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-

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?

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

A=2p(2−(p^2 /8)) (dA/dp)=4−((3p^2 )/4)=0 ⇒p=(4/( (√3)))=((4(√3))/3) B=2p=((8(√3))/3) H=2−((16×3)/(8×9))=2−(2/3)=(4/3) A_(max) =((8(√3))/3)×(4/3)=((32(√3))/9)

$${A}=\mathrm{2}{p}\left(\mathrm{2}−\frac{{p}^{\mathrm{2}} }{\mathrm{8}}\right) \\ $$$$\frac{{dA}}{{dp}}=\mathrm{4}−\frac{\mathrm{3}{p}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$$\Rightarrow{p}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${B}=\mathrm{2}{p}=\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${H}=\mathrm{2}−\frac{\mathrm{16}×\mathrm{3}}{\mathrm{8}×\mathrm{9}}=\mathrm{2}−\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${A}_{{max}} =\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{3}}×\frac{\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{9}} \\ $$

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

$$\mathrm{nice}! \\ $$

Commented by liberty last updated on 17/Mar/21

$${i}\:{got}\:\frac{\mathrm{32}}{\mathrm{3}\sqrt{\mathrm{3}}}\:{sir} \\ $$

Commented by liberty last updated on 18/Mar/21

$${haha}..{your}\:{typo}\:{sir}\:{in}\:\mathrm{2}^{{nd}} \:{line} \\ $$

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

$${yes}.\:{i}\:{have}\:{fixed}. \\ $$

Answered by liberty last updated on 17/Mar/21

let A(−x,y) and B(x,y) two point at parabola the area of rectangle is A=2x(2−y) A(x)= 2x(2−(x^2 /8))=4x−(x^3 /4) A ′(x)= 4−(3/4)x^2 =0⇒x=(4/( (√3))) A(x)_(max) = (8/( (√3))) (2−((16)/(24)))=(8/( (√3)))(2−(2/3)) = ((32)/(3(√3)))

$${let}\:{A}\left(−{x},{y}\right)\:{and}\:{B}\left({x},{y}\right)\:{two} \\ $$$${point}\:{at}\:{parabola}\: \\ $$$${the}\:{area}\:{of}\:{rectangle}\:{is}\:{A}=\mathrm{2}{x}\left(\mathrm{2}−{y}\right) \\ $$$${A}\left({x}\right)=\:\mathrm{2}{x}\left(\mathrm{2}−\frac{{x}^{\mathrm{2}} }{\mathrm{8}}\right)=\mathrm{4}{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{4}} \\ $$$${A}\:'\left({x}\right)=\:\mathrm{4}−\frac{\mathrm{3}}{\mathrm{4}}{x}^{\mathrm{2}} =\mathrm{0}\Rightarrow{x}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}} \\ $$$${A}\left({x}\right)_{{max}} =\:\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}\:\left(\mathrm{2}−\frac{\mathrm{16}}{\mathrm{24}}\right)=\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}\left(\mathrm{2}−\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$$=\:\frac{\mathrm{32}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$

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