Finding the height and top radius of cone so that volume is maximum & Finding the angle so that the...











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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.



Can anyone help me, thanks!






calculus optimization volume







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    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.



    Can anyone help me, thanks!






    calculus optimization volume







    share|cite|improve this question
























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      favorite
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      1





      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.



      Can anyone help me, thanks!






      calculus optimization volume







      share|cite|improve this question













      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.



      Can anyone help me, thanks!






      calculus optimization volume




      calculus optimization volume






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      asked Nov 28 at 1:06









      ShadyAF

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          $V = frac{1}{3}pi R^2cdot h$



          $Cx$ is the remaining circumference after removing a sector.



          The base radius then becomes $frac{Cx}{2pi} = frac{xcdot 2picdot r}{2pi} = xr$



          So $V = frac{1}{3}pi(rx)^2cdot sqrt{r^2-(rx)^2}$



          $$frac{dV}{dx} = frac{2}{3}picdot r^2 xcdot sqrt{r^2-(rx)^2}+frac{1}{3}pi(rx)^2cdot frac{-r^2x}{sqrt{r^2-(rx)^2}}$$



          Max volume occurs when $frac{dV}{dx} = 0$



          Then $$frac{2}{3}picdot r^2 xcdot sqrt{r^2-(rx)^2} = frac{1}{3}pi(rx)^2cdot frac{-r^2x}{sqrt{r^2-(rx)^2}}$$



          Which reduces to $2r^2 = 3(rx)^2$



          And $x = sqrt{frac{2}{3}}$



          $theta = 360 - 360cdot sqrt{frac{2}{3}}$ is the angle of the sector cutout and....



          $R = 3cdot sqrt{frac{2}{3}}$ is the base radius and .....



          $h = sqrt{9 - (3cdot sqrt{frac{2}{3}})^2} = sqrt 3$ is the height






          share|cite|improve this answer




























            up vote
            0
            down vote













            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 $6pi - 3alpha$ (i.e., subtracting the arc length from the original circumference)



            Then the radius of the base is $frac{6pi-3alpha}{2pi}$. 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 $frac{dV}{dalpha}=0$.






            share|cite|improve this answer























            • By alpha do you mean the angle theta?
              – ShadyAF
              Nov 28 at 2:48










            • Yes, $alpha$ is the angle of the wedge removed.
              – user25959
              Nov 28 at 2:59











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            2 Answers
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            active

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            2 Answers
            2






            active

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            active

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            active

            oldest

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            up vote
            1
            down vote



            accepted










            $V = frac{1}{3}pi R^2cdot h$



            $Cx$ is the remaining circumference after removing a sector.



            The base radius then becomes $frac{Cx}{2pi} = frac{xcdot 2picdot r}{2pi} = xr$



            So $V = frac{1}{3}pi(rx)^2cdot sqrt{r^2-(rx)^2}$



            $$frac{dV}{dx} = frac{2}{3}picdot r^2 xcdot sqrt{r^2-(rx)^2}+frac{1}{3}pi(rx)^2cdot frac{-r^2x}{sqrt{r^2-(rx)^2}}$$



            Max volume occurs when $frac{dV}{dx} = 0$



            Then $$frac{2}{3}picdot r^2 xcdot sqrt{r^2-(rx)^2} = frac{1}{3}pi(rx)^2cdot frac{-r^2x}{sqrt{r^2-(rx)^2}}$$



            Which reduces to $2r^2 = 3(rx)^2$



            And $x = sqrt{frac{2}{3}}$



            $theta = 360 - 360cdot sqrt{frac{2}{3}}$ is the angle of the sector cutout and....



            $R = 3cdot sqrt{frac{2}{3}}$ is the base radius and .....



            $h = sqrt{9 - (3cdot sqrt{frac{2}{3}})^2} = sqrt 3$ is the height






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              $V = frac{1}{3}pi R^2cdot h$



              $Cx$ is the remaining circumference after removing a sector.



              The base radius then becomes $frac{Cx}{2pi} = frac{xcdot 2picdot r}{2pi} = xr$



              So $V = frac{1}{3}pi(rx)^2cdot sqrt{r^2-(rx)^2}$



              $$frac{dV}{dx} = frac{2}{3}picdot r^2 xcdot sqrt{r^2-(rx)^2}+frac{1}{3}pi(rx)^2cdot frac{-r^2x}{sqrt{r^2-(rx)^2}}$$



              Max volume occurs when $frac{dV}{dx} = 0$



              Then $$frac{2}{3}picdot r^2 xcdot sqrt{r^2-(rx)^2} = frac{1}{3}pi(rx)^2cdot frac{-r^2x}{sqrt{r^2-(rx)^2}}$$



              Which reduces to $2r^2 = 3(rx)^2$



              And $x = sqrt{frac{2}{3}}$



              $theta = 360 - 360cdot sqrt{frac{2}{3}}$ is the angle of the sector cutout and....



              $R = 3cdot sqrt{frac{2}{3}}$ is the base radius and .....



              $h = sqrt{9 - (3cdot sqrt{frac{2}{3}})^2} = sqrt 3$ is the height






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                $V = frac{1}{3}pi R^2cdot h$



                $Cx$ is the remaining circumference after removing a sector.



