Question-174982

Question Number 174982 by MikeH last updated on 15/Aug/22

Answered by Rasheed.Sindhi last updated on 15/Aug/22

a+b≤9 ∧ a≥2_((i)) ∧ b≥−2_((ii)) (ii)−(i)⇒ b−a≥−2−2=−4 Minimum value of b−a is −4

$${a}+{b}\leqslant\mathrm{9}\:\wedge\:\underset{\left({i}\right)} {\underbrace{{a}\geqslant\mathrm{2}}}\:\wedge\:\underset{\left({ii}\right)} {\underbrace{{b}\geqslant−\mathrm{2}}} \\ $$$$\left({ii}\right)−\left({i}\right)\Rightarrow\:\:{b}−{a}\geqslant−\mathrm{2}−\mathrm{2}=−\mathrm{4} \\ $$$${Minimum}\:{value}\:{of}\:{b}−{a}\:{is}\:−\mathrm{4} \\ $$

Commented by JDamian last updated on 15/Aug/22

that is not correct, sir. if a=3 and b=-2, then b-a = -5

Commented by MikeH last updated on 15/Aug/22

I got an answer which I′m not sure of that′s why I posted. it goes as follows a + b ≤ 9 and a ≥2 and b ≥−2 now b−a is minimum when b is minimum b is minimum when b = −2 ⇒ a−2 ≤ 9 ⇒ a ≤ 11 now min(b−a) = −2−11 = −13. is that correct sir?

$$\mathrm{I}\:\mathrm{got}\:\mathrm{an}\:\mathrm{answer}\:\mathrm{which}\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{of} \\ $$$$\mathrm{that}'\mathrm{s}\:\mathrm{why}\:\mathrm{I}\:\mathrm{posted}.\:\mathrm{it}\:\mathrm{goes}\:\mathrm{as}\:\mathrm{follows} \\ $$$$ \\ $$$${a}\:+\:{b}\:\leqslant\:\mathrm{9}\:\mathrm{and}\:{a}\:\geqslant\mathrm{2}\:\mathrm{and}\:{b}\:\geqslant−\mathrm{2} \\ $$$$\mathrm{now}\:{b}−{a}\:\mathrm{is}\:\mathrm{minimum}\:\mathrm{when}\:{b}\:\mathrm{is}\:\mathrm{minimum} \\ $$$${b}\:\mathrm{is}\:\mathrm{minimum}\:\mathrm{when}\:{b}\:=\:−\mathrm{2} \\ $$$$\Rightarrow\:{a}−\mathrm{2}\:\leqslant\:\mathrm{9}\:\Rightarrow\:{a}\:\leqslant\:\mathrm{11} \\ $$$$\mathrm{now}\:\mathrm{min}\left({b}−{a}\right)\:=\:−\mathrm{2}−\mathrm{11}\:=\:−\mathrm{13}. \\ $$$$\mathrm{is}\:\mathrm{that}\:\mathrm{correct}\:\mathrm{sir}? \\ $$

Commented by Rasheed.Sindhi last updated on 16/Aug/22

$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\mathrm{JDamian}\:\mathrm{sir}! \\ $$

Answered by Frix last updated on 15/Aug/22

a+b≤9∧a≥2∧b≥−2 a=2+p^2 ∧b=−2+q^2 a+b≤9 ⇒ p^2 +q^2 ≤9 ⇒ q^2 ≤9−p^2 b−a≥m ⇒ −p^2 +q^2 −4≥m ⇒ q^2 ≥m+p^2 +4 m+p^2 +4≤9−p^2 m≤5−2p^2 m is min when p^2 is max p^2 +q^2 ≤9 ⇒ max(p^2 )=9 ⇒ q^2 =0 m=−13

$${a}+{b}\leqslant\mathrm{9}\wedge{a}\geqslant\mathrm{2}\wedge{b}\geqslant−\mathrm{2} \\ $$$${a}=\mathrm{2}+{p}^{\mathrm{2}} \wedge{b}=−\mathrm{2}+{q}^{\mathrm{2}} \\ $$$${a}+{b}\leqslant\mathrm{9}\:\Rightarrow\:{p}^{\mathrm{2}} +{q}^{\mathrm{2}} \leqslant\mathrm{9}\:\Rightarrow\:{q}^{\mathrm{2}} \leqslant\mathrm{9}−{p}^{\mathrm{2}} \\ $$$${b}−{a}\geqslant{m}\:\Rightarrow\:−{p}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{4}\geqslant{m}\:\Rightarrow\:{q}^{\mathrm{2}} \geqslant{m}+{p}^{\mathrm{2}} +\mathrm{4} \\ $$$${m}+{p}^{\mathrm{2}} +\mathrm{4}\leqslant\mathrm{9}−{p}^{\mathrm{2}} \\ $$$${m}\leqslant\mathrm{5}−\mathrm{2}{p}^{\mathrm{2}} \\ $$$${m}\:\mathrm{is}\:\mathrm{min}\:\mathrm{when}\:{p}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{max} \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} \leqslant\mathrm{9}\:\Rightarrow\:\mathrm{max}\left({p}^{\mathrm{2}} \right)=\mathrm{9} \\ $$$$\Rightarrow\:{q}^{\mathrm{2}} =\mathrm{0} \\ $$$${m}=−\mathrm{13} \\ $$

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