

Substituting this in the formula for volume, `V = (1/3)*pi*r^2*sqrt(625/((pi)^2*r^2) - r^2)`

To maximize the volume solve V' = 0 for r

=> `-(2*pi^2*r^6+r^2*(pi^2*r^4+625)-1250*r^2)/(3*r*sqrt(25-pi*r^2)*sqrt(pi*r^2+25)*r) = 0`

=> `2*pi^2*r^4+(pi^2*r^4+625)-1250= 0`

Solving the equation gives the positive, real root  `r =5/(3^(1/4)*sqrt(pi)) ~~ 2.143 `

The height of the cone is approximately 3.031 cm. This gives the maximum volume as 14.57 cm^3

The maximum volume of the cone that can be constructed is 14.57 cm^3

aruv's profile pic

Posted on (Answer #2)

Surface area of the cone= area of the paper.


Surface area of the cone= `pirl`

r=radius of the cone and

l= lateral height of the cone.




`` Let height of the cone is h,


`` `h^2=(25/(pir))^2-r^2`



Thus volume of the cone will be



For getting maximum volume,differentiate V wrt r and equate derivative to 0.


`` `=(1/6){2(625-pi^2r^4)-4pi^2r^4}/sqrt(625-pi^2r^4)`

V'=0 if




`r=(1250/(6pi^2))^(1/4)`  ,length is always positive so negative value ruled out.

Check V'' at r=`(1250/(6pi^2))^(1/4)`




`` Thus  we will get maximum volume when r=`(1250/(6pi^2))^(1/4)`

`` and

`V=(25/3)sqrt(2/3)(1250/(6pi^2))^(1/4)` cubic cm.

jeew-m's profile pic

Posted on (Answer #3)

Surface area of a cone(A) having radius r and lateral height l is given by;

`A = pirl`

If the height of the cone is h the volume(V) of the cone is given by;

`V = 1/3pir^2h`


Now let us move to our question.

`25 = pirl`

`625 = pi^2r^2l^2`


using Pythagorean law for cone;

`l^2 = r^2+h^2`

`pi^2r^2l^2 = pi^2r^4+pi^2r^2h^2`

`625 = pi^2r^4+pi^2r^2h^2`

`h^2 = (625-pi^2r^4)/(pi^2r^2)`


`V = 1/3xxpixxr^2xxh`

`V^2 = 1/9xxpi^2xxr^4xxh^2`

`V^2 = 1/9xxpi^2xxr^4xx(625-pi^2r^4)/(pi^2r^2)`

`V^2 = 1/9xxr^2(625-pi^2r^4)`

`9V^2 = 625r^2-pi^2r^6`


For maximum or minimum volume V' = 0

`9V^2 = 625r^2-pi^2r^6`

`9xx2VxxV' = 2xx625r-pi^2xx6xxr^5`

When V' = 0;

`0 = 2r(625-3pi^2r^4)`

Since `r !=0` ;

`625-3pi^2r^4 = 0`

`r = (625/(3pi^2))^(1/4)`

`r = 0.68pi`


If at r = 0.68pi volume is maximum then `(V'')_(r = 0.68pi) <0`

`9xx2VxxV' = 2xx625r-pi^2xx6xxr^5`

`18(VxxV''+(V')^2) = 1250-pi^2xx6xx5r^4`

When `r = 0.68pi;`

`18VxxV''+18xx0 = 1250- pi^2xx6xx5(0.68pi)^4 = -4916.747`

`18VxxV'' = -4916.747`

Since V>0 we can say V''<0

So we have a maximum for the volume of cone.


`r = 0.68pi`

`h = sqrt((625-pi^2r^4)/(pi^2r^2)) = 0.97pi`


Maximum Volume of cone

`= 1/3xxpixx(0.68pi)^2xx0.97pi`

`= 0.149pi^4`

`= 14.56cm^3`


So the maximum volume of the cone is `14.56cm^3` .




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