Question 63 of 82advanced🔧 ApplyLong Answer6 marks

Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin⁡−1(1/3)\sin^{-1}(1/3).

Correct Answer

Let the right circular cone have radius rr, height hh, slant height ll, and semi-vertical angle θ\theta.

Given: Total surface area S=πr2+πrl=constantS = \pi r^2 + \pi r l = \text{constant}

Step 1: Express variables in terms of θ\theta

From the geometry of cone:

  • sin⁡θ=r/l\sin \theta = r/l, so r=lsin⁡θr = l \sin \theta
  • cos⁡θ=h/l\cos \theta = h/l, so h=lcos⁡θh = l \cos \theta

Step 2: Express surface area in terms of ll and θ\theta

S=πr2+πrl=πl2sin⁡2θ+πl2sin⁡θ=πl2(sin⁡2θ+sin⁡θ)S = \pi r^2 + \pi r l = \pi l^2 \sin^2 \theta + \pi l^2 \sin \theta = \pi l^2(\sin^2 \theta + \sin \theta)

Let S/π=kS/\pi = k (constant)

Therefore,

l2=ksin⁡2θ+sin⁡θl^2 = \frac{k}{\sin^2 \theta + \sin \theta}

Step 3: Express volume in terms of θ\theta

V=13πr2h=13πl2sin⁡2θ⋅lcos⁡θ=13πl3sin⁡2θcos⁡θV = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi l^2 \sin^2 \theta \cdot l \cos \theta = \frac{1}{3}\pi l^3 \sin^2 \theta \cos \theta

Substituting l3=[ksin⁡2θ+sin⁡θ]3/2l^3 = \left[\frac{k}{\sin^2 \theta + \sin \theta}\right]^{3/2}:

V=13πk3/2⋅sin⁡2θcos⁡θ(sin⁡2θ+sin⁡θ)3/2V = \frac{1}{3}\pi k^{3/2} \cdot \frac{\sin^2 \theta \cos \theta}{(\sin^2 \theta + \sin \theta)^{3/2}}

Step 4: Let sin⁡θ=t\sin \theta = t and simplify

Let sin⁡θ=t\sin \theta = t, where 0<t<10 < t < 1

V=13πk3/2⋅t21−t2(t2+t)3/2V = \frac{1}{3}\pi k^{3/2} \cdot \frac{t^2 \sqrt{1-t^2}}{(t^2 + t)^{3/2}} V=13πk3/2⋅t(1−t)1+tV = \frac{1}{3}\pi k^{3/2} \cdot \frac{\sqrt{t(1-t)}}{1+t}

Step 5: Maximize V2V^2

V2=19π2k3⋅t(1−t)(1+t)2V^2 = \frac{1}{9}\pi^2 k^3 \cdot \frac{t(1-t)}{(1+t)^2}

Let f(t)=t(1−t)(1+t)2f(t) = \frac{t(1-t)}{(1+t)^2}

f′(t)=(1−2t)(1+t)2−t(1−t)⋅2(1+t)(1+t)4f'(t) = \frac{(1-2t)(1+t)^2 - t(1-t) \cdot 2(1+t)}{(1+t)^4} f′(t)=1−3t(1+t)3f'(t) = \frac{1-3t}{(1+t)^3}

Step 6: Find critical point

For maximum, f′(t)=0f'(t) = 0

1−3t=01 - 3t = 0

giving t=1/3t = 1/3

Step 7: Verify maximum using second derivative

f′′(t)=6(t−1)(1+t)4f''(t) = \frac{6(t-1)}{(1+t)^4}

At t=1/3t = 1/3:

f′′(1/3)=6(1/3−1)(1+1/3)4<0f''(1/3) = \frac{6(1/3-1)}{(1+1/3)^4} < 0

Since f′′(1/3)<0f''(1/3) < 0, t=1/3t = 1/3 gives maximum volume.

Conclusion:

sin⁡θ=t=13\sin \theta = t = \frac{1}{3}

Therefore, θ=sin⁡−1(1/3)\theta = \sin^{-1}(1/3)

Hence, the semi-vertical angle of a right circular cone of given surface area and maximum volume is sin⁡−1(1/3)\sin^{-1}(1/3).

Exercise: EXERCISE 6.3 | Q: 26 | (Page 30)
For More Understanding

Explanation

This problem requires optimization using calculus. The key steps are:

  1. Express all variables in terms of the semi-vertical angle θ\theta using trigonometric relations.
  2. Use the constraint of constant surface area to find ll in terms of θ\theta.
  3. Express volume as a function of θ\theta.
  4. Substitute sin⁡θ=t\sin \theta = t to simplify the expression.
  5. Find the derivative and set it to zero to find critical points.
  6. Use the second derivative test to confirm maximum.

The solution demonstrates that when sin⁡θ=1/3\sin \theta = 1/3, the volume is maximized for a given surface area.

Solution Steps

  1. Step 1: Express rr and hh in terms of ll and θ\theta using sin⁡θ=r/l\sin \theta = r/l and cos⁡θ=h/l\cos \theta = h/l

  2. Step 2: Use surface area constraint S=πl2(sin⁡2θ+sin⁡θ)=constantS = \pi l^2(\sin^2 \theta + \sin \theta) = \text{constant} to find l2l^2

  3. Step 3: Express volume V=13πl3sin⁡2θcos⁡θV = \frac{1}{3}\pi l^3 \sin^2 \theta \cos \theta in terms of θ\theta

  4. Step 4: Substitute sin⁡θ=t\sin \theta = t and simplify to get VV proportional to t(1−t)/(1+t)\sqrt{t(1-t)}/(1+t)

  5. Step 5: Maximize V2V^2 by finding derivative of f(t)=t(1−t)(1+t)2f(t) = \frac{t(1-t)}{(1+t)^2}

  6. Step 6: Set f′(t)=0f'(t) = 0 to get t=1/3t = 1/3

  7. Step 7: Verify maximum using second derivative test (f′′(1/3)<0f''(1/3) < 0)

  8. Step 8: Conclude θ=sin⁡−1(1/3)\theta = \sin^{-1}(1/3)

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