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Question Number 184535 by CrispyXYZ last updated on 08/Jan/23
If a, b>0 such that 2a+b=2, then find the minimum value of: 1) (4a^2 +1)(b^2 +1) 2) ((2a^2 −b+4)/(a+1))+((b^2 −2a−2)/(b+4))
$$\mathrm{If}\:{a},\:{b}>\mathrm{0}\:\mathrm{such}\:\mathrm{that}\:\mathrm{2}{a}+{b}=\mathrm{2}, \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}: \\ $$$$\left.\mathrm{1}\right)\:\:\left(\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:\:\frac{\mathrm{2}{a}^{\mathrm{2}} −{b}+\mathrm{4}}{{a}+\mathrm{1}}+\frac{{b}^{\mathrm{2}} −\mathrm{2}{a}−\mathrm{2}}{{b}+\mathrm{4}} \\ $$
Answered by cortano1 last updated on 08/Jan/23
(1) { ((b=2−2a)),((f(a)=(4a^2 +1)(4a^2 −8a+5))) :} f(a)=16a^4 −32a^3 +24a^2 −8a+5 f ′(a)=64a^3 −96a^2 +48a−8=0 ⇒a=(1/2) ; b=2−1=1 min {(4a^2 +1)(b^2 +1)}= 2×2=4
$$\left(\mathrm{1}\right)\:\begin{cases}{{b}=\mathrm{2}−\mathrm{2}{a}}\\{{f}\left({a}\right)=\left(\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{4}{a}^{\mathrm{2}} −\mathrm{8}{a}+\mathrm{5}\right)}\end{cases} \\ $$$$\:{f}\left({a}\right)=\mathrm{16}{a}^{\mathrm{4}} −\mathrm{32}{a}^{\mathrm{3}} +\mathrm{24}{a}^{\mathrm{2}} −\mathrm{8}{a}+\mathrm{5} \\ $$$$\:{f}\:'\left({a}\right)=\mathrm{64}{a}^{\mathrm{3}} −\mathrm{96}{a}^{\mathrm{2}} +\mathrm{48}{a}−\mathrm{8}=\mathrm{0} \\ $$$$\:\:\:\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}}\:;\:{b}=\mathrm{2}−\mathrm{1}=\mathrm{1} \\ $$$$\:{min}\:\left\{\left(\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}} +\mathrm{1}\right)\right\}=\:\mathrm{2}×\mathrm{2}=\mathrm{4} \\ $$
Answered by mr W last updated on 08/Jan/23
(1) a+(b/2)=1 a=cos^2 θ b=2 sin^2 θ (4 cos^4 θ+1)(4 sin^4 θ+1) =1+4(cos^4 θ+sin^4 θ)+16 (cos θ sin θ)^4 =1+4−2(2 cos θ sin θ)^2 +(2 cos θ sin θ)^4 =5−2 sin^2 2θ+ sin^4 2θ =4+(1−sin^2 2θ)^2 =4+cos^4 2θ ≥ 4=minimum
$$\left(\mathrm{1}\right) \\ $$$${a}+\frac{{b}}{\mathrm{2}}=\mathrm{1} \\ $$$${a}=\mathrm{cos}^{\mathrm{2}} \:\theta \\ $$$${b}=\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\theta \\ $$$$\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{4}} \:\theta+\mathrm{1}\right)\left(\mathrm{4}\:\mathrm{sin}^{\mathrm{4}} \:\theta+\mathrm{1}\right) \\ $$$$=\mathrm{1}+\mathrm{4}\left(\mathrm{cos}^{\mathrm{4}} \:\theta+\mathrm{sin}^{\mathrm{4}} \:\theta\right)+\mathrm{16}\:\left(\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\right)^{\mathrm{4}} \\ $$$$=\mathrm{1}+\mathrm{4}−\mathrm{2}\left(\mathrm{2}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} +\left(\mathrm{2}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\right)^{\mathrm{4}} \\ $$$$=\mathrm{5}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta+\:\mathrm{sin}^{\mathrm{4}} \:\mathrm{2}\theta \\ $$$$=\mathrm{4}+\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta\right)^{\mathrm{2}} \\ $$$$=\mathrm{4}+\mathrm{cos}^{\mathrm{4}} \:\mathrm{2}\theta\:\geqslant\:\mathrm{4}={minimum} \\ $$
