Question Number 186265 by mnjuly1970 last updated on 02/Feb/23
$$ \\ $$$$\:\:\:{f}\left({x}\right)=\:\sqrt{\:{x}\:−{a}}\:\:+\:\sqrt{\mathrm{3}{a}\:−{x}}\:\:\:{with}\:\left(\:{a}>\mathrm{0}\right) \\ $$$$\:\:\:\:{is}\:\:{given}\:.{If}\:\:,\:\:{f}_{\:{max}} \:.\:{f}_{\:{min}} \:=\:\sqrt{\mathrm{32}} \\ $$$$\:\:\:\:\:{find}\:\:,\:\:\:\:\:\:\:''\:\:\:{a}\:\:''\:\:=\:? \\ $$
Answered by mahdipoor last updated on 02/Feb/23
![(df/dx)=(1/(2(√(x−a))))+((−1)/(2(√(3a−x))))=0⇒(√(3a−x))=(√(x−a)) ⇒x=2a D_f =[a,3a] f(a)=f(3a)=(√(2a)) f(2a)=2(√a) f_(min) .f_(max) =(√(8a^2 ))=(√(32)) ⇒a=2 way 2: f^2 =2a+2(√(4ax−x^2 −3a^2 )) g(x)= (√(4ax−x^2 −3a^2 )) { ((min=0)),((max=g(((−4a)/(−1×2)))=a)) :} min f^2 ×max f^2 =2a×4a=((√(32)))^2 ⇒a=2](https://www.tinkutara.com/question/Q186268.png)
$$\frac{{df}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}−{a}}}+\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}{a}−{x}}}=\mathrm{0}\Rightarrow\sqrt{\mathrm{3}{a}−{x}}=\sqrt{{x}−{a}} \\ $$$$\Rightarrow{x}=\mathrm{2}{a}\: \\ $$$${D}_{{f}} =\left[{a},\mathrm{3}{a}\right] \\ $$$${f}\left({a}\right)={f}\left(\mathrm{3}{a}\right)=\sqrt{\mathrm{2}{a}}\:\:\:\:\:\:\:\:\:\:{f}\left(\mathrm{2}{a}\right)=\mathrm{2}\sqrt{{a}} \\ $$$${f}_{{min}} .{f}_{{max}} =\sqrt{\mathrm{8}{a}^{\mathrm{2}} }=\sqrt{\mathrm{32}}\:\:\:\Rightarrow{a}=\mathrm{2} \\ $$$${way}\:\mathrm{2}: \\ $$$${f}^{\mathrm{2}} =\mathrm{2}{a}+\mathrm{2}\sqrt{\mathrm{4}{ax}−{x}^{\mathrm{2}} −\mathrm{3}{a}^{\mathrm{2}} } \\ $$$${g}\left({x}\right)=\:\sqrt{\mathrm{4}{ax}−{x}^{\mathrm{2}} −\mathrm{3}{a}^{\mathrm{2}} }\:\:\begin{cases}{{min}=\mathrm{0}}\\{{max}={g}\left(\frac{−\mathrm{4}{a}}{−\mathrm{1}×\mathrm{2}}\right)={a}}\end{cases} \\ $$$${min}\:{f}^{\mathrm{2}} ×{max}\:{f}^{\mathrm{2}} =\mathrm{2}{a}×\mathrm{4}{a}=\left(\sqrt{\mathrm{32}}\right)^{\mathrm{2}} \Rightarrow{a}=\mathrm{2} \\ $$