Respuesta :
Using the area of rectangle, the dimensions of the rectangle with maximum area are 5√2 and 5/√2.
In the given question we have to find the dimensions of the rectangle with maximum area.
Given Semicircle Radius is 5.
As we know that standard equation of circle is [tex]x^2+y^2=a^2[/tex]
a=5, so the equation is [tex]x^2+y^2=(5)^2[/tex]
[tex]x^2+y^2=25[/tex]
Therefore, Half circle y=[tex]\sqrt{25-x^2}[/tex]
Area of Rectangle(A) =xy
Because Half of the rectangle is lies in first quadrant and half lies in second quadrant
Therefore,
Area of Rectangle(A) = 2xy
Now putting the value of y
Area(A) = 2x[tex]\sqrt{25-x^2}[/tex]
Here we need to maximize the area of Rectangle, so find derivative and set equal to zero.
A'= [tex]\frac{d}{dx}(2x\sqrt{25-x^2})[/tex]
A'= 2[tex][x\frac{d}{dx}\sqrt{25-x^2}+\sqrt{25-x^2}\frac{d}{dx}x][/tex]
A'= 2[tex][x\cdot\frac{-x}{\sqrt{25-x^2}}+\sqrt{25-x^2}\cdot1][/tex]
A'= 2[tex][\frac{-x^2}{\sqrt{25-x^2}}+\sqrt{25-x^2}][/tex]
A'= 2[tex](\frac{-x^2+25-x^2}{\sqrt{25-x^2}})[/tex]
A'= 2[tex](\frac{25-2x^2}{\sqrt{25-x^2}})[/tex]
For critical Number A' = 0
2[tex](\frac{25-2x^2}{\sqrt{25-x^2}})[/tex] = 0
25-2x^2 = 0
2x^2=25
x^2=25/2
x=±√25/2
x=±5/√2
Therefore,
Base of the Rectangle = 2x
Base of the Rectangle = 2×5/√2
Base of the Rectangle = 5√2
and
height of rectangle = [tex]\sqrt{25-x^2}[/tex]
height of rectangle = [tex]\sqrt{25-(5/\sqrt{2})^2}[/tex]
height of rectangle = [tex]\sqrt{25-(25/2)}[/tex]
height of rectangle = √(50-25)/2
height of rectangle = √25/2
height of rectangle = 5/√2
Hence, the dimensions of the rectangle with maximum area are 5√2 and 5/√2.
To learn more about area of rectangle link is here
brainly.com/question/20693059
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