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

Let k be a real number such that the inequality (√(x−3))+(√(6−x))≥k has a solution. Find the maximum value of k.

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

Answered by Her_Majesty last updated on 30/Aug/20

f(x)=(√(x−3))+(√(6−x)) ⇒ 3≤x≤6 minima at the borders f(3)=f(6)=(√3)=k

$${f}\left({x}\right)=\sqrt{{x}−\mathrm{3}}+\sqrt{\mathrm{6}−{x}}\:\Rightarrow\:\mathrm{3}\leqslant{x}\leqslant\mathrm{6} \\ $$$${minima}\:{at}\:{the}\:{borders} \\ $$$${f}\left(\mathrm{3}\right)={f}\left(\mathrm{6}\right)=\sqrt{\mathrm{3}}={k} \\ $$

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

Thanks but what do you mean by minima at the borders?

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

Commented by Her_Majesty last updated on 30/Aug/20

the borders of definition my dear, or do you see other relevant borders?

$${the}\:{borders}\:{of}\:{definition}\:{my}\:{dear},\:{or}\:{do} \\ $$$${you}\:{see}\:{other}\:{relevant}\:{borders}? \\ $$

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

Is minima at the borders true for all functions?

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

Commented by Her_Majesty last updated on 30/Aug/20

no. but you always have to check the borders here: y=(√(x−3))+(√(6−x)) y′=(1/( 2(√(x−3))))−(1/( 2(√(6−x))))=(((√(6−x))−(√(x−3)))/(2(√(x−3))(√(6−x))))=0 ⇔ (√(6−x))=(√(x−3)) ⇔ 6−x=x−3 ⇒ x=9/2 is the only extreme but it′s the maximum at the borders y′ is undefined so you cannot solve the problem in the usual way

$${no}.\:{but}\:{you}\:{always}\:{have}\:{to}\:{check}\:{the}\:{borders} \\ $$$$ \\ $$$${here}: \\ $$$${y}=\sqrt{{x}−\mathrm{3}}+\sqrt{\mathrm{6}−{x}} \\ $$$${y}'=\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{{x}−\mathrm{3}}}−\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{6}−{x}}}=\frac{\sqrt{\mathrm{6}−{x}}−\sqrt{{x}−\mathrm{3}}}{\mathrm{2}\sqrt{{x}−\mathrm{3}}\sqrt{\mathrm{6}−{x}}}=\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$$\sqrt{\mathrm{6}−{x}}=\sqrt{{x}−\mathrm{3}} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{6}−{x}={x}−\mathrm{3}\:\Rightarrow\:{x}=\mathrm{9}/\mathrm{2} \\ $$$${is}\:{the}\:{only}\:{extreme}\:{but}\:{it}'{s}\:{the}\:{maximum} \\ $$$${at}\:{the}\:{borders}\:{y}'\:{is}\:{undefined}\:{so}\:{you}\:{cannot} \\ $$$${solve}\:{the}\:{problem}\:{in}\:{the}\:{usual}\:{way} \\ $$

Commented by Her_Majesty last updated on 30/Aug/20

sorry I misread the question. if we′re looking for at least 1 solution of the inequality we indeed get k=(√6) because f(9/2)=(√6)

$${sorry}\:{I}\:{misread}\:{the}\:{question}. \\ $$$${if}\:{we}'{re}\:{looking}\:{for}\:{at}\:{least}\:\mathrm{1}\:{solution}\:{of} \\ $$$${the}\:{inequality}\:{we}\:{indeed}\:{get}\:{k}=\sqrt{\mathrm{6}}\:{because} \\ $$$${f}\left(\mathrm{9}/\mathrm{2}\right)=\sqrt{\mathrm{6}} \\ $$$$ \\ $$

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

Apply inequality AM−GM we have ( (√(x−3))+(√(6−x)))^2 ≤(1^2 +1^2 )(x−3+6−x)=6 ⇒0< (√(x−3))+(√(6−x))≤(√6).(1)Hence,put f(x)= (√(x−3))+(√(6−x)) Take k=(√6) then the inequality (√(x−3))+(√(6−x))≥k has at least one root k=(√6) is greatest value because if ∃k_0 >(√6) such that (√(x−3))+(√(6−x))≥k_0 then this inequality has no root due to (1)

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

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