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

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xetaalpha | Student | Honors

Posted June 12, 2013 at 4:32 AM via web

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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 | College Teacher | (Level 2) Distinguished Educator

Posted June 12, 2013 at 4:51 AM (Answer #1)

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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

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aruv | High School Teacher | Valedictorian

Posted June 12, 2013 at 5:25 AM (Answer #2)

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Surface area of the cone= area of the paper.

But

Surface area of the cone= `pirl`

r=radius of the cone and

l= lateral height of the cone.

so 

25=`pirl`

`l=25/(pir)`

`` Let height of the cone is h,

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

`` `h^2=(25/(pir))^2-r^2`

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

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

Thus volume of the cone will be

`V=(1/3)pir^2sqrt(625-pi^2r^4)/(pir)`

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

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

`V'=(1/3){sqrt(625-pi^2r^4)+r/(2sqrt(625-pi^2r^4))(-4pi^2r^3)}`

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

V'=0 if

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

`1250-6pi^2r^4=0`

`r^4=1250/(6pi^2)`

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

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

`V'=(1/6)(1250-6pi^2r^4)/sqrt(625-pi^2r^4)`

`V''=(1/6){sqrt(625-pi^2r^4)(-24pi^2r^3)-(1250-6pi^2r^4)(1/2)(-4pi^2r^3)/sqrt(625-pi^2r^4)}/(625-pi^2r^4)`

`V''}_{r=(1250/(6pi^2))^(1/4)}=-4pi^2r^3/sqrt(625-pi^2r^4)}_{r=(1250/(6pi^2))^(1/4)}<0`

`` 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.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted June 12, 2013 at 6:49 AM (Answer #3)

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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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