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Question Number 173908 by ali009 last updated on 20/Jul/22

find the value of b so that the line y=b  divides the region bound by the graphs of  the two functinos , into two regions of equal  area.  f(x)=9−x^2  and g(x)=0

$${find}\:{the}\:{value}\:{of}\:{b}\:{so}\:{that}\:{the}\:{line}\:{y}={b} \\ $$$${divides}\:{the}\:{region}\:{bound}\:{by}\:{the}\:{graphs}\:{of} \\ $$$${the}\:{two}\:{functinos}\:,\:{into}\:{two}\:{regions}\:{of}\:{equal} \\ $$$${area}. \\ $$$${f}\left({x}\right)=\mathrm{9}−{x}^{\mathrm{2}} \:{and}\:{g}\left({x}\right)=\mathrm{0} \\ $$

Commented by mr W last updated on 20/Jul/22

(((9−b)/9))^(3/2) =(1/2)  ⇒b=9(1−(1/( (4)^(1/3) )))=(((2−(2)^(1/3) )×9)/2)≈3.33

$$\left(\frac{\mathrm{9}−{b}}{\mathrm{9}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{b}=\mathrm{9}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{4}}}\right)=\frac{\left(\mathrm{2}−\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)×\mathrm{9}}{\mathrm{2}}\approx\mathrm{3}.\mathrm{33} \\ $$

Commented by mr W last updated on 20/Jul/22

Commented by mr W last updated on 20/Jul/22

Commented by mr W last updated on 21/Jul/22

(b_2 /b_1 )=((a_2 /a_1 ))^2  ⇒(a_2 /a_1 )=((b_2 /b_1 ))^(1/2)   A_1 =(2/3)a_1 b_1   A_2 =(2/3)a_2 b_2   (A_2 /A_1 )=((a_2 b_2 )/(a_1 b_1 ))=((a_2 /a_1 ))((b_2 /b_1 ))=((b_2 /b_1 ))^(1/2) ((b_2 /b_1 ))=((b_2 /b_1 ))^(3/2)

$$\frac{{b}_{\mathrm{2}} }{{b}_{\mathrm{1}} }=\left(\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }\right)^{\mathrm{2}} \:\Rightarrow\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }=\left(\frac{{b}_{\mathrm{2}} }{{b}_{\mathrm{1}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${A}_{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}}{a}_{\mathrm{1}} {b}_{\mathrm{1}} \\ $$$${A}_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}}{a}_{\mathrm{2}} {b}_{\mathrm{2}} \\ $$$$\frac{{A}_{\mathrm{2}} }{{A}_{\mathrm{1}} }=\frac{{a}_{\mathrm{2}} {b}_{\mathrm{2}} }{{a}_{\mathrm{1}} {b}_{\mathrm{1}} }=\left(\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }\right)\left(\frac{{b}_{\mathrm{2}} }{{b}_{\mathrm{1}} }\right)=\left(\frac{{b}_{\mathrm{2}} }{{b}_{\mathrm{1}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{{b}_{\mathrm{2}} }{{b}_{\mathrm{1}} }\right)=\left(\frac{{b}_{\mathrm{2}} }{{b}_{\mathrm{1}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$

Commented by Erikyatusabes last updated on 21/Jul/22

  (((9−b)/9))^(3/2) =(1/2)  ⇒b=9(1−(1/( (4)^(1/3) )))=(((2−(2)^(1/3) )×9)/2)≈3.33

$$ \\ $$$$\left(\frac{\mathrm{9}−{b}}{\mathrm{9}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{b}=\mathrm{9}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{4}}}\right)=\frac{\left(\mathrm{2}−\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)×\mathrm{9}}{\mathrm{2}}\approx\mathrm{3}.\mathrm{33} \\ $$

Commented by Tawa11 last updated on 21/Jul/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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