Question-97068

Question Number 97068 by bemath last updated on 06/Jun/20

Commented by bemath last updated on 06/Jun/20

find the dimension of the rectangle and radius of the circle so area rectangle + area circle is maximum

$$\mathrm{find}\:\mathrm{the}\:\mathrm{dimension}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle} \\ $$$$\mathrm{and}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{so}\:\mathrm{area}\:\mathrm{rectangle} \\ $$$$+\:\mathrm{area}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{maximum} \\ $$

Answered by john santu last updated on 06/Jun/20

Commented by john santu last updated on 06/Jun/20

the radius R = (((a−x))/(2+(√2))) let A(R,x) = πR^2 +x(a−x) ⇔A(x) = ((π(a−x)^2 )/((2+(√2))^2 )) + x(a−x) let p = (π/((2+(√2))^2 )) ⇔A(x)= p(a−x)^2 +x(a−x) A(x) = (a−x){p(a−x)+x} A(x)= (a−x)(x−px+ap) then A ′(x)= −(x−px+ap)+(a−x)(1−p)=0 −x+px−ap+a−ap−x+px= 0 2px−2x = 2ap−a x= ((2ap−a)/(2p−2)) = ((2a((π/(6+4(√2))))−a)/(((2π)/(6+4(√2)))−2)) x = (((π−3−2(√2))a)/(π−4(√2)−6)) & R = (((2+(√2))a)/(12+8(√2)−π))

$$\mathrm{the}\:\mathrm{radius}\:\mathrm{R}\:=\:\frac{\left({a}−{x}\right)}{\mathrm{2}+\sqrt{\mathrm{2}}} \\ $$$$\mathrm{let}\:\mathrm{A}\left(\mathrm{R},{x}\right)\:=\:\pi\mathrm{R}^{\mathrm{2}} +{x}\left({a}−{x}\right) \\ $$$$\Leftrightarrow\mathrm{A}\left({x}\right)\:=\:\frac{\pi\left({a}−{x}\right)^{\mathrm{2}} }{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\:+\:{x}\left({a}−{x}\right) \\ $$$$\mathrm{let}\:{p}\:=\:\frac{\pi}{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\Leftrightarrow\mathrm{A}\left({x}\right)=\:{p}\left({a}−{x}\right)^{\mathrm{2}} +{x}\left({a}−{x}\right) \\ $$$$\mathrm{A}\left({x}\right)\:=\:\left({a}−{x}\right)\left\{{p}\left({a}−{x}\right)+{x}\right\} \\ $$$$\mathrm{A}\left({x}\right)=\:\left({a}−{x}\right)\left({x}−{px}+{ap}\right) \\ $$$$\mathrm{then}\:\mathrm{A}\:'\left({x}\right)=\:−\left({x}−{px}+{ap}\right)+\left({a}−{x}\right)\left(\mathrm{1}−{p}\right)=\mathrm{0} \\ $$$$−{x}+{px}−{ap}+{a}−{ap}−{x}+{px}=\:\mathrm{0} \\ $$$$\mathrm{2}{px}−\mathrm{2}{x}\:=\:\mathrm{2}{ap}−{a} \\ $$$${x}=\:\frac{\mathrm{2}{ap}−{a}}{\mathrm{2}{p}−\mathrm{2}}\:=\:\frac{\mathrm{2}{a}\left(\frac{\pi}{\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}}\right)−{a}}{\frac{\mathrm{2}\pi}{\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}}−\mathrm{2}} \\ $$$${x}\:=\:\frac{\left(\pi−\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right){a}}{\pi−\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{6}}\:\&\: \\ $$$$\mathrm{R}\:=\:\frac{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){a}}{\mathrm{12}+\mathrm{8}\sqrt{\mathrm{2}}−\pi} \\ $$$$ \\ $$

Commented by bemath last updated on 06/Jun/20

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\: \\ $$

Commented by bobhans last updated on 06/Jun/20

coolll

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