Let A(x, y) be the position of airplane and man is at B(3,4)
Distance between man and airplane
AB =
√(x−3)2+(y−4)2 ⇒ AB =
√(x−2)2+(x2+4−4)2 [Since y =
x2 + 4]
⇒
(AB)2 =
x =
(x−3)2+x4 w =
(x−3)2+x4 Distance AB will be maximum or minimum according as 'w' is maximum or minimum
Now, w =
(x−3)2+x4 = 2(𝑥 − 3) +
4𝑥3 ,
=
12x2+2 Critical points of 'w' are given by
= 0
2(𝑥 − 3) +
4𝑥3 = 0
2𝑥3 + 𝑥 − 3 = 0
⇒ (𝑥 − 1)(
2𝑥2 + 2𝑥 + 3) = 0
𝑥 = 1
[
2𝑥2 + 2𝑥 + 3 = 0 gives imaginary value of 𝑥]
𝑁𝑜𝑤,⋅
()x=1 =
12x2 + 2 = 14 > 0
So, x = 1, will give minimum value of 'w'
Hence, distance between man and airplane AB will also be minimum when x = 1
(AB)min =
√(x−3)2+x4 at x = 1
=
√(1−3)2+14 =
√5