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The-value-of-k-which-minimizes-F-k-0-4-x-4-x-k-dx-

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Question Number 46624 by rahul 19 last updated on 29/Oct/18
The value of k which minimizes F(k)= ∫_0 ^4 ∣x(4−x)−k∣dx = ?
ThevalueofkwhichminimizesF(k)=∫04∣x(4−x)−k∣dx=?
Commented by rahul 19 last updated on 29/Oct/18
Ans. is 3.
Ans.is3.
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)=∫04∣4−k−(x−2)2∣dxlet4−k−(α−2)2=0…(i)⇒−1−2(α−2)dαdk=0…(ii)=∫04{(x−2)2−(4−k)}dx+4∫α2{(4−k−(x−2)2}dxF(k)=c−4(4−k)+4(4−k)(2−α)+43(α−2)3F′(k)=4−4(2−α)−4(4−k)dαdk+4(α−2)2dαdkWhenF(k)isminimumF′(k)=0⇒1−(2−α)=dαdk[4−k−(α−2)2]=0using(i)&(ii)1−(2−α)=0⇒α=1hencefrom(i)4−k−(1−2)2=0ork=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 4x−x^2 −k −x^2 +4x−4+4−k −(x^2 −4x+4)+4−k 4−k−(x−2)^2 as the value of x increases so the value of 4−k−(x−2)^2 decreases hence it has no minimum value when x=2 4−k−(x−2)^2 the maximum value of 4−(x−2)^2 is 4−k ∫_0 ^2 ∣4x−x^2 −k∣dx+∫_2 ^4 ∣4x−x^2 −k∣dx wait pls...
x(4−x)−k4x−x2−k−x2+4x−4+4−k−(x2−4x+4)+4−k4−k−(x−2)2asthevalueofxincreasessothevalueof4−k−(x−2)2decreaseshenceithasnominimumvaluewhenx=24−k−(x−2)2themaximumvalueof4−(x−2)2is4−k∫02∣4x−x2−k∣dx+∫24∣4x−x2−k∣dxwaitpls…

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