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

Maxima & Minima.

Q.23. The sum of three positive numbers is 26. The second number is thrice as large as the first. If the sum of squares of these numbers is least, find the numbers.

Solution :

Let the numbers be x, y and z such that x + y + z = 26 and y = 3x.
Then x + y + z = x + 3x + z = 26=> z = 26 – 4x ,
And let S = x² + y² + z² = x² + (3x)² + (26 – 4x)² = 26 x²– 208 x + 676 .
Therefore, dS/dx = 52 x – 208 = 0 [for maxima or minima]
Or, x = 4.
And d²S/dx² = 52 > 0 => S is minimum.
Therefore , numbers are x = 4 , y = 3x = 3 × 4 = 12 , z = = 26 – (4 + 12) = 10. [Ans.]

Q.24. A box is to be constructed from a square metal sheet of side 60 cm by cutting out identical squares from the four corners and turning up the sides. Find the length of the side of the square to be cut out so that the box has maximum volume.

Solution :

Fig.

Volume of the box = V = l × b × h = (60 – 2x) × (60 – 2x) × x
= 3600 x – 240 x² + 4 x² .
dV/dx = 3600 – 480 x + 12 x² = 0 [for maxima or minima]
Or, (x – 30)(x – 10) = 0 => x = 30 or 10 .
As x = 30 is not possible , then x = 10.
Also, d²V/dx² = 24 x – 480 = 24 × 10 – 480 < 0 [at x = 10]
Therefore, V is maximum when x = 10 and maximum volume = (60 – 2x)² × x
= (60 – 20 )² × 10 = 16000 cm3. [Ans.]

Q.25. Find the shortest distance of the point C (0.c) from the parabola y = x², c > 1/2.

Solution :

Fig

Paper By Mr. M.P.Keshari
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