 # You have a 6-inch diameter circle of paper which you want to form into a drinking cup by removing a pie-shaped wedge with central angle theta and forming the remaining paper into a cone. You are given that 3 is the slant height a) Find the height and top radius of the cone so that the volume of the cup is as large as possible. b) What is the angle theta of the arc of the paper that is cut out to create the cone of maximum volume? Cheyanne Jefferson 2022-08-13 Answered
Finding the height and top radius of cone so that volume is maximum and Finding the angle so that the volume is maximum
You have a 6-inch diameter circle of paper which you want to form into a drinking cup by removing a pie-shaped wedge with central angle theta and forming the remaining paper into a cone. - You are given that 3 is the slant height
a) Find the height and top radius of the cone so that the volume of the cup is as large as possible.
b) What is the angle theta of the arc of the paper that is cut out to create the cone of maximum volume?
I know how to do related rates with volume, but I can't seem to figure it out with angles being cut out from a circle.
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Step 1
Perhaps the easiest way to relate $\alpha$ (the angle of the wedge removed) to the rest of the cone's dimensions is to note that the perimeter of the base is $6\pi -3\alpha$ (i.e., subtracting the arc length from the original circumference)
Step 2
Then the radius of the base is $\frac{6\pi -3\alpha }{2\pi }$. Together with the slant height you can find the height using the pythagorean theorem
Then use the formula for the volume of a cone $V=\frac{1}{3}\pi {r}^{2}h$ where both r and h depend on $\alpha$, and solve $\frac{dV}{d\alpha }=0$.
###### Not exactly what you’re looking for? sarahkobearab4
Step 1
$V=\frac{1}{3}\pi {R}^{2}\cdot h$
Cx is the remaining circumference after removing a sector.
The base radius then becomes $\frac{Cx}{2\pi }=\frac{x\cdot 2\pi \cdot r}{2\pi }=xr$.
So $V=\frac{1}{3}\pi \left(rx{\right)}^{2}\cdot \sqrt{{r}^{2}-\left(rx{\right)}^{2}}$
$\frac{dV}{dx}=\frac{2}{3}\pi \cdot {r}^{2}x\cdot \sqrt{{r}^{2}-\left(rx{\right)}^{2}}+\frac{1}{3}\pi \left(rx{\right)}^{2}\cdot \frac{-{r}^{2}x}{\sqrt{{r}^{2}-\left(rx{\right)}^{2}}}$
Max volume occurs when $\frac{dV}{dx}=0$
Then $\frac{2}{3}\pi \cdot {r}^{2}x\cdot \sqrt{{r}^{2}-\left(rx{\right)}^{2}}=\frac{1}{3}\pi \left(rx{\right)}^{2}\cdot \frac{-{r}^{2}x}{\sqrt{{r}^{2}-\left(rx{\right)}^{2}}}$
Step 2
Which reduces to $2{r}^{2}=3\left(rx{\right)}^{2}$
And $x=\sqrt{\frac{2}{3}}$
$\theta =360-360\cdot \sqrt{\frac{2}{3}}$ is the angle of the sector cutout and
$R=3\cdot \sqrt{\frac{2}{3}}$ is the base radius and
$h=\sqrt{9-\left(3\cdot \sqrt{\frac{2}{3}}{\right)}^{2}}=\sqrt{3}$ is the height