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

Maxima & Minima.

Q.19. Prove that the area of right-angled triangle of a given hypotenuse is maximum when the triangle is isosceles.

Solution :

Fig.

Let ABC is a right angled triangle right angled at B , such that LB = 90º , LACB = θ and AC = x cm.
Therefore , BC = x cos θ , AB = x sin θ .
Area of Δ ABC , A = 1/2 AB × BC
Or ,A = 1/2 x sin θ . x cos θ = 1/2 x² sin θ cos θ
Therefore , dA/dθ = 1/2 x².( cos²θ – sin²θ) = 0 [For maxima or minima, as 1/2 x² ≠ 0]. Or, cos²θ = sin²θ => tan²θ = 1 => tanθ = ± 1.
Also , d²A/dθ²= 1/2 x2 [– 2 cosθ .sinθ – 2 sinθ .cos θ] = – 2 x2 sinθ.cosθ < 0 at tanθ = 1. Hence , area is maximum when tan θ = 1 i.e. θ = π/4 = 45º.
In ΔABC if LACB = 45º , LB = 90º , LA 45º.
Therefore Δ ABC is an isosceles for maximum area. [Proved.]

Q.20. A closed circular cylinder has a volume of 2156 c.c. What will be the radius of its base so that its total surface area is minimum. Find the height of the cylinder when its total surface area is minimum.

Solution :

Let radius of the base be r, and height h , then
Volume = π r² h = 2156 c. c. -------------------------- (1)
Total surface area = A = 2πr² + 2πrh
From (1) h = 2156/πr²
A = 2πr² + 2πr (2156/πr²) = 2πr²+ 431² r –1
Then dA/dr = 4πr – 4312/r2 = 0 [For minimum]
Or, 4πr = 4312/r²
Or,r³ = 1078/π = 1078 ×7/2² = 343.
Or, r = 7 cm.
Therefore , πr²h = 2156
Or, h = 2156/πr² = 2156cm3/[(22/7) × (7cm)2] = 14 cm.
Hence , height of the cylinder = 14 cm. [Ans.]

Or

Show that the height of the closed right circular cylinder, of given volume and minimum total surface area, is equal to its diameter.

Paper By Mr. M.P.Keshari
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