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Question Number 203557 by ajfour last updated on 21/Jan/24

Commented by ajfour last updated on 21/Jan/24

$$Findmaimumblueshadedarea.$$

Answered by mr W last updated on 22/Jan/24

Commented by mr W last updated on 22/Jan/24

$$H=\frac{3\times 4}{5}=\frac{12}{5}$$$$\frac{a}{5}=\frac{H-h}{H}\Rightarrow a=5(1-\frac{5h}{12})$$$$\frac{b}{a}=\frac{H-h}{H}\Rightarrow b=5{(1-\frac{5h}{12})}^2$$$$A=\frac{(a+b)h}{2}=6\times \frac{5h}{12}(1-\frac{5h}{12})(2-\frac{5h}{12})$$$$A=6x(1-x)(2-x)withx=\frac{5h}{12}<1$$$$\frac{dA}{dx}=6(1-x)(2-x)-6x(2-x)-6x(1-x)=0$$$$3x^2-6x+2=0$$$$\Rightarrow x=1-\frac{\sqrt{3}}{3}(1+\frac{\sqrt{3}}{3}rejected)$$$$A_{max}=6(1-\frac{\sqrt{3}}{3})\left(\frac{\sqrt{3}}{3}\right)(1+\frac{\sqrt{3}}{3})=\frac{4\sqrt{3}}{3}✓$$

Commented by ajfour last updated on 22/Jan/24

$$ThanksSir.I^{\prime }lltrysoontoo.$$

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