Answered by greougoury555 last updated on 08/Jan/23
(1) f(a,b,λ)=(4a^2 +1)(b^2 +1)+λ(2a+b−2) (∂f/∂a) = 8a(b^2 +1)+2λ=0 (∂f/∂b) = 2b(4a^2 +1)+λ=0 (∂f/∂λ) = 2a+b−2=0⇒b=2−2a ⇒ −4a(b^2 +1)=−2b(4a^2 +1) ⇒a(4a^2 −8a+5)=(1−a)(4a^2 +1) ⇒4a^3 −8a^2 +5a=−4a^3 +4a^2 −a+1 ⇒8a^3 −12a^2 +6a−1=0 ⇒(2a−1)^3 =0 ; a=(1/2) ∴ minimum (4a^2 +1)(b^2 +1)= 4
$$\left(\mathrm{1}\right)\:{f}\left({a},{b},\lambda\right)=\left(\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}} +\mathrm{1}\right)+\lambda\left(\mathrm{2}{a}+{b}−\mathrm{2}\right) \\ $$$$\:\frac{\partial{f}}{\partial{a}}\:=\:\mathrm{8}{a}\left({b}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{2}\lambda=\mathrm{0} \\ $$$$\:\frac{\partial{f}}{\partial{b}}\:=\:\mathrm{2}{b}\left(\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}\right)+\lambda=\mathrm{0} \\ $$$$\:\frac{\partial{f}}{\partial\lambda}\:=\:\mathrm{2}{a}+{b}−\mathrm{2}=\mathrm{0}\Rightarrow{b}=\mathrm{2}−\mathrm{2}{a} \\ $$$$\Rightarrow\:−\mathrm{4}{a}\left({b}^{\mathrm{2}} +\mathrm{1}\right)=−\mathrm{2}{b}\left(\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\Rightarrow{a}\left(\mathrm{4}{a}^{\mathrm{2}} −\mathrm{8}{a}+\mathrm{5}\right)=\left(\mathrm{1}−{a}\right)\left(\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{4}{a}^{\mathrm{3}} −\mathrm{8}{a}^{\mathrm{2}} +\mathrm{5}{a}=−\mathrm{4}{a}^{\mathrm{3}} +\mathrm{4}{a}^{\mathrm{2}} −{a}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{8}{a}^{\mathrm{3}} −\mathrm{12}{a}^{\mathrm{2}} +\mathrm{6}{a}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}{a}−\mathrm{1}\right)^{\mathrm{3}} =\mathrm{0}\:;\:{a}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\therefore\:{minimum}\:\left(\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}} +\mathrm{1}\right)=\:\mathrm{4} \\ $$
Answered by ajfour last updated on 08/Jan/23
1) 2a=t t+b=2 (t^2 +1)(b^2 +1) is minimum for t=b=1 , so the minimum is 2×2=4 2) f(t,b)=((t^2 −2b+8)/(t+2))+((b^2 −t−2)/(b+4)) let t+2=p b+4=q p+q=8 f(p,q)=((p^2 −4p−2q+20)/p)+((q^2 −8q−p+16)/q) =p−4+((20)/p)−((2q)/p)+q−8−(p/q)+((16)/q) =−4+((20)/p)+((16)/q)−((2q)/p)−(p/q) =−4+((2(10−q))/p)+(((16−p))/q) =−4+((2(2+p))/p)+(((8+q))/q) =−1+(4/p)+(8/q) f(p)=−1+(4/p)+(8/(8−p)) (df/dp)=−(4/p^2 )+(8/((8−p)^2 )) =0 ⇒ ((8−p)/p)=±(√2) p=(8/(1+(√2))) or p=−(8/(((√2)−1))) f_1 =−1+(4/p)+(8/(8−p))=((1+(√2))/2)+(1/(1−(1/(1+(√2))))) =−1+((1+(√2))/2)+((1+(√2))/( (√2)))=(1/2)+(√2) f_2 =−1−((((√2)−1))/2)+(1/(1+(1/( (√2)−1)))) =−1+((1−(√2)+2−(√2))/2)=(1/2)−(√2)
