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

Maxima & Minima.

Let radius of the base, height and slant height of the cone be r, h and l respectively and semi-vertical angle be θ, such that sin θ = r/l.
Total surface area, S = πrl + πr2 = πr (l + r) => l = S/(πr) – r ----------------- (i)
Volume of the cone, V = 1/3 π r2h
Or, V² = 1/9 π² r4h² = 1/9 π2r4(l2 – r²)
= 1/9π² r² [{S/(πr) – r}² – r²] [by (i)]
= 1/9 π² r4[S²/(πr)² – 2S/π]
= 1/9 S (Sr² – 2πr²) = f(r), say.
Then, f’(r) = 1/9 S(2Sr – 8πr³) = 2/9S(Sr – 4πr³).
And f”(r) = 2/9S(S – 12πr2).
Now, f’(r) = 0 => Sr – 4πr³ = 0
Or, r = √(S/4π) and f”{√(S/4π)} = 2/9S(S – 3S) = – 4S²/9 < 0.
Therefore, f(r) is maximum when r = √(S/4π)
Or, V² is maximum when S = 4πr²
Or, V is maximum when l + r = 4r
Or, l = 3r => r/l = 1/3 i.e. sin θ = 1/3 i.e. when θ = sin –1(1/3).
Hence volume of the cone is maximum when its semi-vertical angle is sin –1(1/3).
[Proved.]

Q.9. Show that a rectangle of maximum perimeter which can be inscribed in a circle of radius r is a square of side √2 r.

Solution :

Fig.

Let ABCD be the rectangle inscribed in a circle of radius r and centre O. BD is the diameter = 2r. Let LOBA = θ, 0 < θ < π/2.
Now, AB = 2r cos θ and AD = 2r sin θ.
Perimeter of the rectangle, p = 2(AB + CD) = 2(2r cos θ + 2r sin θ)
= 4r (cos θ + sin θ)
Therefore, dp/dθ = 4r (– sin θ + cos θ)
and d2p/dθ2 = 4r (– cos θ – sin θ) = – 4r(cos θ + sin θ).
Now, dp/dθ = 0 => 4r (– sin θ + cos θ) = 0
Or, tan θ = 1 => θ = π/4. [As, 0 < θ < π/2]
Also [d2p/dθ2]θ = π/4 = – 4r ( sin π/4 + cos π/4 )
= – 4r (1/√2 + 1/√2 ) = – 4r.2/√2 = – 4√2 r < 0.
Therefore, p is maximum when θ = π/4.
That is when BC = 2r sin π/4 = 2r. 1/√2 = √2 r and AB = 2r cos π/4 = 2r. 1/√2 = √2 r.
AB and BC are adjacent sides, hence ABCD is a square.
Hence, perimeter of ABCD is maximum when it is a square. [Proved.]

Paper By Mr. M.P.Keshari
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