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If-a-b-gt-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-

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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))
Ifa,b>0suchthat2a+b=2,thenfindtheminimumvalueof:1)(4a2+1)(b2+1)2)2a2−b+4a+1+b2−2a−2b+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
(1){b=2−2af(a)=(4a2+1)(4a2−8a+5)f(a)=16a4−32a3+24a2−8a+5f′(a)=64a3−96a2+48a−8=0⇒a=12;b=2−1=1min{(4a2+1)(b2+1)}=2×2=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
(1)a+b2=1a=cos2θb=2sin2θ(4cos4θ+1)(4sin4θ+1)=1+4(cos4θ+sin4θ)+16(cosθsinθ)4=1+4−2(2cosθsinθ)2+(2cosθsinθ)4=5−2sin22θ+sin42θ=4+(1−sin22θ)2=4+cos42θ⩾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
(1)f(a,b,λ)=(4a2+1)(b2+1)+λ(2a+b−2)∂f∂a=8a(b2+1)+2λ=0∂f∂b=2b(4a2+1)+λ=0∂f∂λ=2a+b−2=0⇒b=2−2a⇒−4a(b2+1)=−2b(4a2+1)⇒a(4a2−8a+5)=(1−a)(4a2+1)⇒4a3−8a2+5a=−4a3+4a2−a+1⇒8a3−12a2+6a−1=0⇒(2a−1)3=0;a=12∴minimum(4a2+1)(b2+1)=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)
1)2a=tt+b=2(t2+1)(b2+1)isminimumfort=b=1,sotheminimumis2×2=42)f(t,b)=t2−2b+8t+2+b2−t−2b+4lett+2=pb+4=qp+q=8f(p,q)=p2−4p−2q+20p+q2−8q−p+16q=p−4+20p−2qp+q−8−pq+16q=−4+20p+16q−2qp−pq=−4+2(10−q)p+(16−p)q=−4+2(2+p)p+(8+q)q=−1+4p+8qf(p)=−1+4p+88−pdfdp=−4p2+8(8−p)2=0⇒8−pp=±2p=81+2orp=−8(2−1)f1=−1+4p+88−p=1+22+11−11+2=−1+1+22+1+22=12+2f2=−1−(2−1)2+11+12−1=−1+1−2+2−22=12−2
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.
pleaserecheckhere:f(t,b)=t2/2−2b+8t+2+b2−t−2b+4ithinkfor(2)thereisnonicelookingsolution.
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))
nosirf(t,b)=t2−2b+8t+2=4a2−2b+82a+2
Commented by mr W last updated on 08/Jan/23
yes, you are right.
yes,youareright.
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)
2)b=2−2a⇒−a2+7(a+1)(a−3)d[−a2+7(a+1)(a−3)]da=2(a2+10a−7)(a+1)2(a−3)2=0⇒a=−5±42localmax=12−2ata=−5−42localmin=12+2ata=−5+42
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)
(2)2a+b=2⇒0<a<1,0<b<22a2+2a−2+4a+1+b2+b−2−2b+4=2(a2+a+1)a+1+b2+b−4b+4=(2a)22a+2+b2−8b+4+3=(2−b)24−b+b2−84+b+3=4(b2−b−4)16−b2+3=4(12−b)16−b2−1=4(14−b+24+bf(b))−1f′(b)=1(4−b)2−2(4+b)2=0(4+b)2−2(4−b)2=0(b−2b+4+42)(b+2b+4−42)=0⇒b=4(2−1)2+1=4(3−22)or⇒b=4(2+1)2−1=4(3+22)(>2,rejected)atb=4(3−22)minimum=2+12

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