$$\left.\mathrm{1}\right) \\ $$$$\mathrm{2}{a}={t} \\ $$$${t}+{b}=\mathrm{2} \\ $$$$\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}} +\mathrm{1}\right)\:\:\:{is}\:{minimum} \\ $$$${for}\:{t}={b}=\mathrm{1}\:\:,\:{so}\:\: \\ $$$${the}\:{minimum}\:{is}\:\mathrm{2}×\mathrm{2}=\mathrm{4} \\ $$$$\left.\mathrm{2}\right) \\ $$$${f}\left({t},{b}\right)=\frac{{t}^{\mathrm{2}} −\mathrm{2}{b}+\mathrm{8}}{{t}+\mathrm{2}}+\frac{{b}^{\mathrm{2}} −{t}−\mathrm{2}}{{b}+\mathrm{4}} \\ $$$${let}\:\:{t}+\mathrm{2}={p}\:\:\:\:\:{b}+\mathrm{4}={q} \\ $$$${p}+{q}=\mathrm{8} \\ $$$${f}\left({p},{q}\right)=\frac{{p}^{\mathrm{2}} −\mathrm{4}{p}−\mathrm{2}{q}+\mathrm{20}}{{p}}+\frac{{q}^{\mathrm{2}} −\mathrm{8}{q}−{p}+\mathrm{16}}{{q}} \\ $$$$\:\:={p}−\mathrm{4}+\frac{\mathrm{20}}{{p}}−\frac{\mathrm{2}{q}}{{p}}+{q}−\mathrm{8}−\frac{{p}}{{q}}+\frac{\mathrm{16}}{{q}} \\ $$$$\:\:=−\mathrm{4}+\frac{\mathrm{20}}{{p}}+\frac{\mathrm{16}}{{q}}−\frac{\mathrm{2}{q}}{{p}}−\frac{{p}}{{q}} \\ $$$$\:\:=−\mathrm{4}+\frac{\mathrm{2}\left(\mathrm{10}−{q}\right)}{{p}}+\frac{\left(\mathrm{16}−{p}\right)}{{q}} \\ $$$$\:\:=−\mathrm{4}+\frac{\mathrm{2}\left(\mathrm{2}+{p}\right)}{{p}}+\frac{\left(\mathrm{8}+{q}\right)}{{q}} \\ $$$$\:\:=−\mathrm{1}+\frac{\mathrm{4}}{{p}}+\frac{\mathrm{8}}{{q}} \\ $$$${f}\left({p}\right)=−\mathrm{1}+\frac{\mathrm{4}}{{p}}+\frac{\mathrm{8}}{\mathrm{8}−{p}} \\ $$$$\frac{{df}}{{dp}}=−\frac{\mathrm{4}}{{p}^{\mathrm{2}} }+\frac{\mathrm{8}}{\left(\mathrm{8}−{p}\right)^{\mathrm{2}} }\:=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{8}−{p}}{{p}}=\pm\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:{p}=\frac{\mathrm{8}}{\mathrm{1}+\sqrt{\mathrm{2}}}\:\:\:{or}\:\:\:{p}=−\frac{\mathrm{8}}{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)} \\ $$$${f}_{\mathrm{1}} =−\mathrm{1}+\frac{\mathrm{4}}{{p}}+\frac{\mathrm{8}}{\mathrm{8}−{p}}=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}}}} \\ $$$$\:\:\:\:=−\mathrm{1}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{2}} \\ $$$$\:\:{f}_{\mathrm{2}} =−\mathrm{1}−\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}} \\ $$$$\:\:\:\:\:=−\mathrm{1}+\frac{\mathrm{1}−\sqrt{\mathrm{2}}+\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}−\sqrt{\mathrm{2}} \\ $$$$ \\ $$
Commented by mr W last updated on 08/Jan/23
please recheck here: f(t,b)=((t^2 /2−2b+8)/(t+2))+((b^2 −t−2)/(b+4)) i think for (2) there is no nice looking solution.
