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Question-203557




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

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