CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Maxima & Minima.

Q.17. Show that the semi-vertical angle of a cone of maximum volume and of given slant height is tan –1(√2).

Solution :

Fig.

Let semi-vertical angle of the cone be α , height be h , radius be r and slant height be l.
Then sin α = r/l => r = l sin α , and cos α = h/l => h = l cos α
Therefore , V = 1/3 π r2 h = 1/3 π (l sin α)2(l cos α)
= 1/3 (π l3 sin² α cos α)
Therefore , dV/dα = 1/3 π l3 [– sin³α + 2cos α . sin α. cos α]
Thus dV/dα = 0 ,gives sin α [– sin² α + 2 cos²α] = 0
Or, – sin² α + 2 – 2 sin²α = 0[sin α ≠ 0]
Or, sin α = √(2/3) => tan α √2 , cos α 1/√3 .
Now, d2V/dα² = 1/3 π l3 [– 3 sin² α cos α + 2 cos α cos 2α + sin 2α (– sin α)]
= 1/3 π l3[0 – 3 ×2/3×1/√3 + 2/√3(2×1/3 –1) + 2×√(2/3) ×1/√3{–√(2/3)}
= 1/3 π l3 [0 – 2/√3 – 2/3√3 – 4/3√3] < 0
Therefore , for maximum volume tanα √2 .
Or, α = tan –1 (√2) . [Proved.]

Q.18. Find the volume of the largest cone that can be inscribed in a sphere of radius R.

Solution :


Fig.

Let base radius of the cone be r and height h and radius of the sphere is R.
In fig. CA = h, In Δ OAB, R2 = (h – R)2 + r2
Or, r² = R² – h² + 2hR – R² = 2hR – h² .
Volume of the cone, V = 1/3 π r² h = 1/3 π (2hR – h²) h
Or, V = 1/3 π (2h²R – h³)
Therefore, dV/dh = 1/3 π (4hR – 3h2) = 0 [For maximum volume]
Or, 4hR – 3h² = 0 => h = 4R/3 [As, h ≠ 0]
Andd²V/dh² = 1/3 π (4R – 6h)
= 1/3 π (4R – 6 × 4R/3) < 0 , [At h = 4R/3]
Therefore , volume is maximum for h = 4R/3, r² = 2hR – h²
= 2R ×4R/3 – 16R2/9
= 8R2/9.
Volume of the cone = 1/3 π × 8R2/9 × 4R/3.
Hence , maximum volume of the cone = (32/81)πR3 cu. units. [Ans.]

Or

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere. [Vol. of cone = (32/81)πR³= (8/27).4πR³]

Paper By Mr. M.P.Keshari
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