$${please}\:{recheck}\:{here}: \\ $$$${f}\left({t},{b}\right)=\frac{{t}^{\mathrm{2}} /\mathrm{2}−\mathrm{2}{b}+\mathrm{8}}{{t}+\mathrm{2}}+\frac{{b}^{\mathrm{2}} −{t}−\mathrm{2}}{{b}+\mathrm{4}} \\ $$$${i}\:{think}\:{for}\:\left(\mathrm{2}\right)\:{there}\:{is}\:{no}\:{nice}\:{looking} \\ $$$${solution}. \\ $$
Commented by ajfour last updated on 08/Jan/23
no sir f(t,b)=((t^2 −2b+8)/(t+2)) =((4a^2 −2b+8)/(2a+2))
$${no}\:\:{sir}\:\:\:\:{f}\left({t},{b}\right)=\frac{{t}^{\mathrm{2}} −\mathrm{2}{b}+\mathrm{8}}{{t}+\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{2}{b}+\mathrm{8}}{\mathrm{2}{a}+\mathrm{2}} \\ $$
Commented by mr W last updated on 08/Jan/23
yes, you are right.
$${yes},\:{you}\:{are}\:{right}. \\ $$
Answered by Frix last updated on 08/Jan/23
2) b=2−2a ⇒ −((a^2 +7)/((a+1)(a−3))) ((d[−((a^2 +7)/((a+1)(a−3)))])/da)=((2(a^2 +10a−7))/((a+1)^2 (a−3)^2 ))=0 ⇒ a=−5±4(√2) local max=(1/2)−(√2) at a=−5−4(√2) local min=(1/2)+(√2) at a=−5+4(√2)
$$\left.\mathrm{2}\right) \\ $$$${b}=\mathrm{2}−\mathrm{2}{a} \\ $$$$\Rightarrow \\ $$$$−\frac{{a}^{\mathrm{2}} +\mathrm{7}}{\left({a}+\mathrm{1}\right)\left({a}−\mathrm{3}\right)} \\ $$$$\frac{{d}\left[−\frac{{a}^{\mathrm{2}} +\mathrm{7}}{\left({a}+\mathrm{1}\right)\left({a}−\mathrm{3}\right)}\right]}{{da}}=\frac{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{10}{a}−\mathrm{7}\right)}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} \left({a}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${a}=−\mathrm{5}\pm\mathrm{4}\sqrt{\mathrm{2}} \\ $$$$\mathrm{local}\:\mathrm{max}=\frac{\mathrm{1}}{\mathrm{2}}−\sqrt{\mathrm{2}}\:\mathrm{at}\:{a}=−\mathrm{5}−\mathrm{4}\sqrt{\mathrm{2}} \\ $$$$\mathrm{local}\:\mathrm{min}=\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{2}}\:\mathrm{at}\:{a}=−\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}} \\ $$
Answered by mr W last updated on 08/Jan/23
(2) 2a+b=2 ⇒0<a<1, 0<b<2 ((2a^2 +2a−2+4)/(a+1))+((b^2 +b−2−2)/(b+4)) =((2(a^2 +a+1))/(a+1))+((b^2 +b−4)/(b+4)) =(((2a)^2 )/(2a+2))+((b^2 −8)/(b+4))+3 =(((2−b)^2 )/(4−b))+((b^2 −8)/(4+b))+3 =((4(b^2 −b−4))/(16−b^2 ))+3 =((4(12−b))/(16−b^2 ))−1 =4((1/(4−b))+(2/(4+b))_(f(b)) )−1 f′(b)=(1/((4−b)^2 ))−(2/((4+b)^2 ))=0 (4+b)^2 −2(4−b)^2 =0 (b−(√2)b+4+4(√2))(b+(√2)b+4−4(√2))=0 ⇒b=((4((√2)−1))/( (√2)+1))=4(3−2(√2)) or ⇒b=((4((√2)+1))/( (√2)−1))=4(3+2(√2)) (>2, rejected) at b=4(3−2(√2)) minimum=(√2)+(1/2)
