The dimensions of the rectangle having greatest area in the piece of plexiglass which is a semicircle is [tex]2\sqrt{2}[/tex] and [tex]\sqrt{2}[/tex].
Given radius of piece of plexiglass which is a semi circle equal to 2m.
We have to find the dimensions of the rectangle which is inscribed in the semi circle.
Draw a rectangle in a semicircle.
Suppose the length of rectangle be x and width be y.
Draw a line from the center to the vertex of the rectangle which forms a right angled triangle with sides x,y and 2.
[tex](x/2)^{2} +y^{2} =2^{2}[/tex]
y=[tex]\sqrt{16-x^{2} }/2[/tex]
Area of rectangle=2xy
=x[tex]\sqrt{16-x^{2} }[/tex]/2
Finding the derivative of area
d A/dx=1/2[x(-2x)/2[tex]\sqrt{16-x^{2} }[/tex]+[tex]\sqrt{16-x^{2} }[/tex]]
d A/dx=0
1/2[[tex]-2x^{2} /2\sqrt{16-x^{2} }+\sqrt{16-x^{2} } ][/tex]=0
Solving for x
[tex]\sqrt{16-x^{2}[/tex]=[tex]x^{2} /\sqrt{16-x^{2} }[/tex]
16-[tex]x^{2}[/tex]=[tex]x^{2}[/tex]
16=2[tex]x^{2}[/tex]
[tex]x^{2}[/tex]=8
x=2[tex]\sqrt{2}[/tex]
put the value of x in y=[tex]\sqrt{16-x^{2} }/2[/tex] to get the value of y
y=[tex]\sqrt{16-8} /2[/tex]
=[tex]\sqrt{8} /2[/tex]
= 2[tex]\sqrt{2} /2[/tex]
=[tex]\sqrt{2}[/tex]
Hence the dimensions of rectangle which is in a semi circle of radius 2 is [tex]\sqrt{2} and 2\sqrt{2}[/tex].
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