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

Maxima & Minima.

Q.13. Find the altitude of a right circular cone of maximum curved surface which can be inscribed in a sphere of radius r.

Solution :

Fig

Let radius and height of the inscribed right circular cone be x and h respectively. By Pythagoras theorem, OM2 + MC2 = OC2
Or,(h – r)² + x² = r² [As, OA = r]
Or, x² = r² – (h – r)² = 2hr – h². ----------------- (i)
Let curved surface area of the cone be S, then
S = π.x l = π x √(h2 + x2)
Or, S² = π2 x2 (h² + x²) = π2 (2hr – h2) (h2 + 2hr – h2)
= π2 (2hr – h2) .2hr
= 2π2 r (2h2 r – h3).
As, S > 0, therefore, S is maximum if and only if S2 is maximum. So, we need to find the value of h for which S2 is maximum. Writing S2 as f(h).
Or, f(h) = 2π² r (2h²r – h³), 0 < h < 2r.
Therefore, f’(h) = 2π² r(4hr – 3h²) and f”(h) = 2π² r(4r – 6h).
Now, f’(h) = 0 => 2π² r(4hr – 3h²) = 0 => 4hr – 3h² = 0 => h = 4r/3. [0 < h < 2r]
Also, f”(4r/3) = 2π2 r (4r – 6.4r/3) = – 8 π2 r2 < 0.
Therefore, f(h) is maximum when h = 4r/3 => S2 is maximum when h = 4r/3 =>
S is maximum when h = 4r/3.
Hence, curved surface area of the cone will be maximum when its altitude is 4r/3. [Proved.]

Q.14. A wire of length 20 m is available to fence off a flower bed in the form of a sector of a circle. What must be the radius of the circle, if we wish to have a flower bed with the greatest possible area?

Solution :

Fig.

As per question we have, 2r + l = 20 => l = 20 – 2r,
A = 1/2 rl = 1/2 r(20 – 2r) = 10r – r2.
Therefore, dA/dr = 10 – 2r = 0
Or, r = 5.
d2A/dr2 = – 2 < 0
Therefore, A is maximum when r = 5 m. [Ans.]

Paper By Mr. M.P.Keshari
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