Question Number 57283 by ajfour last updated on 01/Apr/19
Commented by ajfour last updated on 01/Apr/19
$$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{cut}\:\mathrm{out}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{and} \\ $$$$\mathrm{rectangle}\:\mathrm{from}\:\mathrm{an}\:\mathrm{equilateral}\: \\ $$$$\mathrm{triangular}\:\mathrm{sheet}\:\mathrm{of}\:\mathrm{paper},\:\mathrm{edges}\:\mathrm{a}. \\ $$
Answered by mr W last updated on 01/Apr/19
$${let}\:{x}={width}\:{of}\:{rectangle} \\ $$$${heigth}\:{of}\:{rectangle}\:{is}\:\left(\mathrm{1}−\frac{{x}}{{a}}\right)\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}} \\ $$$${Area}\:{rectangle}=\left(\mathrm{1}−\frac{{x}}{{a}}\right)\frac{\sqrt{\mathrm{3}}{ax}}{\mathrm{2}} \\ $$$${radius}\:{of}\:{circle}\:{is}\:\frac{\sqrt{\mathrm{3}}{x}}{\mathrm{6}} \\ $$$${Area}\:{of}\:{circle}=\frac{\pi{x}^{\mathrm{2}} }{\mathrm{12}} \\ $$$${A}_{\Box+ } =\left(\mathrm{1}−\frac{{x}}{{a}}\right)\frac{\sqrt{\mathrm{3}}{ax}}{\mathrm{2}}+\frac{\pi{x}^{\mathrm{2}} }{\mathrm{12}}=\frac{\sqrt{\mathrm{3}}{ax}}{\mathrm{2}}−\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\pi}{\mathrm{12}}\right){x}^{\mathrm{2}} \\ $$$$\frac{{dA}}{{dx}}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}−\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\pi}{\mathrm{12}}\right){x}=\mathrm{0} \\ $$$$\left(\mathrm{6}\sqrt{\mathrm{3}}−\pi\right){x}=\mathrm{3}\sqrt{\mathrm{3}}{a} \\ $$$$\Rightarrow{x}=\frac{\mathrm{3}\sqrt{\mathrm{3}}{a}}{\mathrm{6}\sqrt{\mathrm{3}}−\pi}=\mathrm{0}.\mathrm{717}\:{a} \\ $$$${A}_{{max}} =... \\ $$
Commented by ajfour last updated on 02/Apr/19
$$\mathrm{thanks}\:\mathrm{sir}.\mathrm{its}\:\mathrm{alright}. \\ $$