Question Number 45994 by peter frank last updated on 19/Oct/18
Answered by MrW3 last updated on 20/Oct/18
$${we}\:{know}\:\frac{{x}+{y}}{\mathrm{2}}\geqslant\sqrt{{xy}}\:{when}\:{x},{y}\geqslant\mathrm{0}. \\ $$$$ \\ $$$$\mathrm{2}\left({a}+{b}\right)={L}\Rightarrow{a}+{b}=\frac{{L}}{\mathrm{2}} \\ $$$${A}_{{rectangular}} ={ab}\leqslant\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\frac{{L}}{\mathrm{4}}\right)^{\mathrm{2}} ={area}\:{of}\:{square}\:{with}\:{side}\:{length}\:\frac{{L}}{\mathrm{4}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Oct/18
$${L}=\mathrm{2}\left({a}+{b}\right) \\ $$$${A}={ab} \\ $$$${A}={a}\left(\frac{{L}}{\mathrm{2}}−{a}\right)=\frac{{aL}}{\mathrm{2}}−{a}^{\mathrm{2}} \\ $$$$\frac{{dA}}{{da}}=\frac{{L}}{\mathrm{2}}−\mathrm{2}{a} \\ $$$${for}\:{max}/{min}\:\:\frac{{dA}}{{da}}=\mathrm{0}\:\:\:\:\:{so}\:\frac{{L}}{\mathrm{2}}−\mathrm{2}{a}=\mathrm{0}\:\:\:{a}=\frac{{L}}{\mathrm{4}} \\ $$$$\frac{{d}^{\mathrm{2}} {A}}{{da}^{\mathrm{2}} }=−\mathrm{2}\:\:<\mathrm{0}\:\:{since}\:\frac{{d}^{\mathrm{2}} {A}}{{da}^{\mathrm{2}} }\:{is}−{ve} \\ $$$${so}\:{at}\:{a}=\frac{{L}}{\mathrm{2}}\:\:{the}\:{rectangle}\:{has}\:{max}\:{area} \\ $$$$\mathrm{2}{a}+\mathrm{2}{b}={L} \\ $$$$\mathrm{2}×\frac{{L}}{\mathrm{4}}+\mathrm{2}{b}={L} \\ $$$$\mathrm{2}{b}={L}−\frac{{L}}{\mathrm{2}}=\frac{{L}}{\mathrm{2}}\:\:\:{so}\:{b}=\frac{{L}}{\mathrm{4}} \\ $$$${since}\:{a}={b}=\frac{{L}}{\mathrm{4}}\:{so}\:{rectangle}\:{is}\:{square} \\ $$$${area}\:{of}\:{square}=\frac{{L}}{\mathrm{4}}×\frac{{L}}{\mathrm{4}}=\frac{{L}^{\mathrm{2}} }{\mathrm{16}} \\ $$$$ \\ $$$$ \\ $$