Question 912071
There is a relationship between {{{r}}} and {{{h}}}.
When {{{r=5}}}, {{{h=0}}}
When {{{r=0}}}, {{{h=10}}}.
So then using the point-slope form of a line,
{{{h-10=((10-0)/(0-5))(r-0)}}}
{{{h-10=-2r}}}
{{{h=10-2r}}}
So the volume of the cylinder is,
{{{V=pi*r^2*h}}}
{{{V=pi*r^2(10-2r)}}}
To maximize the volume, take the derivative of the volume and set it equal to zero (use the chain rule).
{{{dV/dt=pi*(r^2(-2)+(10-2r)(2r))}}}
{{{dV/dt=pi*(-2r^2+(20r-4r^2))}}}
{{{dV/dt=pi*(20r-6r^2)}}}
{{{dV/dt=2pi*r(10-3r)}}}
So then,
{{{10-3r=0}}}
{{{-3r=-10}}}
{{{r=10/3}}}
Then,
{{{h=10-2(10/3)}}}
{{{h=10-20/3}}}
{{{h=30/3-20/3}}}
{{{h=10/3}}}
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.
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{{{V[max]=pi*(10/3)^2*(10/3)}}}
{{{V[max]=(1000/27)pi}}}