Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 120895 by ajfour last updated on 03/Nov/20

Commented by ajfour last updated on 05/Nov/20

Find side of maximum area square. Parabola is y=x^2 .

$${Find}\:{side}\:{of}\:{maximum} \\ $$$${area}\:{square}.\:{Parabola}\:{is}\:{y}={x}^{\mathrm{2}} . \\ $$

Commented by mr W last updated on 07/Nov/20

if square and C on y−axis, i think there is only one solution: A=O, B(1,1), C(0,2)

$${if}\:{square}\:{and}\:{C}\:{on}\:{y}−{axis},\:{i}\:{think} \\ $$$${there}\:{is}\:{only}\:{one}\:{solution}: \\ $$$${A}={O},\:{B}\left(\mathrm{1},\mathrm{1}\right),\:{C}\left(\mathrm{0},\mathrm{2}\right) \\ $$

Commented by ajfour last updated on 07/Nov/20

yes sir, you must be correct.

$${yes}\:{sir},\:{you}\:{must}\:{be}\:{correct}. \\ $$

Answered by MJS_new last updated on 03/Nov/20

there′s no maximum. let C= ((0),(h) ) ⇒ A= ((((√(h−2))/2)),(((h−2)/4)) ) B= (((((√(h−2))+(√(9h−2)))/4)),(((5h−2+(√(9h−2))(√(h−2)))/8)) ) D= (((((√(h−2))−(√(9h−2)))/4)),(((5h−2−(√(9h−2))(√(h−2)))/8)) ) ⇒ the area is a=((h(√(9h−2)))/4); lim_(h→∞) a =∞

$$\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{maximum}.\:\mathrm{let}\:{C}=\begin{pmatrix}{\mathrm{0}}\\{{h}}\end{pmatrix}\:\Rightarrow \\ $$$${A}=\begin{pmatrix}{\frac{\sqrt{{h}−\mathrm{2}}}{\mathrm{2}}}\\{\frac{{h}−\mathrm{2}}{\mathrm{4}}}\end{pmatrix}\: \\ $$$${B}=\begin{pmatrix}{\frac{\sqrt{{h}−\mathrm{2}}+\sqrt{\mathrm{9}{h}−\mathrm{2}}}{\mathrm{4}}}\\{\frac{\mathrm{5}{h}−\mathrm{2}+\sqrt{\mathrm{9}{h}−\mathrm{2}}\sqrt{{h}−\mathrm{2}}}{\mathrm{8}}}\end{pmatrix} \\ $$$${D}=\begin{pmatrix}{\frac{\sqrt{{h}−\mathrm{2}}−\sqrt{\mathrm{9}{h}−\mathrm{2}}}{\mathrm{4}}}\\{\frac{\mathrm{5}{h}−\mathrm{2}−\sqrt{\mathrm{9}{h}−\mathrm{2}}\sqrt{{h}−\mathrm{2}}}{\mathrm{8}}}\end{pmatrix} \\ $$$$\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{is}\:{a}=\frac{{h}\sqrt{\mathrm{9}{h}−\mathrm{2}}}{\mathrm{4}};\:\underset{{h}\rightarrow\infty} {\mathrm{lim}}\:{a}\:=\infty \\ $$

Commented by ajfour last updated on 03/Nov/20

sir how do you obtain A(x_A , x_A ^2 ) after C(0,h) ?

$${sir}\:{how}\:{do}\:{you}\:{obtain}\:{A}\left({x}_{{A}} ,\:{x}_{{A}} ^{\mathrm{2}} \right) \\ $$$${after}\:\:{C}\left(\mathrm{0},{h}\right)\:\:? \\ $$

Commented by MJS_new last updated on 03/Nov/20

I started with P= ((x_P ),(x_P ^2 ) ) for A, B, D and C= ((0),(h) ) . then the lines must be parallel / right angled in pairs. it′s not hard to solve from there, h>0∧x_B >x_A >0∧x_D <0 and we don′t have to take 2 different solutions in any step. sorry I have no time now to write it out in detail

$$\mathrm{I}\:\mathrm{started}\:\mathrm{with}\:{P}=\begin{pmatrix}{{x}_{{P}} }\\{{x}_{{P}} ^{\mathrm{2}} }\end{pmatrix}\:\mathrm{for}\:{A},\:{B},\:{D}\:\mathrm{and} \\ $$$${C}=\begin{pmatrix}{\mathrm{0}}\\{{h}}\end{pmatrix}\:.\:\mathrm{then}\:\mathrm{the}\:\mathrm{lines}\:\mathrm{must}\:\mathrm{be}\:\mathrm{parallel}\:/ \\ $$$$\mathrm{right}\:\mathrm{angled}\:\mathrm{in}\:\mathrm{pairs}.\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{solve} \\ $$$$\mathrm{from}\:\mathrm{there},\:{h}>\mathrm{0}\wedge{x}_{{B}} >{x}_{{A}} >\mathrm{0}\wedge{x}_{{D}} <\mathrm{0}\:\mathrm{and} \\ $$$$\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{to}\:\mathrm{take}\:\mathrm{2}\:\mathrm{different}\:\mathrm{solutions} \\ $$$$\mathrm{in}\:\mathrm{any}\:\mathrm{step}.\:\mathrm{sorry}\:\mathrm{I}\:\mathrm{have}\:\mathrm{no}\:\mathrm{time}\:\mathrm{now}\:\mathrm{to} \\ $$$$\mathrm{write}\:\mathrm{it}\:\mathrm{out}\:\mathrm{in}\:\mathrm{detail} \\ $$

Commented by ajfour last updated on 03/Nov/20

thanks for as much, Sir.

$${thanks}\:{for}\:{as}\:{much},\:{Sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com