The value of k which minimizes F(k)= ∫_0 ^4 ∣x(4−x)−k∣dx = ?


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 46624 by rahul 19 last updated on 29/Oct/18

$T h e v a l u e o f k w h i c h m i n i m i z e s$ $F (k) = \int_{0}^{4} ∣ x (4 - x) - k ∣ d x = ?$
Commented by rahul 19 last updated on 29/Oct/18

$A n s . i s 3 .$
Commented by ajfour last updated on 30/Oct/18

Commented by ajfour last updated on 30/Oct/18
$F(k) = ∫_0 ^( 4) ∣4−k−(x−2)^2 ∣dx let 4−k−(α−2)^2 =0 ...(i) ⇒ −1−2(α−2)(dα/dk) = 0 ...(ii) = ∫_0 ^( 4) {(x−2)^2 −(4−k)}dx +4∫_α ^( 2) {(4−k−(x−2)^2 }dx F (k)= c−4(4−k) +4(4−k)(2−α)+(4/3)(α−2)^3 F ′(k)= 4−4(2−α)−4(4−k)(dα/dk) +4(α−2)^2 (dα/dk) When F(k) is minimum F ′(k)=0 ⇒ 1−(2−α)= (dα/dk)[4−k−(α−2)^2 ]=0 using (i) & (ii) 1−(2−α) = 0 ⇒ α = 1 hence from (i) 4−k−(1−2)^2 = 0 or k = 3 .$
$F (k) = \int_{0}^{4} ∣ 4 - k - {(x - 2)}^{2} ∣ d x$ $l e t 4 - k - {(α - 2)}^{2} = 0 . . . (i)$ $\Rightarrow - 1 - 2 (α - 2) \frac{d α}{d k} = 0 . . . (i i)$ $= \int_{0}^{4} {{(x - 2)}^{2} - (4 - k)} d x$ $+ 4 \int_{α}^{2} {(4 - k - {(x - 2)}^{2}} d x$ $F (k) = c - 4 (4 - k)$ $+ 4 (4 - k) (2 - α) + \frac{4}{3} {(α - 2)}^{3}$ $F^{'} (k) = 4 - 4 (2 - α) - 4 (4 - k) \frac{d α}{d k}$ $+ 4 {(α - 2)}^{2} \frac{d α}{d k}$ $W h e n F (k) i s m i n i m u m F^{'} (k) = 0$ $\Rightarrow 1 - (2 - α) = \frac{d α}{d k} [4 - k - {(α - 2)}^{2}] = 0$ $u s i n g (i) & (i i)$ $1 - (2 - α) = 0$ $\Rightarrow α = 1$ $h e n c e f r o m (i)$ $4 - k - {(1 - 2)}^{2} = 0$ $o r k = 3 .$
Commented by rahul 19 last updated on 31/Oct/18
thank you sir!
	Answered by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18

	$x (4 - x) - k$ $4 x - x^{2} - k$ $- x^{2} + 4 x - 4 + 4 - k$ $- (x^{2} - 4 x + 4) + 4 - k$ $4 - k - {(x - 2)}^{2}$ $a s t h e v a l u e o f x i n c r e a s e s s o t h e v a l u e o f$ $4 - k - {(x - 2)}^{2} d e c r e a s e s h e n c e i t h a s n o m i n i m u m$ $v a l u e$ $w h e n x = 2$ $4 - k - {(x - 2)}^{2} t h e m a x i m u m v a l u e o f$ $4 - {(x - 2)}^{2} i s 4 - k$ $\int_{0}^{2} ∣ 4 x - x^{2} - k ∣ d x + \int_{2}^{4} ∣ 4 x - x^{2} - k ∣ d x$ $w a i t p l s . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum