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Question Number 3830 by Rasheed Soomro last updated on 21/Dec/15

Draw a rectangle of maximum perimeter,  by ruler and compass,when area is ab.   (AB =a,CD=b are given.)

$$\mathcal{D}{raw}\:{a}\:{rectangle}\:{of}\:{maximum}\:{perimeter}, \\ $$$${by}\:{ruler}\:{and}\:{compass},{when}\:{area}\:{is}\:\boldsymbol{\mathrm{ab}}.\: \\ $$$$\left(\boldsymbol{\mathrm{AB}}\:=\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{CD}}=\boldsymbol{\mathrm{b}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{given}}.\right) \\ $$

Commented by prakash jain last updated on 22/Dec/15

Area ab, let us say one side is x  Perimeter =2(((ab)/x)+x)  lim_(x→0)  (((ab)/x)+x)→∞  There is no maximum perimeter for given  area.

$$\mathrm{Area}\:{ab},\:\mathrm{let}\:\mathrm{us}\:\mathrm{say}\:\mathrm{one}\:\mathrm{side}\:\mathrm{is}\:{x} \\ $$$$\mathrm{Perimeter}\:=\mathrm{2}\left(\frac{{ab}}{{x}}+{x}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{{ab}}{{x}}+{x}\right)\rightarrow\infty \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{maximum}\:\mathrm{perimeter}\:\mathrm{for}\:\mathrm{given} \\ $$$$\mathrm{area}. \\ $$

Commented by Yozzii last updated on 22/Dec/15

P(x)=2((c/x)+x)⇒(dP/dx)=2(1−(c/x^2 ))=2(1−cx^(−2) )  (dP/dx)=0⇒ x=±(√c) ∵x>0⇒x=(√c)  (d^2 P/dx^2 )=2(2cx^(−3) )>0 if x>0  At x=(√c)=(√(ab)), (d^2 P/dx^2 )>0⇒minimum pt.  for P exists only since x>0.

$${P}\left({x}\right)=\mathrm{2}\left(\frac{{c}}{{x}}+{x}\right)\Rightarrow\frac{{dP}}{{dx}}=\mathrm{2}\left(\mathrm{1}−\frac{{c}}{{x}^{\mathrm{2}} }\right)=\mathrm{2}\left(\mathrm{1}−{cx}^{−\mathrm{2}} \right) \\ $$$$\frac{{dP}}{{dx}}=\mathrm{0}\Rightarrow\:{x}=\pm\sqrt{{c}}\:\because{x}>\mathrm{0}\Rightarrow{x}=\sqrt{{c}} \\ $$$$\frac{{d}^{\mathrm{2}} {P}}{{dx}^{\mathrm{2}} }=\mathrm{2}\left(\mathrm{2}{cx}^{−\mathrm{3}} \right)>\mathrm{0}\:{if}\:{x}>\mathrm{0} \\ $$$${At}\:{x}=\sqrt{{c}}=\sqrt{{ab}},\:\frac{{d}^{\mathrm{2}} {P}}{{dx}^{\mathrm{2}} }>\mathrm{0}\Rightarrow{minimum}\:{pt}. \\ $$$${for}\:{P}\:{exists}\:{only}\:{since}\:{x}>\mathrm{0}. \\ $$$$ \\ $$

Commented by Rasheed Soomro last updated on 22/Dec/15

Thank^S !

$$\mathbb{T}\mathrm{han}\Bbbk^{\mathcal{S}} ! \\ $$

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