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Maximizing the volume of a cone formed by revolving a right triangle

leprofece

Member
Jan 23, 2014
241
A triangle hypotenuse given rectangle is rotated around one of their legs to generate a right circular cone?
find the cone of greater volume.
resp V= (2Sqrt(3)pi L^3)/27

It says hypotenuse given but it has no value According to the answer you can name it L
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 3

Hello leprofece,

We require that you show your work, as I have stated once before. Our desire is to help you solve the problem, rather than work it for you. This way you learn more, by being an active participant in the process of solving the problem.

To being the problem, can you state the objective function, that is, the function you wish to optimize, along with the constraint on the variables?

The objective function will be the volume of a right circular cone, and the constraint will be the relationship between the radius and height of the cone with its slant height, which you have labeled as $L$.
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 3

Hello leprofece,

We require that you show your work, as I have stated once before. Our desire is to help you solve the problem, rather than work it for you. This way you learn more, by being an active participant in the process of solving the problem.

To being the problem, can you state the objective function, that is, the function you wish to optimize, along with the constraint on the variables?

The objective function will be the volume of a right circular cone, and the constraint will be the relationship between the radius and height of the cone with its slant height, which you have labeled as $L$.
R+ H =L

is that right??'

- - - Updated - - -

R2+ H2 =L2

is that right??'
R+ H =L
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 3

Yes, the two legs of the right triangle are the radius and height of the resulting cone, and they are indeed related to the slant height (the hypotenuse of the right triangle) by the Pythagorean theorem:

\(\displaystyle R^2+H^2=L^2\)

So this is the constraint...what is the objective function, i.e., the volume of the cone?
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 3

yes, the two legs of the right triangle are the radius and height of the resulting cone, and they are indeed related to the slant height (the hypotenuse of the right triangle) by the pythagorean theorem:

\(\displaystyle r^2+h^2=l^2\)

so this is the constraint...what is the objective function, i.e., the volume of the cone?
r2= l2-h2
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 3

r2= l2-h2
the V pi(L2-h2[/QUOTE]*H/3and derive with respect to L
L2h-h2[/QUOTE]
hL2-h^3/3
derivating I got 2hL-3h^2/3
and This is equal to L =3h2/(2h)
That is right isnot it????
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 3

Your objective function is the volume of the cone:

\(\displaystyle V=\frac{1}{3}\pi R^2H\)

Using:

\(\displaystyle R^2=L^2-H^2\)

This becomes:

\(\displaystyle V(H)=\frac{1}{3}\pi \left(L^2-H^2 \right)H=\frac{1}{3}\pi\left(HL^2-H^3 \right)\)

Differentiating with respect to $H$ and equating the result to zero, we obtain:

\(\displaystyle V'(H)=\frac{1}{3}\pi\left(L^2-3H^2 \right)=0\)

Recall $L$ is a constant.

This implies:

\(\displaystyle L^2-3H^2=0\implies H=\frac{L}{\sqrt{3}}\)

We observe that the second derivative of the volume function is negative at this critical value which means we have a maximum. And so we may conclude:

\(\displaystyle V_{\max}=V\left(\frac{L}{\sqrt{3}} \right)=\frac{1}{3}\pi\left(\left(\frac{L}{\sqrt{3}} \right)L^2-\left(\frac{L}{\sqrt{3}} \right)^3 \right)=\frac{1}{3}\pi \frac{L^3}{\sqrt{3}}\left(1-\frac{1}{3} \right)=\frac{2\sqrt{3}}{27}\pi L^3\)