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Maximizing the volume of a cone formed by revolving a right triangle
- Thread starter leprofece
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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 $L$.
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 $L$.
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- #3
Re: max and min 3
is that right??'
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R+ H =LHello 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 $L$.
is that right??'
- - - Updated - - -
R+ H =LR2+ H2 =L2
is that right??'
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- #4
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?
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?
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- #5
Re: max and min 3
r2= l2-h2yes, 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?
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- #6
Re: max and min 3
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????
the V pi(L2-h2[/QUOTE]*H/3and derive with respect to Lr2= l2-h2
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????
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- #7
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:
\(\displaystyle V_{\max}=V\left(\frac{L}{\sqrt{3}} \right)=\frac{1}{3}\pi\left(\left(\frac{L}{\sqrt{3}} \right)L^2-\left(\frac{L}{\sqrt{3}} \right)^3 \right)=\frac{1}{3}\pi \frac{L^3}{\sqrt{3}}\left(1-\frac{1}{3} \right)=\frac{2\sqrt{3}}{27}\pi L^3\)
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:
\(\displaystyle V_{\max}=V\left(\frac{L}{\sqrt{3}} \right)=\frac{1}{3}\pi\left(\left(\frac{L}{\sqrt{3}} \right)L^2-\left(\frac{L}{\sqrt{3}} \right)^3 \right)=\frac{1}{3}\pi \frac{L^3}{\sqrt{3}}\left(1-\frac{1}{3} \right)=\frac{2\sqrt{3}}{27}\pi L^3\)