Question

3. Find the maximum volume of a right - circular cylinder that can be inscribed in a right-circular cone of
radius R = 3 and height H = 9.

Answer 1

$A\>right\>circular\>cylinder\>is\>inscribed\>in\>a\>cone\>with\>height\>h\>and\>base\>radius\>r\\Find\>the\>largest\>volume\>of\>such\>a\>case\>\\h=height\>of\>cone\\r=radius\>of\>cone\\volume\>of\>cylinder\>changes\>as\>a\>function\>of\>cylinder’s\>radius,\>x\>\\volume\>of\>cylinder\>=(Area\>of\>cylinder)(height\>of\>cylinder)\\volume\>of\>cylinder=(π\left(x^{\smash{2}}\right))\>(height\>of\>cylinder)\\h-y=height\>of\>cylinder\\y=height\>between\>height\>of\>cylinder\>and\>to\>of\>cone\\use\>the\>diagram\>of\>similar\>triangles\>at\>right\\{y\over\>x}={h\over\>r}\\y=\>{hx\over\>r}\\volume\>of\>cylinder=(π\left(x^{\smash{2}}\right))(h-y)=\>(π\left(x^{\smash{2}}\right))(h-\>hx/r)=\>(π\left(x^{\smash{2}}\right)h-π\left(x^{\smash{3}}\right)h/r)\>\>-->eq(1)\\v’(x)=2πxh-3πhπ\left(x^{\smash{2}}\right)/r=πxh(2-3x/r)\\so\>x=0,\>x=2r/3\\v’’(x)=2πh-6πhx/r\\if\>x=2r/3\>\>\>\>\>\>----->\>second\>derivative\>test\>gives\>maximum\>volume\>\\v’’(2r/3)=-2π<0----->eq(1)\\v(2r/3)=\>π(2r/3)2h-πh/r(2r/3)3=\>{4π\left(r^{\smash{2}}\right)h\>\over\>27}\>\\Maximum\>volume\>of\>cylinder\>inscribed\>in\>a\>cone\>of\>radius\>r\>and\>height\>h\>=\>{4π\left(r^{\smash{2}}\right)h\>\over\>27}\>$