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Question Number 45994 by peter frank last updated on 19/Oct/18

Answered by MrW3 last updated on 20/Oct/18

$$weknow\frac{x+y}{2}⩾\sqrt{xy}whenx,y⩾0.$$$$$$$$2(a+b)=L\Rightarrow a+b=\frac{L}{2}$$$$A_{rectangular}=ab⩽{\left(\frac{a+b}{2}\right)}^2={\left(\frac{L}{4}\right)}^2=areaofsquarewithsidelength\frac{L}{4}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Oct/18

$$L=2(a+b)$$$$A=ab$$$$A=a(\frac{L}{2}-a)=\frac{aL}{2}-a^2$$$$\frac{dA}{da}=\frac{L}{2}-2a$$$$formax/min\frac{dA}{da}=0so\frac{L}{2}-2a=0a=\frac{L}{4}$$$$\frac{d^{2}A}{{da}^2}=-2<0since\frac{d^{2}A}{{da}^2}is-ve$$$$soata=\frac{L}{2}therectanglehasmaxarea$$$$2a+2b=L$$$$2\times \frac{L}{4}+2b=L$$$$2b=L-\frac{L}{2}=\frac{L}{2}sob=\frac{L}{4}$$$$sincea=b=\frac{L}{4}sorectangleissquare$$$$areaofsquare=\frac{L}{4}\times \frac{L}{4}=\frac{L^2}{16}$$$$$$$$$$

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