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Question Number 110673 by Aina Samuel Temidayo last updated on 30/Aug/20

$$$$$$\mathrm{Let}\mathrm{k}\mathrm{be}\mathrm{a}\mathrm{real}\mathrm{number}\mathrm{such}\mathrm{that}\mathrm{the}$$$$\mathrm{inequality}\sqrt{\mathrm{x}-3}+\sqrt{6-\mathrm{x}}⩾\mathrm{k}\mathrm{has}\mathrm{a}$$$$\mathrm{solution}.\mathrm{Find}\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}$$$$\mathrm{k}.$$

Answered by Her_Majesty last updated on 30/Aug/20

$$f\left(x\right)=\sqrt{x-3}+\sqrt{6-x}\Rightarrow 3⩽x⩽6$$$$minimaattheborders$$$$f\left(3\right)=f\left(6\right)=\sqrt{3}=k$$

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{Thanks}\mathrm{but}\mathrm{what}\mathrm{do}\mathrm{you}\mathrm{mean}\mathrm{by}$$$$\mathrm{minima}\mathrm{at}\mathrm{the}\mathrm{borders}?$$

Commented by Her_Majesty last updated on 30/Aug/20

$$thebordersofdefinitionmydear,ordo$$$$youseeotherrelevantborders?$$

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{Is}\mathrm{minima}\mathrm{at}\mathrm{the}\mathrm{borders}\mathrm{true}\mathrm{for}$$$$\mathrm{all}\mathrm{functions}?$$

Commented by Her_Majesty last updated on 30/Aug/20

$$no.butyoualwayshavetochecktheborders$$$$$$$$here:$$$$y=\sqrt{x-3}+\sqrt{6-x}$$$$y^{\prime }=\frac{1}{2\sqrt{x-3}}-\frac{1}{2\sqrt{6-x}}=\frac{\sqrt{6-x}-\sqrt{x-3}}{2\sqrt{x-3}\sqrt{6-x}}=0$$$$\iff$$$$\sqrt{6-x}=\sqrt{x-3}$$$$\iff$$$$6-x=x-3\Rightarrow x=9/2$$$$istheonlyextremebut{it}^{\prime }sthemaximum$$$$attheborders{y}^{\prime }isundefinedsoyoucannot$$$$solvetheproblemintheusualway$$

Commented by Her_Majesty last updated on 30/Aug/20

$$sorryImisreadthequestion.$$$$if{we}^{\prime }relookingforatleast1solutionof$$$$theinequalityweindeedgetk=\sqrt{6}because$$$$f\left(9/2\right)=\sqrt{6}$$$$$$

Answered by 1549442205PVT last updated on 30/Aug/20

$$\mathrm{Apply}\mathrm{inequality}\mathrm{AM}-\mathrm{GM}\mathrm{we}\mathrm{have}$$$${(\sqrt{\mathrm{x}-3}+\sqrt{6-\mathrm{x}})}^2⩽(1^2+1^2)(\mathrm{x}-3+6-\mathrm{x})=6$$$$\Rightarrow 0<\sqrt{\mathrm{x}-3}+\sqrt{6-\mathrm{x}}⩽\sqrt{6}.\left(1\right)\mathrm{Hence},\mathrm{put}$$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}-3}+\sqrt{6-\mathrm{x}}$$$$\mathrm{Take}\mathrm{k}=\sqrt{6}\mathrm{then}\mathrm{the}\mathrm{inequality}$$$$\sqrt{\mathrm{x}-3}+\sqrt{6-\mathrm{x}}⩾\mathrm{k}\mathrm{has}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{root}$$$$\mathrm{k}=\sqrt{6}\mathrm{is}\mathrm{greatest}\mathrm{value}\mathrm{because}\mathrm{if}$$$$\mathrm{\exists }{\mathrm{k}}_0>\sqrt{6}\mathrm{such}\mathrm{that}$$$$\sqrt{\mathrm{x}-3}+\sqrt{6-\mathrm{x}}⩾{\mathrm{k}}_0\mathrm{then}\mathrm{this}\mathrm{inequality}$$$$\mathrm{has}\mathrm{no}\mathrm{root}\mathrm{due}\mathrm{to}\left(1\right)$$

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