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Question Number 131196 by john_santu last updated on 02/Feb/21
The minimum value of the expression B = ∣z∣^2 +∣z−3∣^2 +∣z−6i∣^2 is p. What the value of (p/(10)) .?
$$\:{The}\:{minimum}\:{value}\:{of}\:{the}\: \\ $$$${expression}\:{B}\:=\:\mid{z}\mid^{\mathrm{2}} +\mid{z}−\mathrm{3}\mid^{\mathrm{2}} +\mid{z}−\mathrm{6}{i}\mid^{\mathrm{2}} \\ $$$${is}\:{p}.\:{What}\:{the}\:{value}\:{of}\:\frac{{p}}{\mathrm{10}}\:.? \\ $$
Answered by liberty last updated on 02/Feb/21
z=a+bi → { ((∣z∣^2 =a^2 +b^2 )),((∣z−3∣^2 = (a−3)^2 +b^2 )),((∣z−6i∣^2 = a^2 +b^2 +36−12b)) :} now we find B=3a^2 +3b^2 −6a−12b+45 B=3(a−1)^2 +3(b−2)^2 +30 so the minimum value of B is 30 when { ((a=1)),((b=2)) :} ⇒ z=1+2i then p = 30 and (p/(10)) = ((30)/(10)) = 3.
$${z}={a}+{bi}\:\rightarrow\begin{cases}{\mid{z}\mid^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\\{\mid{z}−\mathrm{3}\mid^{\mathrm{2}} =\:\left({a}−\mathrm{3}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \:}\\{\mid{z}−\mathrm{6}{i}\mid^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{36}−\mathrm{12}{b}}\end{cases} \\ $$$${now}\:{we}\:{find}\: \\ $$$${B}=\mathrm{3}{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} −\mathrm{6}{a}−\mathrm{12}{b}+\mathrm{45}\: \\ $$$${B}=\mathrm{3}\left({a}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\left({b}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{30} \\ $$$${so}\:{the}\:{minimum}\:{value}\:{of}\:{B}\:\mathrm{is}\:\mathrm{30}\:\mathrm{when} \\ $$$$\:\begin{cases}{{a}=\mathrm{1}}\\{{b}=\mathrm{2}}\end{cases}\:\Rightarrow\:{z}=\mathrm{1}+\mathrm{2}{i}\:{then}\:{p}\:=\:\mathrm{30}\: \\ $$$${and}\:\frac{{p}}{\mathrm{10}}\:=\:\frac{\mathrm{30}}{\mathrm{10}}\:=\:\mathrm{3}. \\ $$$$ \\ $$
Commented by john_santu last updated on 02/Feb/21
gave kudos
$${gave}\:{kudos}\: \\ $$

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