Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 168558 by ajfour last updated on 13/Apr/22

Commented by ajfour last updated on 13/Apr/22

$$IftheshadedareabeA.Findmax$$$$of{y}_P(takecurvetobey=x^2).$$

Answered by mr W last updated on 13/Apr/22

$$P(p,p^2)$$$$tangentatP:$$$$y=p^2+2p(x-p)=2px-p^2=0\Rightarrow x=\frac{p}{2}$$$$A=\frac{p\times p^2}{3}-\frac{1}{2}\times \frac{p}{2}\times p=\frac{p^3}{12}$$$$p=\sqrt[3]{12A}$$$$y_{P,max}=p^2$$

Commented by Tawa11 last updated on 13/Apr/22

$$\mathrm{Nice}\mathrm{sir}.$$

Commented by ajfour last updated on 16/Apr/22

$$Thankusir.$$

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

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

$$whenPisgiven,areaofOPQAis$$$$minimum,whenPQistangent.$$$$similarlywhenareaAisgiven,P$$$$ishighestwhenPQistangent.$$

Answered by mindispower last updated on 13/Apr/22

$$P(p,p^2)$$$$y=ax+b,beethelignwithe$$$$ap+b=p^2$$$$let{x}_0,{ax}_0+b=0,x_0>0$$$$A={\int }_0^p{x}^{2}dx+\frac{1}{2}{p}^2(x_0-p)$$$$A\left(p\right)=-\frac{p^3}{6}+\frac{x_0{p}^2}{2}$$$$A^{\prime }\left(p\right)=0\Rightarrow -\frac{p^2}{2}+x_{0}p=0\Rightarrow x_0=\frac{p}{2}$$$$A\left(p\right)=\frac{p^3}{12}\Rightarrow p=\sqrt[3]{12A}$$$$$$$$$$$$$$$$$$

Commented by ajfour last updated on 16/Apr/22

$$Thankyousir.$$

Commented by mindispower last updated on 19/Apr/22

$$withePleasurallgoodthingsfor[yousir$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com