$$\left(\mathrm{2}\right) \\ $$$$\mathrm{2}{a}+{b}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{0}<{a}<\mathrm{1},\:\mathrm{0}<{b}<\mathrm{2} \\ $$$$ \\ $$$$\frac{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{2}+\mathrm{4}}{{a}+\mathrm{1}}+\frac{{b}^{\mathrm{2}} +{b}−\mathrm{2}−\mathrm{2}}{{b}+\mathrm{4}} \\ $$$$=\frac{\mathrm{2}\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)}{{a}+\mathrm{1}}+\frac{{b}^{\mathrm{2}} +{b}−\mathrm{4}}{{b}+\mathrm{4}} \\ $$$$=\frac{\left(\mathrm{2}{a}\right)^{\mathrm{2}} }{\mathrm{2}{a}+\mathrm{2}}+\frac{{b}^{\mathrm{2}} −\mathrm{8}}{{b}+\mathrm{4}}+\mathrm{3} \\ $$$$=\frac{\left(\mathrm{2}−{b}\right)^{\mathrm{2}} }{\mathrm{4}−{b}}+\frac{{b}^{\mathrm{2}} −\mathrm{8}}{\mathrm{4}+{b}}+\mathrm{3} \\ $$$$=\frac{\mathrm{4}\left({b}^{\mathrm{2}} −{b}−\mathrm{4}\right)}{\mathrm{16}−{b}^{\mathrm{2}} }+\mathrm{3} \\ $$$$=\frac{\mathrm{4}\left(\mathrm{12}−{b}\right)}{\mathrm{16}−{b}^{\mathrm{2}} }−\mathrm{1} \\ $$$$=\mathrm{4}\left(\underset{{f}\left({b}\right)} {\frac{\mathrm{1}}{\mathrm{4}−{b}}+\frac{\mathrm{2}}{\mathrm{4}+{b}}}\right)−\mathrm{1} \\ $$$${f}'\left({b}\right)=\frac{\mathrm{1}}{\left(\mathrm{4}−{b}\right)^{\mathrm{2}} }−\frac{\mathrm{2}}{\left(\mathrm{4}+{b}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\left(\mathrm{4}+{b}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{4}−{b}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({b}−\sqrt{\mathrm{2}}{b}+\mathrm{4}+\mathrm{4}\sqrt{\mathrm{2}}\right)\left({b}+\sqrt{\mathrm{2}}{b}+\mathrm{4}−\mathrm{4}\sqrt{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow{b}=\frac{\mathrm{4}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{2}}+\mathrm{1}}=\mathrm{4}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\:{or} \\ $$$$\Rightarrow{b}=\frac{\mathrm{4}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\:\sqrt{\mathrm{2}}−\mathrm{1}}=\mathrm{4}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\:\left(>\mathrm{2},\:{rejected}\right) \\ $$$${at}\:{b}=\mathrm{4}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\:{minimum}=\sqrt{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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