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Question Number 167321 by infinityaction last updated on 13/Mar/22

Commented by mr W last updated on 13/Mar/22

what′s the question?

$${what}'{s}\:{the}\:{question}? \\ $$

Commented by infinityaction last updated on 13/Mar/22

maximum value of x+y

$${maximum}\:{value}\:{of}\:{x}+{y} \\ $$

Commented by mr W last updated on 13/Mar/22

x=(√((a+c cos θ)^2 +(c sin θ)^2 )) x=(√(a^2 +c^2 +2ac cos θ)) y=(√(b^2 +c^2 −2bc cos θ)) ((d(x+y))/dθ)=−((ac sin θ)/( (√(a^2 +c^2 +2ac cos θ))))+((bc sin θ)/( (√(b^2 +c^2 −2bc cos θ))))=0 (a/( (√(a^2 +c^2 +2ac cos θ))))=(b/( (√(b^2 +c^2 −2bc cos θ)))) (a^2 /( a^2 +c^2 +2ac cos θ))=(b^2 /( b^2 +c^2 −2bc cos θ)) 2ab cos θ=(a−b)c ⇒cos θ=(((a−b)c)/(2ab)) (x+y)_(max) =(√(a^2 +(1+(a/b))c^2 ))+(√(b^2 +(1+(b/a))c^2 ))

$${x}=\sqrt{\left({a}+{c}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left({c}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} } \\ $$$${x}=\sqrt{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}\:\mathrm{cos}\:\theta} \\ $$$${y}=\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\:\mathrm{cos}\:\theta} \\ $$$$\frac{{d}\left({x}+{y}\right)}{{d}\theta}=−\frac{{ac}\:\mathrm{sin}\:\theta}{\:\sqrt{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}\:\mathrm{cos}\:\theta}}+\frac{{bc}\:\mathrm{sin}\:\theta}{\:\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\:\mathrm{cos}\:\theta}}=\mathrm{0} \\ $$$$\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}\:\mathrm{cos}\:\theta}}=\frac{{b}}{\:\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\:\mathrm{cos}\:\theta}} \\ $$$$\frac{{a}^{\mathrm{2}} }{\:{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}\:\mathrm{cos}\:\theta}=\frac{{b}^{\mathrm{2}} }{\:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\:\mathrm{cos}\:\theta} \\ $$$$\mathrm{2}{ab}\:\mathrm{cos}\:\theta=\left({a}−{b}\right){c} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\left({a}−{b}\right){c}}{\mathrm{2}{ab}} \\ $$$$\left({x}+{y}\right)_{{max}} =\sqrt{{a}^{\mathrm{2}} +\left(\mathrm{1}+\frac{{a}}{{b}}\right){c}^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} +\left(\mathrm{1}+\frac{{b}}{{a}}\right){c}^{\mathrm{2}} } \\ $$

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