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Question Number 76872 by john santu last updated on 31/Dec/19

the product of 3 integer x,y,z is 192  . z = 4 and t is equal to average?  of x and y . what is the minimum   posible value of t?

$${the}\:{product}\:{of}\:\mathrm{3}\:{integer}\:{x},{y},{z}\:{is}\:\mathrm{192} \\ $$$$.\:{z}\:=\:\mathrm{4}\:{and}\:{t}\:{is}\:{equal}\:{to}\:{average}? \\ $$$${of}\:{x}\:{and}\:{y}\:.\:{what}\:{is}\:{the}\:{minimum}\: \\ $$$${posible}\:{value}\:{of}\:{t}? \\ $$

Commented by mr W last updated on 31/Dec/19

t=((x+y)/2)=(1/2)(x+((48)/x))  t_(min) =t(−1)=−((49)/2)

$${t}=\frac{{x}+{y}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\frac{\mathrm{48}}{{x}}\right) \\ $$$${t}_{{min}} ={t}\left(−\mathrm{1}\right)=−\frac{\mathrm{49}}{\mathrm{2}} \\ $$

Commented by jagoll last updated on 31/Dec/19

this problem applies  to negative  integer sir?

$${this}\:{problem}\:{applies}\:\:{to}\:{negative} \\ $$$${integer}\:{sir}? \\ $$

Commented by mr W last updated on 31/Dec/19

if only positive integers are allowed,  then t_(min) =t(6)=t(8)=(1/2)(6+((48)/6))=7

$${if}\:{only}\:{positive}\:{integers}\:{are}\:{allowed}, \\ $$$${then}\:{t}_{{min}} ={t}\left(\mathrm{6}\right)={t}\left(\mathrm{8}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{6}+\frac{\mathrm{48}}{\mathrm{6}}\right)=\mathrm{7} \\ $$

Commented by john santu last updated on 31/Dec/19

thanks sir

$${thanks}\:{sir} \\ $$

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