Question #1be57

Nov 16, 2016

The cylinder has radius, $r = \frac{11}{3}$ and $h = 1.5$

Explanation:

If you put the center of the cone on the y axis and look at the cross-section of the cone in the x-y plane, it is a triangle that intersects the y axis at 4.5 and the x axis at 5.5; the equation of that line is:

$y = - \frac{9}{11} x + 4.5 \text{ [1]}$

We want to find the point where the cylinder intersects this line such that its volume is maximized:

$V = \pi {r}^{2} h$

But we are looking at these solid objects as cross-sections in the x-y plane so $x = r \mathmr{and} y = h$:

$V = \pi {x}^{2} y \text{ [2]}$

Substitute the right side of equation [1] for y in equation [2]:

$V = \pi {x}^{2} \left(- \frac{9}{11} x + 4.5\right)$

$V = \frac{- 9 \pi}{11} {x}^{3} + \frac{9 \pi}{2} {x}^{2}$

Compute the first derivative with respect to x:

$\frac{\mathrm{dV}}{\mathrm{dx}} = \frac{- 27 \pi}{11} {x}^{2} + \frac{18 \pi}{2} x$

Set the first derivative equal to zero:

$\frac{- 27 \pi}{11} {x}^{2} + 9 \pi x = 0$

Divide by $9 \pi x$:

$\frac{- 3}{11} x + 1 = 0$

(Please notice that we got rid of the root x = 0 but that is clearly a minimum)

Solve for x:

$x = \frac{11}{3}$

Do the second derivative test:

$\frac{{d}^{2} V}{\mathrm{dx}} ^ 2 = \frac{- 74 \pi}{11} x + 9 \pi {|}_{x = \frac{11}{3}} = - \frac{47 \pi}{3}$

It is clearly a maximum.

Substitute $x = \frac{11}{3}$ into equation [1]:

$y = - \frac{9}{11} \left(\frac{11}{3}\right) + 4.5$

$y = 1.5$

Translating back to r and h:

The cylinder has radius, $r = \frac{11}{3}$ and $h = 1.5$