Question Number 203557 by ajfour last updated on 21/Jan/24
Commented by ajfour last updated on 21/Jan/24
$${Find}\:{maimum}\:{blue}\:{shaded}\:{area}. \\ $$
Answered by mr W last updated on 22/Jan/24
Commented by mr W last updated on 22/Jan/24
$${H}=\frac{\mathrm{3}×\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\frac{{a}}{\mathrm{5}}=\frac{{H}−{h}}{{H}}\:\Rightarrow{a}=\mathrm{5}\left(\mathrm{1}−\frac{\mathrm{5}{h}}{\mathrm{12}}\right) \\ $$$$\frac{{b}}{{a}}=\frac{{H}−{h}}{{H}}\:\Rightarrow{b}=\mathrm{5}\left(\mathrm{1}−\frac{\mathrm{5}{h}}{\mathrm{12}}\right)^{\mathrm{2}} \\ $$$${A}=\frac{\left({a}+{b}\right){h}}{\mathrm{2}}=\mathrm{6}×\frac{\mathrm{5}{h}}{\mathrm{12}}\left(\mathrm{1}−\frac{\mathrm{5}{h}}{\mathrm{12}}\right)\left(\mathrm{2}−\frac{\mathrm{5}{h}}{\mathrm{12}}\right) \\ $$$${A}=\mathrm{6}{x}\left(\mathrm{1}−{x}\right)\left(\mathrm{2}−{x}\right)\:{with}\:{x}=\frac{\mathrm{5}{h}}{\mathrm{12}}<\mathrm{1} \\ $$$$\frac{{dA}}{{dx}}=\mathrm{6}\left(\mathrm{1}−{x}\right)\left(\mathrm{2}−{x}\right)−\mathrm{6}{x}\left(\mathrm{2}−{x}\right)−\mathrm{6}{x}\left(\mathrm{1}−{x}\right)=\mathrm{0} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:{rejected}\right) \\ $$$${A}_{{max}} =\mathrm{6}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\checkmark \\ $$
Commented by ajfour last updated on 22/Jan/24
$${Thanks}\:{Sir}.\:{I}'\:{ll}\:{try}\:{soon}\:{too}. \\ $$