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Question-10464




Question Number 10464 by ABD last updated on 10/Feb/17
Answered by mrW1 last updated on 10/Feb/17
for ∣x−2∣=0 ⇒x=2, ∣3x−8∣=2  for ∣3x−8∣=0 ⇒x=(8/3), ∣x−2∣=(2/3)<2  A_(max) =((24)/((2/3)+0))=36
$${for}\:\mid{x}−\mathrm{2}\mid=\mathrm{0}\:\Rightarrow{x}=\mathrm{2},\:\mid\mathrm{3}{x}−\mathrm{8}\mid=\mathrm{2} \\ $$$${for}\:\mid\mathrm{3}{x}−\mathrm{8}\mid=\mathrm{0}\:\Rightarrow{x}=\frac{\mathrm{8}}{\mathrm{3}},\:\mid{x}−\mathrm{2}\mid=\frac{\mathrm{2}}{\mathrm{3}}<\mathrm{2} \\ $$$${A}_{{max}} =\frac{\mathrm{24}}{\frac{\mathrm{2}}{\mathrm{3}}+\mathrm{0}}=\mathrm{36} \\ $$

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