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Question Number 26917 by Joel578 last updated on 31/Dec/17

Given a^2  + b^2  = 1 and c^2  + d^2  = 1  The minimum value of ac + bd − 2 is ...

$$\mathrm{Given}\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:=\:\mathrm{1}\:\mathrm{and}\:{c}^{\mathrm{2}} \:+\:{d}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\mathrm{The}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:{ac}\:+\:{bd}\:−\:\mathrm{2}\:\mathrm{is}\:... \\ $$

Commented by prakash jain last updated on 31/Dec/17

(a+c)^2 +(b+d)^2 ≥0  a^2 +c^2 +b^2 +d^2 +2(ac+bd)≥0  2+2(ac+bd)≥0  (ac+bd)≥−1

$$\left({a}+{c}\right)^{\mathrm{2}} +\left({b}+{d}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${a}^{\mathrm{2}} +{c}^{\mathrm{2}} +{b}^{\mathrm{2}} +{d}^{\mathrm{2}} +\mathrm{2}\left({ac}+{bd}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{2}+\mathrm{2}\left({ac}+{bd}\right)\geqslant\mathrm{0} \\ $$$$\left({ac}+{bd}\right)\geqslant−\mathrm{1} \\ $$

Commented by Joel578 last updated on 31/Dec/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Answered by ajfour last updated on 31/Dec/17

let  a=cos θ,  b=sin θ         c=cos φ,  d=sin φ  ac+bd−2=cos (θ−φ)−2  min. value =−3 .

$${let}\:\:{a}=\mathrm{cos}\:\theta,\:\:{b}=\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:\:\:{c}=\mathrm{cos}\:\phi,\:\:{d}=\mathrm{sin}\:\phi \\ $$$${ac}+{bd}−\mathrm{2}=\mathrm{cos}\:\left(\theta−\phi\right)−\mathrm{2} \\ $$$${min}.\:{value}\:=−\mathrm{3}\:. \\ $$

Commented by Joel578 last updated on 02/Jan/18

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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