Question Number 46624 by rahul 19 last updated on 29/Oct/18
$${The}\:{value}\:{of}\:{k}\:{which}\:{minimizes} \\ $$$${F}\left({k}\right)=\:\int_{\mathrm{0}} ^{\mathrm{4}} \mid{x}\left(\mathrm{4}−{x}\right)−{k}\mid{dx}\:=\:? \\ $$
Commented by rahul 19 last updated on 29/Oct/18
$${Ans}.\:{is}\:\mathrm{3}. \\ $$
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 .](https://www.tinkutara.com/question/Q46690.png)
$${F}\left({k}\right)\:=\:\int_{\mathrm{0}} ^{\:\:\mathrm{4}} \mid\mathrm{4}−{k}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} \mid{dx} \\ $$$$\:{let}\:\:\mathrm{4}−{k}−\left(\alpha−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0}\:\:\:\:\:…\left({i}\right) \\ $$$$\Rightarrow\:\:\:−\mathrm{1}−\mathrm{2}\left(\alpha−\mathrm{2}\right)\frac{{d}\alpha}{{dk}}\:=\:\mathrm{0}\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\:\mathrm{4}} \left\{\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{4}−{k}\right)\right\}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:+\mathrm{4}\int_{\alpha} ^{\:\:\mathrm{2}} \left\{\left(\mathrm{4}−{k}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} \right\}{dx}\right. \\ $$$${F}\:\left({k}\right)=\:{c}−\mathrm{4}\left(\mathrm{4}−{k}\right) \\ $$$$\:\:\:\:\:\:\:\:\:+\mathrm{4}\left(\mathrm{4}−{k}\right)\left(\mathrm{2}−\alpha\right)+\frac{\mathrm{4}}{\mathrm{3}}\left(\alpha−\mathrm{2}\right)^{\mathrm{3}} \\ $$$${F}\:'\left({k}\right)=\:\mathrm{4}−\mathrm{4}\left(\mathrm{2}−\alpha\right)−\mathrm{4}\left(\mathrm{4}−{k}\right)\frac{{d}\alpha}{{dk}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}\left(\alpha−\mathrm{2}\right)^{\mathrm{2}} \:\frac{{d}\alpha}{{dk}}\:\: \\ $$$${When}\:{F}\left({k}\right)\:{is}\:{minimum}\:{F}\:'\left({k}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{1}−\left(\mathrm{2}−\alpha\right)=\:\frac{{d}\alpha}{{dk}}\left[\mathrm{4}−{k}−\left(\alpha−\mathrm{2}\right)^{\mathrm{2}} \right]=\mathrm{0} \\ $$$${using}\:\left({i}\right)\:\&\:\left({ii}\right) \\ $$$$\:\:\mathrm{1}−\left(\mathrm{2}−\alpha\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:\alpha\:=\:\mathrm{1} \\ $$$${hence}\:{from}\:\left({i}\right) \\ $$$$\:\:\:\:\:\mathrm{4}−{k}−\left(\mathrm{1}−\mathrm{2}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$${or}\:\:\:\:\:\boldsymbol{{k}}\:=\:\mathrm{3}\:. \\ $$$$ \\ $$
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
$${x}\left(\mathrm{4}−{x}\right)−{k} \\ $$$$\mathrm{4}{x}−{x}^{\mathrm{2}} −{k} \\ $$$$−{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{4}+\mathrm{4}−{k} \\ $$$$−\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}\right)+\mathrm{4}−{k} \\ $$$$\mathrm{4}−{k}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$${as}\:{the}\:{value}\:{of}\:{x}\:{increases}\:\:{so}\:{the}\:{value}\:{of}\: \\ $$$$\mathrm{4}−{k}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:{decreases}\:{hence}\:{it}\:{has}\:{no}\:{minimum} \\ $$$${value} \\ $$$${when}\:{x}=\mathrm{2}\:\: \\ $$$$\mathrm{4}−{k}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:\:\:\:{the}\:{maximum}\:{value}\:{of} \\ $$$$\mathrm{4}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:\:{is}\:\mathrm{4}−{k} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \mid\mathrm{4}{x}−{x}^{\mathrm{2}} −{k}\mid{dx}+\int_{\mathrm{2}} ^{\mathrm{4}} \mid\mathrm{4}{x}−{x}^{\mathrm{2}} −{k}\mid{dx} \\ $$$${wait}\:{pls}… \\ $$$$ \\ $$