Question #1be57

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Question #1be57

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Answer 1 · 2016-11-16T22:08:22

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

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:

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

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

$V = π r^{2} h$ $V = pir^2h$

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

$V = π x^{2} y [2]$ $V = pix^2y" [2]"$

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

$V = π x^{2} (- \frac{9}{11} x + 4.5)$ $V = pix^2(-9/11x + 4.5)$

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

Compute the first derivative with respect to x:

$\frac{d V}{d x} = \frac{- 27 π}{11} x^{2} + \frac{18 π}{2} x$ $(dV)/dx = (-27pi)/11x^2 + (18pi)/2x$

Set the first derivative equal to zero:

$\frac{- 27 π}{11} x^{2} + 9 π x = 0$ $(-27pi)/11x^2 + 9pix = 0$

Divide by $9 π x$ $9pix$ :

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

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

Solve for x:

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

Do the second derivative test:

$\frac{d^{2} V}{{d x}^{2}} = \frac{- 74 π}{11} x + 9 π ∣_{x = \frac{11}{3}} = - \frac{47 π}{3}$ $(d^2V)/dx^2 = (-74pi)/11x + 9pi|_(x=11/3) = -(47pi)/3$

It is clearly a maximum.

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

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

$y = 1.5$ $y = 1.5$

Translating back to r and h:

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

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Question #1be57

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