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Question Number 184535 by CrispyXYZ last updated on 08/Jan/23

$$\mathrm{If}a,b>0\mathrm{such}\mathrm{that}2a+b=2,$$ $$\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}:$$ $$1)(4a^2+1)(b^2+1)$$ $$2)\frac{2a^2-b+4}{a+1}+\frac{b^2-2a-2}{b+4}$$

Answered by cortano1 last updated on 08/Jan/23

$$\left(1\right)\{\begin{array}{l}b=2-2a\\ f\left(a\right)=(4a^2+1)(4a^2-8a+5)\end{array}$$ $$f\left(a\right)=16a^4-32a^3+24a^2-8a+5$$ $$f{}^{\prime }\left(a\right)=64a^3-96a^2+48a-8=0$$ $$\Rightarrow a=\frac{1}{2};b=2-1=1$$ $$min\left\{(4a^2+1)(b^2+1)\right\}=2\times 2=4$$

Answered by mr W last updated on 08/Jan/23

$$\left(1\right)$$ $$a+\frac{b}{2}=1$$ $$a={\cos }^2\theta$$ $$b=2{\sin }^2\theta$$ $$(4{\cos }^4\theta +1)(4{\sin }^4\theta +1)$$ $$=1+4({\cos }^4\theta +{\sin }^4\theta )+16{\left(\cos \theta \sin \theta \right)}^4$$ $$=1+4-2{\left(2\cos \theta \sin \theta \right)}^2+{\left(2\cos \theta \sin \theta \right)}^4$$ $$=5-2{\sin }^{2}2\theta +{\sin }^{4}2\theta$$ $$=4+{(1-{\sin }^{2}2\theta )}^2$$ $$=4+{\cos }^{4}2\theta ⩾4=minimum$$

Answered by greougoury555 last updated on 08/Jan/23

$$\left(1\right)f(a,b,\lambda )=(4a^2+1)(b^2+1)+\lambda (2a+b-2)$$ $$\frac{\partial f}{\partial a}=8a(b^2+1)+2\lambda =0$$ $$\frac{\partial f}{\partial b}=2b(4a^2+1)+\lambda =0$$ $$\frac{\partial f}{\partial \lambda }=2a+b-2=0\Rightarrow b=2-2a$$ $$\Rightarrow -4a(b^2+1)=-2b(4a^2+1)$$ $$\Rightarrow a(4a^2-8a+5)=(1-a)(4a^2+1)$$ $$\Rightarrow 4a^3-8a^2+5a=-4a^3+4a^2-a+1$$ $$\Rightarrow 8a^3-12a^2+6a-1=0$$ $$\Rightarrow {(2a-1)}^3=0;a=\frac{1}{2}$$ $$\therefore minimum(4a^2+1)(b^2+1)=4$$

Answered by ajfour last updated on 08/Jan/23

$$1)$$ $$2a=t$$ $$t+b=2$$ $$(t^2+1)(b^2+1)isminimum$$ $$fort=b=1,so$$ $$theminimumis2\times 2=4$$ $$2)$$ $$f(t,b)=\frac{t^2-2b+8}{t+2}+\frac{b^2-t-2}{b+4}$$ $$lett+2=pb+4=q$$ $$p+q=8$$ $$f(p,q)=\frac{p^2-4p-2q+20}{p}+\frac{q^2-8q-p+16}{q}$$ $$=p-4+\frac{20}{p}-\frac{2q}{p}+q-8-\frac{p}{q}+\frac{16}{q}$$ $$=-4+\frac{20}{p}+\frac{16}{q}-\frac{2q}{p}-\frac{p}{q}$$ $$=-4+\frac{2(10-q)}{p}+\frac{(16-p)}{q}$$ $$=-4+\frac{2(2+p)}{p}+\frac{(8+q)}{q}$$ $$=-1+\frac{4}{p}+\frac{8}{q}$$ $$f\left(p\right)=-1+\frac{4}{p}+\frac{8}{8-p}$$ $$\frac{df}{dp}=-\frac{4}{p^2}+\frac{8}{{(8-p)}^2}=0$$ $$\Rightarrow \frac{8-p}{p}=\pm \sqrt{2}$$ $$p=\frac{8}{1+\sqrt{2}}orp=-\frac{8}{(\sqrt{2}-1)}$$ $$f_1=-1+\frac{4}{p}+\frac{8}{8-p}=\frac{1+\sqrt{2}}{2}+\frac{1}{1-\frac{1}{1+\sqrt{2}}}$$ $$=-1+\frac{1+\sqrt{2}}{2}+\frac{1+\sqrt{2}}{\sqrt{2}}=\frac{1}{2}+\sqrt{2}$$ $$f_2=-1-\frac{(\sqrt{2}-1)}{2}+\frac{1}{1+\frac{1}{\sqrt{2}-1}}$$ $$=-1+\frac{1-\sqrt{2}+2-\sqrt{2}}{2}=\frac{1}{2}-\sqrt{2}$$ $$$$

