Question 80 of 82advanced🔧 ApplyLong Answer6 marks

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius RR is 2R3\frac{2R}{\sqrt{3}}. Also find the maximum volume.

Correct Answer

Height =2R3= \dfrac{2R}{\sqrt{3}}, Maximum Volume =4πR339= \dfrac{4\pi R^{3}\sqrt{3}}{9}

Exercise: Miscellaneous Exercise on Chapter 6 | Q: 14 | (Page 38)
For More Understanding

Explanation

This is an optimimization problem using derivatives. Following the methodology shown in Example 26 of the context, we express volume as a function of height, find critical points using first derivative, and confirm maxima using second derivative test.

Solution Steps

  1. Step 1: Let RR be the radius of sphere, rr be the radius of cylinder, and hh be the height of the cylinder inscribed in the sphere.

  2. Step 2: From the geometry, the center of sphere lies at the center of cylinder. Using right triangle:

    r2+(h2)2=R2r^{2} + \left(\frac{h}{2}\right)^{2} = R^{2}

    which gives:

    r2=R2−h24r^{2} = R^{2} - \frac{h^{2}}{4}

  3. Step 3: Volume of cylinder:

    V=πr2h=π(R2−h24)h=π(R2h−h34)V = \pi r^{2} h = \pi \left(R^{2} - \frac{h^{2}}{4}\right) h = \pi \left(R^{2}h - \frac{h^{3}}{4}\right)

  4. Step 4: First derivative:

    dVdh=π(R2−3h24)\frac{dV}{dh} = \pi \left(R^{2} - \frac{3h^{2}}{4}\right)

    Setting dVdh=0\frac{dV}{dh} = 0 gives:

    R2−3h24=0R^{2} - \frac{3h^{2}}{4} = 0

    so:

    h2=4R23h^{2} = \frac{4R^{2}}{3}

    hence h=2R3h = \dfrac{2R}{\sqrt{3}}

  5. Step 5: Second derivative:

    d2Vdh2=−3πh2<0for h>0\frac{d^{2}V}{dh^{2}} = -\frac{3\pi h}{2} < 0 \quad \text{for } h > 0

    This confirms maximum volume at h=2R3h = \dfrac{2R}{\sqrt{3}}.

  6. Step 6: Maximum Volume:

    V=π(R2×2R3−(2R3)34)=π(2R33−2R333)V = \pi \left(R^{2} \times \frac{2R}{\sqrt{3}} - \frac{\left(\frac{2R}{\sqrt{3}}\right)^{3}}{4}\right) = \pi \left(\frac{2R^{3}}{\sqrt{3}} - \frac{2R^{3}}{3\sqrt{3}}\right)

    ∴ 4πR339\dfrac{4\pi R^{3}\sqrt{3}}{9}

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