right circular cylinder is inscribed in cone with height h, base radius r. Find largest possible volume of cylinder.
- Thread starter Lolaploca
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Consider the following diagram, showing one-half of a cross-section through the axis of symmetry for the system:
Our objective function is the volume of the cylinder:
[MATH]V\left(h_C,r_C\right)=\pi r_C^2h_C[/MATH]
And, using the two-intercept form of a line, our constraint is:
[MATH]g\left(h_C,r_C\right)=\frac{r_C}{r}+\frac{h_C}{h}-1=0[/MATH]
Can you proceed?
Our objective function is the volume of the cylinder:
[MATH]V\left(h_C,r_C\right)=\pi r_C^2h_C[/MATH]
And, using the two-intercept form of a line, our constraint is:
[MATH]g\left(h_C,r_C\right)=\frac{r_C}{r}+\frac{h_C}{h}-1=0[/MATH]
Can you proceed?
- Joined
- Nov 24, 2012
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- 3,021
To follow up:
First, let's look at the method that would be used in the first semester of calculus.We will use the constraint to write \(h_C\) as a function of \(r_C\):
[MATH]h_C=h\left(1-\frac{r_C}{r}\right)[/MATH]
Substituting this into our volume function, we now have the volume as a function of one variable:
[MATH]V\left(r_C\right)=\pi hr_C^2\left(1-\frac{r_C}{r}\right)[/MATH]
Now, we may compute the first derivative of the volume function and equate this to zero to determine any critical values:
[MATH]V'\left(r_C\right)=\pi h\left(2r_C\left(1-\frac{r_C}{r}\right)+r_C^2\left(-\frac{1}{r}\right)\right)=\frac{\pi h}{r}r_C\left(2r-3r_C\right)=0[/MATH]
This gives us two critical values:
[MATH]r_C\in\left\{0,\frac{2}{3}r\right\}[/MATH]
Now, to check the nature of the extrema associated with these critical values, let's compute the second derivative of the volume function:
[MATH]V''\left(r_C\right)=\frac{\pi h}{r}\left(\left(2r-3r_C\right)+r_C(-3)\right)=\frac{2\pi h}{r}\left(r-3r_C\right)[/MATH]
We then find:
[MATH]V''(0)=2\pi h>0[/MATH]
[MATH]V''\left(\frac{2}{3}r\right)=-2\pi h[/MATH]
And so, by the second derivative test, we find the critical value that maximizes the volume function is:
[MATH]r_C=\frac{2}{3}r[/MATH]
We may thus state:
[MATH]V_{\max}=V\left(\frac{2}{3}r\right)=\pi h\left(\frac{2}{3}r\right)^2\left(1-\frac{\dfrac{2}{3}r}{r}\right)=\frac{4\pi}{27}hr^2[/MATH]
Next, let's look at a method used in the third semester of calculus, using Lagrange Multipliers. This method gives rise to the system:
[MATH]\pi r_C^2=\lambda\left(\frac{1}{h}\right)[/MATH]
[MATH]2\pi r_Ch_C=\lambda\left(\frac{1}{r}\right)[/MATH]
This implies:
[MATH]\pi r_C^2h=2\pi r_Ch_Cr[/MATH]
Or:
[MATH]r_C\left(hr_C-2h_Cr\right)=0[/MATH]
This in turn implies:
[MATH]r_C=0[/MATH]
[MATH]hr_C-2h_Cr=0\implies r_C=\frac{2h_Cr}{h}[/MATH]
Plugging these value into the constraint gives us two critical points:
[MATH]\left(h_C,r_C\right)=(h,0),\,\left(\frac{1}{3}h,\frac{2}{3}r\right)[/MATH]
Evaluating the volume function at these critical points, we find:
[MATH]V(h,0)=0[/MATH]
[MATH]V\left(\frac{1}{3}h,\frac{2}{3}r\right)=\pi\left(\frac{1}{3}h\right)\left(\frac{2}{3}r\right)^2=\frac{4\pi}{27}hr^2[/MATH]
And so we may conclude:
[MATH]V_{\min}=V(h,0)=0[/MATH]
[MATH]V_{\max}=V\left(\frac{1}{3}h,\frac{2}{3}r\right)=\frac{4\pi}{27}hr^2[/MATH]
First, let's look at the method that would be used in the first semester of calculus.We will use the constraint to write \(h_C\) as a function of \(r_C\):
[MATH]h_C=h\left(1-\frac{r_C}{r}\right)[/MATH]
Substituting this into our volume function, we now have the volume as a function of one variable:
[MATH]V\left(r_C\right)=\pi hr_C^2\left(1-\frac{r_C}{r}\right)[/MATH]
Now, we may compute the first derivative of the volume function and equate this to zero to determine any critical values:
[MATH]V'\left(r_C\right)=\pi h\left(2r_C\left(1-\frac{r_C}{r}\right)+r_C^2\left(-\frac{1}{r}\right)\right)=\frac{\pi h}{r}r_C\left(2r-3r_C\right)=0[/MATH]
This gives us two critical values:
[MATH]r_C\in\left\{0,\frac{2}{3}r\right\}[/MATH]
Now, to check the nature of the extrema associated with these critical values, let's compute the second derivative of the volume function:
[MATH]V''\left(r_C\right)=\frac{\pi h}{r}\left(\left(2r-3r_C\right)+r_C(-3)\right)=\frac{2\pi h}{r}\left(r-3r_C\right)[/MATH]
We then find:
[MATH]V''(0)=2\pi h>0[/MATH]
[MATH]V''\left(\frac{2}{3}r\right)=-2\pi h[/MATH]
And so, by the second derivative test, we find the critical value that maximizes the volume function is:
[MATH]r_C=\frac{2}{3}r[/MATH]
We may thus state:
[MATH]V_{\max}=V\left(\frac{2}{3}r\right)=\pi h\left(\frac{2}{3}r\right)^2\left(1-\frac{\dfrac{2}{3}r}{r}\right)=\frac{4\pi}{27}hr^2[/MATH]
Next, let's look at a method used in the third semester of calculus, using Lagrange Multipliers. This method gives rise to the system:
[MATH]\pi r_C^2=\lambda\left(\frac{1}{h}\right)[/MATH]
[MATH]2\pi r_Ch_C=\lambda\left(\frac{1}{r}\right)[/MATH]
This implies:
[MATH]\pi r_C^2h=2\pi r_Ch_Cr[/MATH]
Or:
[MATH]r_C\left(hr_C-2h_Cr\right)=0[/MATH]
This in turn implies:
[MATH]r_C=0[/MATH]
[MATH]hr_C-2h_Cr=0\implies r_C=\frac{2h_Cr}{h}[/MATH]
Plugging these value into the constraint gives us two critical points:
[MATH]\left(h_C,r_C\right)=(h,0),\,\left(\frac{1}{3}h,\frac{2}{3}r\right)[/MATH]
Evaluating the volume function at these critical points, we find:
[MATH]V(h,0)=0[/MATH]
[MATH]V\left(\frac{1}{3}h,\frac{2}{3}r\right)=\pi\left(\frac{1}{3}h\right)\left(\frac{2}{3}r\right)^2=\frac{4\pi}{27}hr^2[/MATH]
And so we may conclude:
[MATH]V_{\min}=V(h,0)=0[/MATH]
[MATH]V_{\max}=V\left(\frac{1}{3}h,\frac{2}{3}r\right)=\frac{4\pi}{27}hr^2[/MATH]