a cone is constructed by cutting a sector from a circular sheet of metal with radius 22. the cut sheet is…

Explanation:

Step1: Relate cone - radius and slant - height

The slant - height (l) of the cone is equal to the radius of the circular sheet, so (l = 22). Using the Pythagorean theorem for a cone, (h=\sqrt{l^{2}-r^{2}}=\sqrt{484 - r^{2}}), where (h) is the height of the cone and (r) is the radius of the base of the cone. The volume of a cone is (V=\frac{1}{3}\pi r^{2}h). Substituting (h=\sqrt{484 - r^{2}}) into the volume formula, we get (V=\frac{1}{3}\pi r^{2}\sqrt{484 - r^{2}}).

Step2: Differentiate the volume function

Let (u = r^{2}) and (v=\sqrt{484 - r^{2}}=(484 - r^{2})^{\frac{1}{2}}). Then (V=\frac{1}{3}\pi uv). Using the product - rule ((uv)^\prime=u^\prime v+uv^\prime), (u^\prime = 2r) and (v^\prime=\frac{1}{2}(484 - r^{2})^{-\frac{1}{2}}\times(-2r)=\frac{-r}{\sqrt{484 - r^{2}}}). So (V^\prime=\frac{1}{3}\pi\left(2r\sqrt{484 - r^{2}}+r^{2}\times\frac{-r}{\sqrt{484 - r^{2}}}\right)=\frac{\pi}{3}\times\frac{2r(484 - r^{2})-r^{3}}{\sqrt{484 - r^{2}}}=\frac{\pi}{3}\times\frac{968r-2r^{3}-r^{3}}{\sqrt{484 - r^{2}}}=\frac{\pi(968r - 3r^{3})}{3\sqrt{484 - r^{2}}}).

Step3: Find the critical points

Set (V^\prime = 0), then (968r-3r^{3}=r(968 - 3r^{2}) = 0). We have (r = 0) or (968-3r^{2}=0). Since (r = 0) gives a cone with zero volume, we solve (968-3r^{2}=0). So (3r^{2}=968), and (r^{2}=\frac{968}{3}), then (r=\frac{2\sqrt{726}}{3}).

Step4: Find the height

Substitute (r=\frac{2\sqrt{726}}{3}) into the formula (h=\sqrt{l^{2}-r^{2}}). (h=\sqrt{484-\frac{968}{3}}=\sqrt{\frac{1452 - 968}{3}}=\sqrt{\frac{484}{3}}=\frac{22\sqrt{3}}{3}).

Answer:

(r = \frac{2\sqrt{726}}{3}), (h=\frac{22\sqrt{3}}{3})