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

Maxima & Minima.

Q.10. Show that the rectangle of maximum area that can be inscribed in a circle is a square.

Solution :

Do yourself.
[Area = A = AB×AD = (2r cos θ)(2r sin θ) = 4r²cos θ sin θ = 2r²sin 2θ.]

Q.11. A window is in the form of a rectangle surmounted by a semi-circular opening. If the perimeter of the window is 20 m, find the dimensions of the window so that the maximum possible light may be admitted through the whole opening.

Solution :

Do yourself. [Ans. = 20/(4 + π).]

Q.12. Show that the height of a cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3.

Or,

Prove that the height of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3. Also find the maximum volume.
Solution :


Fig

Let height of the and diameter of the cylinder be h and x respectively, then
h² + x² = (2R)² => x² = 4R² – h². -------------- (i)
radius of the cylinder = x/2.
Volume of the cylinder, V = π (x/2)².h = π/4 x2h
= π/4 (4R² – h2)h [By (i)]
= π/4 (4R²h – h³).
Therefore, dV/dh = π/4 (4R² .1 – 3h²) and d²V/dh² = π/4 (0 – 6h) = – 3/2 πh.
Now, dV/dh = 0 => π/4 (4R²– 3h²) = 0 => 3h² = 4R² => h = 2R/√3.
When h = 2R/√3, d2V/dh² = – 3/2π.2/√3R < 0.
Thus V is maximum when h = 2R/√3. [Proved.]

Paper By Mr. M.P.Keshari
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