Commented bymr W last updated on 08/Jan/23

$$pleaserecheckhere:$$ $$f(t,b)=\frac{t^2/2-2b+8}{t+2}+\frac{b^2-t-2}{b+4}$$ $$ithinkfor\left(2\right)thereisnonicelooking$$ $$solution.$$

Commented byajfour last updated on 08/Jan/23

$$nosirf(t,b)=\frac{t^2-2b+8}{t+2}$$ $$=\frac{4a^2-2b+8}{2a+2}$$

Commented bymr W last updated on 08/Jan/23

$$yes,youareright.$$

Answered by Frix last updated on 08/Jan/23

$$2)$$ $$b=2-2a$$ $$\Rightarrow$$ $$-\frac{a^2+7}{(a+1)(a-3)}$$ $$\frac{d[-\frac{a^2+7}{(a+1)(a-3)}]}{da}=\frac{2(a^2+10a-7)}{{(a+1)}^2{(a-3)}^2}=0$$ $$\Rightarrow$$ $$a=-5\pm 4\sqrt{2}$$ $$\mathrm{local}\max =\frac{1}{2}-\sqrt{2}\mathrm{at}a=-5-4\sqrt{2}$$ $$\mathrm{local}\min =\frac{1}{2}+\sqrt{2}\mathrm{at}a=-5+4\sqrt{2}$$

Answered by mr W last updated on 08/Jan/23

$$\left(2\right)$$ $$2a+b=2$$ $$\Rightarrow 0<a<1,0<b<2$$ $$$$ $$\frac{2a^2+2a-2+4}{a+1}+\frac{b^2+b-2-2}{b+4}$$ $$=\frac{2(a^2+a+1)}{a+1}+\frac{b^2+b-4}{b+4}$$ $$=\frac{{\left(2a\right)}^2}{2a+2}+\frac{b^2-8}{b+4}+3$$ $$=\frac{{(2-b)}^2}{4-b}+\frac{b^2-8}{4+b}+3$$ $$=\frac{4(b^2-b-4)}{16-b^2}+3$$ $$=\frac{4(12-b)}{16-b^2}-1$$ $$=4\left(\underset{f\left(b\right)}{\frac{1}{4-b}+\frac{2}{4+b}}\right)-1$$ $$f^{\prime }\left(b\right)=\frac{1}{{(4-b)}^2}-\frac{2}{{(4+b)}^2}=0$$ $${(4+b)}^2-2{(4-b)}^2=0$$ $$(b-\sqrt{2}b+4+4\sqrt{2})(b+\sqrt{2}b+4-4\sqrt{2})=0$$ $$\Rightarrow b=\frac{4(\sqrt{2}-1)}{\sqrt{2}+1}=4(3-2\sqrt{2})or$$ $$\Rightarrow b=\frac{4(\sqrt{2}+1)}{\sqrt{2}-1}=4(3+2\sqrt{2})(>2,rejected)$$ $$atb=4(3-2\sqrt{2})minimum=\sqrt{2}+\frac{1}{2}$$

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