A right triangle whose hypotenuse is StartRoot 7 EndRoot7 m long is revolved about one of its legs to generate a right circular cone. Find the​ radius, height, and volume of the cone of greatest volume that can be made this way.

Answer:The maximum volume of cone is 138.25 m³[tex]Radius, r=7\sqrt{\dfrac{2}{3}}[/tex] m [tex]Height, h=\dfrac{7}{\sqrt{3}}[/tex] m Step-by-step explanation:A right circular cone whose hypotenuse is [tex]\sqrt{7}[/tex] m It is revolved about one leg to generate a right circular cone. Let radius be r m and height be h mFor right angle triangle, [tex]r^2+h^2=7^2[/tex][tex]r^2=49-h^2[/tex]Volume of generated cone [tex]=\dfrac{1}{3}\pi r^2h[/tex] [tex]V=\dfrac{1}{3}\pi (49-h^2)h[/tex] Differentiate w.r.t h [tex]\dfrac{dV}{dh}=\dfrac{1}{3}\pi (49-3h^2)[/tex]For maximum/minimum [tex]\dfrac{dV}{dh}=0[/tex][tex]\dfrac{1}{3}\pi (49-3h^2)=0[/tex][tex]h=\dfrac{7}{\sqrt{3}}[/tex] [tex]r^2=49-\dfrac{49}{3}[/tex][tex]r=7\sqrt{\dfrac{2}{3}}[/tex]Using double derivative test [tex]\dfrac{d^2V}{dh^2}=\dfrac{1}{3}\pi (-6h)[/tex]At [tex]h=\dfrac{7}{\sqrt{3}}[/tex]  [tex]\dfrac{d^2V}{dh^2}<0[/tex] so get maximum volume. Dimension of cone: [tex]Radius, r=7\sqrt{\dfrac{2}{3}}[/tex] m [tex]Height, h=\dfrac{7}{\sqrt{3}}[/tex] m [tex]V=\frac{1}{3}\pi\cdot\left(7\sqrt{\frac{2}{3}}\right)^{2}\cdot\frac{7}{\sqrt{3}}[/tex]The maximum volume of cone is 138.25 m³