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Question Number 205914 by mr W last updated on 02/Apr/24
if a+b+c+d+e+f=10 and a^2 +b^2 +c^2 +d^2 +e^2 +f^2 =25, find a_(min) and f_(max) .
$${if}\:{a}+{b}+{c}+{d}+{e}+{f}=\mathrm{10}\:{and} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} =\mathrm{25},\:{find} \\ $$$${a}_{{min}} \:{and}\:{f}_{{max}} . \\ $$
Commented by Tinku Tara last updated on 03/Apr/24
Question: Why is min=−5 and max=+5 not the solution?
$$\mathrm{Question}:\: \\ $$$$\mathrm{Why}\:\mathrm{is}\:\mathrm{min}=−\mathrm{5}\:\mathrm{and}\:\mathrm{max}=+\mathrm{5} \\ $$$$\mathrm{not}\:\mathrm{the}\:\mathrm{solution}? \\ $$
Commented by mr W last updated on 03/Apr/24
if min=−5, say a=−5, to fulfill a^2 +b^2 +c^2 +d^2 +e^2 +f^2 =25, we must have b=c=d=e=f=0, i.e. a+b+c+d+e+f=−5, but this is contradiction to a+b+c+d+e+f=10, therefore a≠−5.
$${if}\:{min}=−\mathrm{5},\:{say}\:{a}=−\mathrm{5},\:{to}\:{fulfill} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} =\mathrm{25},\:{we}\:{must} \\ $$$${have}\:{b}={c}={d}={e}={f}=\mathrm{0},\:{i}.{e}. \\ $$$${a}+{b}+{c}+{d}+{e}+{f}=−\mathrm{5},\:{but}\:{this}\:{is} \\ $$$${contradiction}\:{to}\:{a}+{b}+{c}+{d}+{e}+{f}=\mathrm{10}, \\ $$$${therefore}\:{a}\neq−\mathrm{5}. \\ $$
Answered by mr W last updated on 05/Apr/24
Method II a=10−(b+c+d+e+f) (10−(b+c+d+e+f))^2 +b^2 +c^2 +d^2 +e^2 +f^2 −25=0 F=a=10−(b+c+d+e+f)+λ[(10−(b+c+d+e+f))^2 +b^2 +c^2 +d^2 +e^2 +f^2 −25] (∂F/∂b)=−1+λ[−2(10−(b+c+d+e+f))+2b]=0 =>−(1/2)+λ[−10+(b+c+d+e+f)+b]=0 ⇒b=(1/2)+10λ−λ(b+c+d+e+f) similarly ⇒c=d=e=f=(1/2)+10λ−λ(b+c+d+e+f) i.e. for a_(min) or a_(max) , b=c=d=e=f=t, say then we get as shown above a_(min) =((10−5(√(10)))/6), a_(max) =((10+5(√(10)))/6)
$$\underline{{Method}\:{II}} \\ $$$${a}=\mathrm{10}−\left({b}+{c}+{d}+{e}+{f}\right) \\ $$$$\left(\mathrm{10}−\left({b}+{c}+{d}+{e}+{f}\right)\right)^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} −\mathrm{25}=\mathrm{0} \\ $$$${F}={a}=\mathrm{10}−\left({b}+{c}+{d}+{e}+{f}\right)+\lambda\left[\left(\mathrm{10}−\left({b}+{c}+{d}+{e}+{f}\right)\right)^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} −\mathrm{25}\right] \\ $$$$\frac{\partial{F}}{\partial{b}}=−\mathrm{1}+\lambda\left[−\mathrm{2}\left(\mathrm{10}−\left({b}+{c}+{d}+{e}+{f}\right)\right)+\mathrm{2}{b}\right]=\mathrm{0} \\ $$$$=>−\frac{\mathrm{1}}{\mathrm{2}}+\lambda\left[−\mathrm{10}+\left({b}+{c}+{d}+{e}+{f}\right)+{b}\right]=\mathrm{0} \\ $$$$\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{10}\lambda−\lambda\left({b}+{c}+{d}+{e}+{f}\right) \\ $$$${similarly} \\ $$$$\Rightarrow{c}={d}={e}={f}=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{10}\lambda−\lambda\left({b}+{c}+{d}+{e}+{f}\right) \\ $$$${i}.{e}.\:{for}\:{a}_{{min}} \:{or}\:{a}_{{max}} ,\: \\ $$$${b}={c}={d}={e}={f}={t},\:{say} \\ $$$${then}\:{we}\:{get}\:{as}\:{shown}\:{above} \\ $$$${a}_{{min}} =\frac{\mathrm{10}−\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{6}},\:{a}_{{max}} =\frac{\mathrm{10}+\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{6}} \\ $$
Answered by mr W last updated on 05/Apr/24
Method III p^→ =(a,b,c,d,e,f) q^→ =(1,1,1,1,1,1) cos θ=((p^→ ∙q^→ )/(∣p∣∣q∣)) =((a+b+c+d+e+f)/( (√(6(a^2 +b^2 +c^2 +d^2 +e^2 +f^2 )))))=(2/( (√6))) a_(min/max) =5 cos (α±θ) with α=cos^(−1) (1/( (√6))) a_(min/max) =5×((1/( (√6)))×(2/( (√6)))∓((√5)/( (√6)))×((√2)/( (√6)))) =((5(2∓(√(10))))/6) ✓
