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

Maxima & Minima.

Let P(x, y) be any point on the given parabola y = x2, then
| CP | = √{(x – 0)² + (y – c)²} = √{y + (y – c)²} [ writing y for x² as, y = x²]
=√{y2 – (2c – 1)y + c2}.
Or, | CP |2 = y² – (2c – 1)y + c²
Now, | CP | is the shortest if and only if | CP | 2 is the shortest.
Writing, | CP |2 as f(y), we get
f(y) = y² – (2c – 1)y + c² ------------ (i)
f’(y) = 2y – (2c – 1) and f”(y) = 2.
Now, f’(y) = 0 => 2y – (2c – 1) = 0
Or, y = (2c – 1)/2.
Hence, f”{(2c – 1)/2} = 2 > 0.
Therefore, f(y) is minimum when y = (2c – 1)/2
i.e. | CP | is minimum when y = (2c – 1)/2
and the minimum value of | CP | = √[{(2c – 1)/2} + {(2c – 1)/2 – c}²]
= √[(2c – 1)/2 + 1/4]
= √[(4c – 1)/2]. [Ans.]

Q.26. An enemy vehicle is moving along the curve y = x² + 2. Find the shortest distance between the vehicle and our artillery located at (3, 2). Also find the co-ordinates of the vehicle when the distance is shortest.

Solution :

Let A (3, 2) be the co-ordinate of artillery and P(x, y) the co-ordinate of enemy vehicle on the curve y = x² + 2, then
| AP | = √[(x – 3)² + (y – 2)²]
= √[(x – 3)² + (x² + 2 – 2)2] [Using y = x² + 2.]
= √(x4 + x² – 6x + 9).
Or, | AP |2 = x4 + x² – 6x + 9.
Now, | AP | is the shortest if and only if | AP |2 is the shortest.
Writing | AP |2 as f(x) we get,
f(x) = x4 + x²– 6x + 9.
Now, f’(x) = 4x³ + 2x – 6 = 2(x – 1)(2x² + 2x + 3), and f”(x) = 12 x² + 2.
And f’(x) = 0 => 2(x – 1)(2x² + 2x + 3) = 0 => x = 1. [for real x, 2x²+ 2x + 3 ≠ 0]
Also f”(1) = 12 .(1)² + 2 = 14 > 0.
Therefore, f(x) is minimum if x = 1. and minimum value = f(1) = 14 + 12 – 6.1 + 9 = 5. Or, minimum value of | AP |2 is 5. Then minimum value of | AP | = √5. Hence the shortest distance is √5.
Also, when x = 1, y = 12 + 2 = 3. Thus the co-ordinates of the vehicle when the distance is the shortest are (1, 3). [Ans.]

Q.27. Given the sum of the perimeters of a square and a circle, that the sum of their areas is least when the side of the square is equal to the diameter of the circle.

Solution :

Let the side of the square be a and the radius of the circle be r. Then
P = 4a + 2πr ---------------- (1)
Let the sum of their areas be A = a² + πr²
= {(P - 2πr)/4}² + πr² [Usin (1)]
dA/dr = 2{(P– 2πr)/4}{(– 2π)/4} + 2πr

= (– πP + 2π²r)/4 + 2πr
= (– πP/4) + (π2/2 + 2π)r
and d²A/dr2 = π2/2 + 2π > 0.
Therefore, for minimum area dA/dr = 0 => – πP/4 + (π2/2 + 2π)r = 0
r = (P/4)/(π/2 + 2) ---------------- (2)

A is minimum for this value of r.
From (2), P = 4r(π/2 + 2)
=> 4a + 2πr = 2πr + 8r [using (1)]
=> 4a = 8
=> a = 2r = diameter of the circle. [Proved.]

Paper By Mr. M.P.Keshari
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