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Question-150056

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Question Number 150056 by ajfour last updated on 09/Aug/21
Commented by ajfour last updated on 09/Aug/21
If the blue region is a square of maximum area, find a and side s of the square.
$${If}\:{the}\:{blue}\:{region}\:{is}\:{a}\:{square} \\ $$$${of}\:{maximum}\:{area},\:{find}\:{a} \\ $$$${and}\:{side}\:{s}\:{of}\:{the}\:{square}.\:\:\: \\ $$
Answered by mr W last updated on 09/Aug/21
y=(x−a)^2 =x x^2 −(2a+1)x+a^2 =0 s=x=((2a+1−(√(4a+1)))/2) (ds/da)=0 a−(2/( (√(4a+1))))=0 a^2 =(4/(4a+1)) (1/a^3 )−(1/(4a))−1=0 (1/a)=((((√(1293))/(72))+(1/2)))^(1/3) −((((√(1297))/(72))−(1/2)))^(1/3) ⇒a≈0.9232
$${y}=\left({x}−{a}\right)^{\mathrm{2}} ={x} \\ $$$${x}^{\mathrm{2}} −\left(\mathrm{2}{a}+\mathrm{1}\right){x}+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${s}={x}=\frac{\mathrm{2}{a}+\mathrm{1}−\sqrt{\mathrm{4}{a}+\mathrm{1}}}{\mathrm{2}} \\ $$$$\frac{{ds}}{{da}}=\mathrm{0} \\ $$$${a}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}{a}+\mathrm{1}}}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{4}{a}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4}{a}}−\mathrm{1}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{a}}=\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{1293}}}{\mathrm{72}}+\frac{\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{1297}}}{\mathrm{72}}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow{a}\approx\mathrm{0}.\mathrm{9232} \\ $$
Commented by ajfour last updated on 09/Aug/21
thank u sir, easy 4 u.
$${thank}\:{u}\:{sir},\:{easy}\:\mathrm{4}\:{u}. \\ $$
Commented by ajfour last updated on 09/Aug/21
no it is not so, sir. for a given a, the area might be max. for a rectangle; And for this maximum area rectangle, to be a square as well , there has to be some specific a value, and hence s value.
$${no}\:{it}\:{is}\:{not}\:{so},\:{sir}. \\ $$$${for}\:{a}\:{given}\:{a},\:{the}\:{area}\:{might} \\ $$$${be}\:{max}.\:{for}\:{a}\:{rectangle}; \\ $$$${And}\:{for}\:{this}\:{maximum}\:{area} \\ $$$${rectangle},\:{to}\:{be}\:{a}\:{square}\:{as}\:{well} \\ $$$$,\:{there}\:{has}\:{to}\:{be}\:{some}\:{specific} \\ $$$$\:{a}\:{value},\:{and}\:{hence}\:{s}\:{value}. \\ $$
Commented by mr W last updated on 09/Aug/21
but in fact there is no absolute maximum or minimum. since s→∞ when a→∞ s→0 when a→0
$${but}\:{in}\:{fact}\:{there}\:{is}\:{no}\:{absolute} \\ $$$${maximum}\:{or}\:{minimum}.\:{since} \\ $$$${s}\rightarrow\infty\:{when}\:{a}\rightarrow\infty \\ $$$${s}\rightarrow\mathrm{0}\:{when}\:{a}\rightarrow\mathrm{0} \\ $$
Commented by mr W last updated on 09/Aug/21
the upper right corner of the square on the parabola is the intersection point of y=(x−a)^2 and y=x. we see the intersection point always exists for any value of a from 0 till ∞.
$${the}\:{upper}\:{right}\:{corner}\:{of}\:{the}\:{square} \\ $$$${on}\:{the}\:{parabola}\:{is}\:{the}\:{intersection} \\ $$$${point}\:{of}\:{y}=\left({x}−{a}\right)^{\mathrm{2}} \:{and}\:{y}={x}. \\ $$$${we}\:{see}\:{the}\:{intersection}\:{point}\:{always} \\ $$$${exists}\:{for}\:{any}\:{value}\:{of}\:{a}\:{from}\:\mathrm{0}\:{till}\:\infty. \\ $$
Commented by mr W last updated on 09/Aug/21
if you mean the case that the square happens to be the maximum rectangle, then there is such a value for a.
$${if}\:{you}\:{mean}\:{the}\:{case}\:{that}\:{the}\:{square} \\ $$$${happens}\:{to}\:{be}\:{the}\:{maximum}\:{rectangle}, \\ $$$${then}\:{there}\:{is}\:{such}\:{a}\:{value}\:{for}\:{a}. \\ $$
Answered by mr W last updated on 09/Aug/21
say the upper right corner of the rectangle is (p,(p−a)^2 ). area of rectangle is A=p(p−a)^2 (dA/dp)=(p−a)^2 +2p(p−a)=0 p=(a/3) such that the maximum rectangle is a square, p=(p−a)^2 (a/3)=((a/3)−a)^2 ⇒a=(3/4)
$${say}\:{the}\:{upper}\:{right}\:{corner}\:{of}\:{the} \\ $$$${rectangle}\:{is}\:\left({p},\left({p}−{a}\right)^{\mathrm{2}} \right). \\ $$$${area}\:{of}\:{rectangle}\:{is} \\ $$$${A}={p}\left({p}−{a}\right)^{\mathrm{2}} \\ $$$$\frac{{dA}}{{dp}}=\left({p}−{a}\right)^{\mathrm{2}} +\mathrm{2}{p}\left({p}−{a}\right)=\mathrm{0} \\ $$$${p}=\frac{{a}}{\mathrm{3}} \\ $$$${such}\:{that}\:{the}\:{maximum}\:{rectangle} \\ $$$${is}\:{a}\:{square}, \\ $$$${p}=\left({p}−{a}\right)^{\mathrm{2}} \\ $$$$\frac{{a}}{\mathrm{3}}=\left(\frac{{a}}{\mathrm{3}}−{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by ajfour last updated on 09/Aug/21
Yes, this is answer to my question, sir. thanks for all efforts. really good.
$${Yes},\:{this}\:{is}\:{answer}\:{to}\:{my} \\ $$$${question},\:{sir}.\:{thanks}\:{for} \\ $$$${all}\:{efforts}.\:{really}\:{good}. \\ $$

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