Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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

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

$${what}^{\prime }sthequestion?$$

Commented by infinityaction last updated on 13/Mar/22

$$maximumvalueofx+y$$

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com