Question Number 118612 by rexfordattacudjoe last updated on 18/Oct/20
Answered by 1549442205PVT last updated on 18/Oct/20
![y=0.4x^2 −36x+1000= (4/(10))(x^2 −90x+2500)=(2/5)[(x−45)^2 +475] ≥((2.475)/5)=190 ⇒y_(min) =190 when x=45 (years−old) It means that numbers of car accidents are lowest at age 45 .That number is 190 cases per 50 million km driven](Q118615.png)
$$\mathrm{y}=\mathrm{0}.\mathrm{4x}^{\mathrm{2}} −\mathrm{36x}+\mathrm{1000}= \\ $$$$\frac{\mathrm{4}}{\mathrm{10}}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{90x}+\mathrm{2500}\right)=\frac{\mathrm{2}}{\mathrm{5}}\left[\left(\mathrm{x}−\mathrm{45}\right)^{\mathrm{2}} +\mathrm{475}\right] \\ $$$$\geqslant\frac{\mathrm{2}.\mathrm{475}}{\mathrm{5}}=\mathrm{190} \\ $$$$\Rightarrow\mathrm{y}_{\mathrm{min}} =\mathrm{190}\:\mathrm{when}\:\mathrm{x}=\mathrm{45}\:\left(\mathrm{years}−\mathrm{old}\right) \\ $$$$\mathrm{It}\:\mathrm{means}\:\mathrm{that}\:\mathrm{numbers}\:\mathrm{of}\:\mathrm{car}\:\mathrm{accidents}\: \\ $$$$\mathrm{are}\:\mathrm{lowest}\:\mathrm{at}\:\mathrm{age}\:\mathrm{45}\:.\mathrm{That}\:\mathrm{number}\:\mathrm{is} \\ $$$$\mathrm{190}\:\mathrm{cases}\:\mathrm{per}\:\mathrm{50}\:\mathrm{million}\:\mathrm{km}\:\mathrm{driven} \\ $$