Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 3830 by Rasheed Soomro last updated on 21/Dec/15

$$\mathcal{D}rawarectangleofmaximumperimeter,$$$$byrulerandcompass,whenareais\mathbf{ab}.$$$$(\mathbf{AB}=\mathbf{a},\mathbf{CD}=\mathbf{b}\mathbf{are}\mathbf{given}.)$$

Commented by prakash jain last updated on 22/Dec/15

$$\mathrm{Area}ab,\mathrm{let}\mathrm{us}\mathrm{say}\mathrm{one}\mathrm{side}\mathrm{is}x$$$$\mathrm{Perimeter}=2(\frac{ab}{x}+x)$$$$\underset{x\to 0}{\lim }(\frac{ab}{x}+x)\to \mathrm{\infty }$$$$\mathrm{There}\mathrm{is}\mathrm{no}\mathrm{maximum}\mathrm{perimeter}\mathrm{for}\mathrm{given}$$$$\mathrm{area}.$$

Commented by Yozzii last updated on 22/Dec/15

$$P\left(x\right)=2(\frac{c}{x}+x)\Rightarrow \frac{dP}{dx}=2(1-\frac{c}{x^2})=2(1-{cx}^{-2})$$$$\frac{dP}{dx}=0\Rightarrow x=\pm \sqrt{c}\because x>0\Rightarrow x=\sqrt{c}$$$$\frac{d^{2}P}{{dx}^2}=2\left(2{cx}^{-3}\right)>0ifx>0$$$$Atx=\sqrt{c}=\sqrt{ab},\frac{d^{2}P}{{dx}^2}>0\Rightarrow minimumpt.$$$$forPexistsonlysincex>0.$$$$$$

Commented by Rasheed Soomro last updated on 22/Dec/15

$$\mathbb{T}\mathrm{han}{\mathbb{k}}^{\mathcal{S}}!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com