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find-values-a-b-c-such-that-1-ax-2-bx-c-1-and-6b-2-8-a-2-3-is-Max-




Question Number 146062 by mnjuly1970 last updated on 10/Jul/21
      find  values  a , b , c  such that:      −1≤ ax^2 +bx +c ≤ 1        and  ((6b^( 2) + 8 a^( 2) )/3) is Max...
$$ \\ $$$$\:\:\:\:{find}\:\:{values}\:\:{a}\:,\:{b}\:,\:{c}\:\:{such}\:{that}: \\ $$$$\:\:\:\:−\mathrm{1}\leqslant\:{ax}\:^{\mathrm{2}} +{bx}\:+{c}\:\leqslant\:\mathrm{1} \\ $$$$\:\:\:\:\:\:{and}\:\:\frac{\mathrm{6}{b}^{\:\mathrm{2}} +\:\mathrm{8}\:{a}^{\:\mathrm{2}} }{\mathrm{3}}\:{is}\:{Max}… \\ $$
Commented by mr W last updated on 10/Jul/21
it′s not possible with a≠0 and b≠0  that  −1≤ ax^2 +bx +c ≤ 1.  that means, such that  −1≤ ax^2 +bx +c ≤ 1  we must have a=b=0.  −1≤c≤1.  since max=((6b^2 +8a^2 )/3) =0 ⇒c=0
$${it}'{s}\:{not}\:{possible}\:{with}\:{a}\neq\mathrm{0}\:{and}\:{b}\neq\mathrm{0} \\ $$$${that}\:\:−\mathrm{1}\leqslant\:{ax}\:^{\mathrm{2}} +{bx}\:+{c}\:\leqslant\:\mathrm{1}. \\ $$$${that}\:{means},\:{such}\:{that} \\ $$$$−\mathrm{1}\leqslant\:{ax}\:^{\mathrm{2}} +{bx}\:+{c}\:\leqslant\:\mathrm{1} \\ $$$${we}\:{must}\:{have}\:{a}={b}=\mathrm{0}. \\ $$$$−\mathrm{1}\leqslant{c}\leqslant\mathrm{1}. \\ $$$${since}\:{max}=\frac{\mathrm{6}{b}^{\mathrm{2}} +\mathrm{8}{a}^{\mathrm{2}} }{\mathrm{3}}\:=\mathrm{0}\:\Rightarrow{c}=\mathrm{0} \\ $$

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