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Question Number 136006 by liberty last updated on 17/Mar/21

$$$$ What are the dimensions of the rectangle of maximum area that can be inscribed in the portion of the parabola x^2=8y intercepted by the line y=2?\n

Answered by mr W last updated on 18/Mar/21

$$A=2p(2-\frac{p^2}{8})$$ $$\frac{dA}{dp}=4-\frac{3p^2}{4}=0$$ $$\Rightarrow p=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}$$ $$B=2p=\frac{8\sqrt{3}}{3}$$ $$H=2-\frac{16\times 3}{8\times 9}=2-\frac{2}{3}=\frac{4}{3}$$ $$A_{max}=\frac{8\sqrt{3}}{3}\times \frac{4}{3}=\frac{32\sqrt{3}}{9}$$

Commented byotchereabdullai@gmail.com last updated on 17/Mar/21

$$\mathrm{nice}!$$

Commented byliberty last updated on 17/Mar/21

$$igot\frac{32}{3\sqrt{3}}sir$$

Commented byliberty last updated on 18/Mar/21

$$haha..yourtyposirin{2}^{nd}line$$

Commented bymr W last updated on 18/Mar/21

$$yes.ihavefixed.$$

Answered by liberty last updated on 17/Mar/21

$$letA(-x,y)andB(x,y)two$$ $$pointatparabola$$ $$theareaofrectangleisA=2x(2-y)$$ $$A\left(x\right)=2x(2-\frac{x^2}{8})=4x-\frac{x^3}{4}$$ $$A{}^{\prime }\left(x\right)=4-\frac{3}{4}{x}^2=0\Rightarrow x=\frac{4}{\sqrt{3}}$$ $$A{\left(x\right)}_{max}=\frac{8}{\sqrt{3}}(2-\frac{16}{24})=\frac{8}{\sqrt{3}}(2-\frac{2}{3})$$ $$=\frac{32}{3\sqrt{3}}$$

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