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Find-the-maximum-and-minimum-of-the-expression-i-1-n-a-i-x-i-with-i-1-n-x-i-b-i-2-c-2-where-a-i-b-i-and-c-are-constants-extracted-and-modified-from-Q83331-

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Question Number 83341 by mr W last updated on 01/Mar/20
Find the maximum and minimum of the expression 𝚺_(i=1) ^n a_i x_i with 𝚺_(i=1) ^n (x_i −b_i )^2 =c^2 , where a_i , b_i and c are constants. (extracted and modified from Q83331)
$${Find}\:{the}\:{maximum}\:{and}\:{minimum} \\ $$$${of}\:{the}\:{expression}\:\underset{\boldsymbol{{i}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\boldsymbol{\sum}}{a}}_{\boldsymbol{{i}}} \boldsymbol{{x}}_{\boldsymbol{{i}}} \:{with} \\ $$$$\underset{\boldsymbol{{i}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\boldsymbol{\sum}}}\left(\boldsymbol{{x}}_{\boldsymbol{{i}}} −\boldsymbol{{b}}_{\boldsymbol{{i}}} \right)^{\mathrm{2}} =\boldsymbol{{c}}^{\mathrm{2}} ,\:{where}\:\boldsymbol{{a}}_{\boldsymbol{{i}}} ,\:\boldsymbol{{b}}_{\boldsymbol{{i}}} \:{and}\:\boldsymbol{{c}}\:{are} \\ $$$${constants}. \\ $$$$ \\ $$$$\left({extracted}\:{and}\:{modified}\:{from}\:{Q}\mathrm{83331}\right) \\ $$
Commented by mr W last updated on 01/Mar/20
Solution: in nD space Σ_(i=1) ^n (x_i −b_i )^2 =c^2 represents a sphere with center at the point B(b_1 ,b_2 ,...,b_n ) and radius c. let Σ_(i=1) ^n a_i x_i =s. Σ_(i=1) ^n a_i x_i =s represents a plane which intercepts the x_i −axis at (a_i /s). when the plane tangents the sphere we′ll get the maximum and minimum value from s. i.e. the distance from point B to the plane should be equal to the radius of the sphere. ((∣Σ_(i=1) ^n a_i b_i −s∣)/( (√(Σ_(i=1) ^n a_i ^2 ))))=c ⇒Σ_(i=1) ^n a_i b_i −s=±c(√(Σ_(i=1) ^n a_i ^2 )) ⇒s=Σ_(i=1) ^n a_i b_i ±c(√(Σ_(i=1) ^n a_i ^2 )) i.e. Σ_(i=1) ^n a_i b_i −c(√(Σ_(i=1) ^n a_i ^2 ))≤Σ_(i=1) ^n a_i x_i ≤Σ_(i=1) ^n a_i b_i +c(√(Σ_(i=1) ^n a_i ^2 )) special case: c=1 and b_i =0 with Σ_(i=1) ^n x_i ^2 =1 −(√(Σ_(i=1) ^n a_i ^2 ))≤Σ_(i=1) ^n a_i x_i ≤(√(Σ_(i=1) ^n a_i ^2 ))
$${Solution}: \\ $$$${in}\:{nD}\:{space}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({x}_{{i}} −{b}_{{i}} \right)^{\mathrm{2}} ={c}^{\mathrm{2}} \:{represents} \\ $$$${a}\:{sphere}\:{with}\:\:{center}\:{at}\:{the}\:{point} \\ $$$${B}\left({b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,…,{b}_{{n}} \right)\:{and}\:{radius}\:{c}. \\ $$$${let}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {x}_{{i}} ={s}.\: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {x}_{{i}} ={s}\:{represents}\:{a}\:{plane}\:{which} \\ $$$${intercepts}\:{the}\:{x}_{{i}} −{axis}\:{at}\:\frac{{a}_{{i}} }{{s}}. \\ $$$${when}\:{the}\:{plane}\:{tangents}\:{the}\:{sphere} \\ $$$${we}'{ll}\:{get}\:{the}\:{maximum}\:{and}\:{minimum} \\ $$$${value}\:{from}\:{s}. \\ $$$${i}.{e}.\:{the}\:{distance}\:{from}\:{point}\:{B}\:{to}\:{the} \\ $$$${plane}\:{should}\:{be}\:{equal}\:{to}\:{the}\:{radius}\:{of} \\ $$$${the}\:{sphere}. \\ $$$$\frac{\mid\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {b}_{{i}} −{s}\mid}{\:\sqrt{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} ^{\mathrm{2}} }}={c} \\ $$$$\Rightarrow\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {b}_{{i}} −{s}=\pm{c}\sqrt{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} ^{\mathrm{2}} } \\ $$$$\Rightarrow{s}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {b}_{{i}} \pm{c}\sqrt{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} ^{\mathrm{2}} } \\ $$$${i}.{e}.\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {b}_{{i}} −{c}\sqrt{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} ^{\mathrm{2}} }\leqslant\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {x}_{{i}} \leqslant\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {b}_{{i}} +{c}\sqrt{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} ^{\mathrm{2}} } \\ $$$$ \\ $$$${special}\:{case}:\:{c}=\mathrm{1}\:{and}\:{b}_{{i}} =\mathrm{0} \\ $$$${with}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{x}_{{i}} ^{\mathrm{2}} =\mathrm{1} \\ $$$$−\sqrt{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} ^{\mathrm{2}} }\leqslant\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {x}_{{i}} \leqslant\sqrt{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} ^{\mathrm{2}} } \\ $$
Answered by M±th+et£s last updated on 01/Mar/20
Commented by mr W last updated on 01/Mar/20
thanks sir! this is the general way. but if possible, i prefer an obvious and visible way, for example a geometrical way.
$${thanks}\:{sir}! \\ $$$${this}\:{is}\:{the}\:{general}\:{way}.\:{but}\:{if}\:{possible}, \\ $$$${i}\:{prefer}\:{an}\:{obvious}\:{and}\:{visible}\:{way}, \\ $$$${for}\:{example}\:{a}\:{geometrical}\:{way}. \\ $$

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