$$\underline{{Method}\:{III}} \\ $$$$\overset{\rightarrow} {\boldsymbol{{p}}}=\left({a},{b},{c},{d},{e},{f}\right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{q}}}=\left(\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1}\right) \\ $$$$\mathrm{cos}\:\theta=\frac{\overset{\rightarrow} {\boldsymbol{{p}}}\centerdot\overset{\rightarrow} {\boldsymbol{{q}}}}{\mid\boldsymbol{{p}}\mid\mid\boldsymbol{{q}}\mid} \\ $$$$\:\:\:\:=\frac{{a}+{b}+{c}+{d}+{e}+{f}}{\left.\:\sqrt{\mathrm{6}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} \right.}\right)}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{6}}} \\ $$$${a}_{{min}/{max}} =\mathrm{5}\:\mathrm{cos}\:\left(\alpha\pm\theta\right) \\ $$$${with}\:\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}} \\ $$$${a}_{{min}/{max}} =\mathrm{5}×\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{6}}}\mp\frac{\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{6}}}×\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{6}}}\right) \\ $$$$\:\:\:=\frac{\mathrm{5}\left(\mathrm{2}\mp\sqrt{\mathrm{10}}\right)}{\mathrm{6}}\:\checkmark \\ $$
Commented by mr W last updated on 05/Apr/24
Answered by mr W last updated on 04/Apr/24
Method I if b+c+d+e+f=s, b^2 +c^2 +d^2 +e^2 +f^2 ≥(((b+c+d+e+f)^2 )/5)=(s^2 /5) equality holds when b=c=d=e=f=(s/5) since a+b+c+d+e+f=10 and a^2 +b^2 +c^2 +d^2 +e^2 +f^2 =25, such that a is as large as possible, b+c+d+e+f should be as small as possible and besides b, c, d, e, f should be equal. say b=c=d=e=f=t, then a+5t=10 ⇒t=((10−a)/5) a^2 +5t^2 =25 ⇒a^2 +5×(((10−a)/5))^2 =25 6a^2 −20a−25=0 a=((10±5(√(10)))/6) i.e. a_(min) =((10−5(√(10)))/6), a_(max) =((10+5(√(10)))/6)
$$\underline{{Method}\:{I}} \\ $$$${if}\:{b}+{c}+{d}+{e}+{f}={s}, \\ $$$${b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} \geqslant\frac{\left({b}+{c}+{d}+{e}+{f}\right)^{\mathrm{2}} }{\mathrm{5}}=\frac{{s}^{\mathrm{2}} }{\mathrm{5}} \\ $$$${equality}\:{holds}\:{when}\:{b}={c}={d}={e}={f}=\frac{{s}}{\mathrm{5}} \\ $$$${since}\:{a}+{b}+{c}+{d}+{e}+{f}=\mathrm{10}\:{and} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} =\mathrm{25},\:{such}\:{that} \\ $$$${a}\:{is}\:{as}\:{large}\:{as}\:{possible},\:{b}+{c}+{d}+{e}+{f} \\ $$$${should}\:{be}\:{as}\:{small}\:{as}\:{possible}\:{and} \\ $$$${besides}\:{b},\:{c},\:{d},\:{e},\:{f}\:{should}\:{be}\:{equal}. \\ $$$${say}\:{b}={c}={d}={e}={f}={t},\:{then} \\ $$$${a}+\mathrm{5}{t}=\mathrm{10}\:\Rightarrow{t}=\frac{\mathrm{10}−{a}}{\mathrm{5}} \\ $$$${a}^{\mathrm{2}} +\mathrm{5}{t}^{\mathrm{2}} =\mathrm{25}\:\Rightarrow{a}^{\mathrm{2}} +\mathrm{5}×\left(\frac{\mathrm{10}−{a}}{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{25} \\ $$$$\mathrm{6}{a}^{\mathrm{2}} −\mathrm{20}{a}−\mathrm{25}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{10}\pm\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{6}} \\ $$$${i}.{e}.\:{a}_{{min}} =\frac{\mathrm{10}−\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{6}},\:{a}_{{max}} =\frac{\mathrm{10}+\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{6}} \\ $$
Answered by mr W last updated on 04/Apr/24
Method IV f(x)=(x−b)^2 +(x−c)^2 +(x−d)^2 +(x−e)^2 +(x−f)^2 ≥0 f(x)=5x^2 −2(b+c+d+e+f)+(b^2 +c^2 +d^2 +e^2 +f^2 )≥0 f(x)=5x^2 −2(10−a)x+(25−a^2 )≥0 ⇒(10−a)^2 −5(25−a^2 )≤0 ⇒6a^2 −20a−25≤0 ⇒((10−5(√(10)))/6)≤a≤((10+5(√(10)))/6)
$$\underline{{Method}\:{IV}} \\ $$$${f}\left({x}\right)=\left({x}−{b}\right)^{\mathrm{2}} +\left({x}−{c}\right)^{\mathrm{2}} +\left({x}−{d}\right)^{\mathrm{2}} +\left({x}−{e}\right)^{\mathrm{2}} +\left({x}−{f}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${f}\left({x}\right)=\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}\left({b}+{c}+{d}+{e}+{f}\right)+\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} \right)\geqslant\mathrm{0} \\ $$$${f}\left({x}\right)=\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{10}−{a}\right){x}+\left(\mathrm{25}−{a}^{\mathrm{2}} \right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{10}−{a}\right)^{\mathrm{2}} −\mathrm{5}\left(\mathrm{25}−{a}^{\mathrm{2}} \right)\leqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{6}{a}^{\mathrm{2}} −\mathrm{20}{a}−\mathrm{25}\leqslant\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{10}−\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{6}}\leqslant{a}\leqslant\frac{\mathrm{10}+\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{6}} \\ $$

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