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

Maxima & Minima.

Q. 1. An open box, with a square base, is to be made out of a given quantity of metal sheet of area C². Show that the maximum volume of the box is C³/6√3.

Solution :

Let the side of the square base be x and height of the open box be y. Then
C² = Surface area of the open box
= x² + 4 xy
Or, y = (C² – x²)/4x.
Now, Volume of the box, V = x².y = x². (C² – x²)/4x
Or, V = 1/4 (C²x – x²) --------------- (1)
Therefore, dV/dx = 1/4 (C² – 3x²) and d2V/dx2 = – 3/2 x ---------- (2)
For V to be maximum or minimum, dV/dx = 0
Or, C² – 3x² = 0 => x = ± C/√3.
As x is not negative being length, so x = C/√3.
Therefore, d2V/dx] x = C/√3 = – 3/2. C/√3 – (√3/2)C < 0 [ As, C > 0]
Therefore, V is maximum.
Maximum value of V = 1/4[C². C/√3 – (C/√3)3] = C3/6√3. [Proved.]

Q.2. A window is in the form of a rectangle surmounted by a semi-circle. If the total perimeter of the window is 30 m, find the dimensions of the window so that maximum light is admitted.

Solution :

Fig

Let AB = 2x and BC = y, then radius of the semicircle = x.
As per question, 2x + y + πx + y = 30
Or, y = 15 – {(π + 2)/2}x ------------------- (1)
Let area of the window be A, then
A = 2x.y + 1/2 πx²
= 2x[15 – {(π + 2)/2}x]
= 30x – (π + 2)x² + 1/2 π x²
= 30x – (π/2 + 2) x².
Therefore, dA/dx = 30 – (π + 4) x and d²A/dx² = – (π + 4) ---------- (2)
For A to be maximum or minimum dA/dx = 0 => x = 30/(π + 4).
[d²A/dx²]x = 30/(π + 4) < 0
Therefore, maximum light will come through the window.
Now, AB = 2x = 60/(π + 4)m ; r = x = 30/(π + 4) m.
BC = y = 15 – {(π + 2)/2}x from (1)
= 15 – {(π + 2)/2}{30/(π + 4)}
= 15(π + 4 – π – 2)/(π + 4)
= 30/(π + 4) m.
Hence, AB = 30/(π + 4) m and BC = 30/(π + 4) m. [Ans.]

Paper By Mr. M.P.Keshari
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