                The base radius then becomes $frac{Cx}{2pi} = frac{xcdot 2picdot r}{2pi} = xr$



                So $V = frac{1}{3}pi(rx)^2cdot sqrt{r^2-(rx)^2}$



                $$frac{dV}{dx} = frac{2}{3}picdot r^2 xcdot sqrt{r^2-(rx)^2}+frac{1}{3}pi(rx)^2cdot frac{-r^2x}{sqrt{r^2-(rx)^2}}$$



                Max volume occurs when $frac{dV}{dx} = 0$



                Then $$frac{2}{3}picdot r^2 xcdot sqrt{r^2-(rx)^2} = frac{1}{3}pi(rx)^2cdot frac{-r^2x}{sqrt{r^2-(rx)^2}}$$



                Which reduces to $2r^2 = 3(rx)^2$



                And $x = sqrt{frac{2}{3}}$



                $theta = 360 - 360cdot sqrt{frac{2}{3}}$ is the angle of the sector cutout and....



                $R = 3cdot sqrt{frac{2}{3}}$ is the base radius and .....



                $h = sqrt{9 - (3cdot sqrt{frac{2}{3}})^2} = sqrt 3$ is the height






                share|cite|improve this answer












                $V = frac{1}{3}pi R^2cdot h$



                $Cx$ is the remaining circumference after removing a sector.



                The base radius then becomes $frac{Cx}{2pi} = frac{xcdot 2picdot r}{2pi} = xr$



                So $V = frac{1}{3}pi(rx)^2cdot sqrt{r^2-(rx)^2}$



                $$frac{dV}{dx} = frac{2}{3}picdot r^2 xcdot sqrt{r^2-(rx)^2}+frac{1}{3}pi(rx)^2cdot frac{-r^2x}{sqrt{r^2-(rx)^2}}$$



                Max volume occurs when $frac{dV}{dx} = 0$



                Then $$frac{2}{3}picdot r^2 xcdot sqrt{r^2-(rx)^2} = frac{1}{3}pi(rx)^2cdot frac{-r^2x}{sqrt{r^2-(rx)^2}}$$



                Which reduces to $2r^2 = 3(rx)^2$



                And $x = sqrt{frac{2}{3}}$



                $theta = 360 - 360cdot sqrt{frac{2}{3}}$ is the angle of the sector cutout and....



                $R = 3cdot sqrt{frac{2}{3}}$ is the base radius and .....



                $h = sqrt{9 - (3cdot sqrt{frac{2}{3}})^2} = sqrt 3$ is the height







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 28 at 2:50









                Phil H

                3,9882312




                3,9882312






















                    up vote
                    0
                    down vote













                    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 $6pi - 3alpha$ (i.e., subtracting the arc length from the original circumference)



                    Then the radius of the base is $frac{6pi-3alpha}{2pi}$. 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 $frac{dV}{dalpha}=0$.






                    share|cite|improve this answer























                    • By alpha do you mean the angle theta?
                      – ShadyAF
                      Nov 28 at 2:48










                    • Yes, $alpha$ is the angle of the wedge removed.
                      – user25959
                      Nov 28 at 2:59















                    up vote
                    0
                    down vote













                    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 $6pi - 3alpha$ (i.e., subtracting the arc length from the original circumference)



                    Then the radius of the base is $frac{6pi-3alpha}{2pi}$. 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 $frac{dV}{dalpha}=0$.






                    share|cite|improve this answer























                    • By alpha do you mean the angle theta?
                      – ShadyAF
                      Nov 28 at 2:48










                    • Yes, $alpha$ is the angle of the wedge removed.
                      – user25959
                      Nov 28 at 2:59













                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    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 $6pi - 3alpha$ (i.e., subtracting the arc length from the original circumference)



                    Then the radius of the base is $frac{6pi-3alpha}{2pi}$. 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 $frac{dV}{dalpha}=0$.






                    share|cite|improve this answer














                    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 $6pi - 3alpha$ (i.e., subtracting the arc length from the original circumference)



                    Then the radius of the base is $frac{6pi-3alpha}{2pi}$. 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 $frac{dV}{dalpha}=0$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 28 at 3:06

























                    answered Nov 28 at 1:24









                    user25959

                    1,559816




                    1,559816












                    • By alpha do you mean the angle theta?
                      – ShadyAF
                      Nov 28 at 2:48










                    • Yes, $alpha$ is the angle of the wedge removed.
                      – user25959
                      Nov 28 at 2:59


















                    • By alpha do you mean the angle theta?
                      – ShadyAF
                      Nov 28 at 2:48










                    • Yes, $alpha$ is the angle of the wedge removed.
                      – user25959
                      Nov 28 at 2:59
















                    By alpha do you mean the angle theta?
                    – ShadyAF
                    Nov 28 at 2:48




                    By alpha do you mean the angle theta?
                    – ShadyAF
                    Nov 28 at 2:48












                    Yes, $alpha$ is the angle of the wedge removed.
                    – user25959
                    Nov 28 at 2:59




                    Yes, $alpha$ is the angle of the wedge removed.
                    – user25959
                    Nov 28 at 2:59


















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