How do you find the dimensions of the rectangle with largest area that can be inscribed in a semicircle of radius #r# ?

1 Answer
Aug 12, 2018

Answer:

The dimensions of the rectangle is #sqrt2r# and #r/sqrt2#

Explanation:

The equation of the semicircle is

#x^2+y^2=r^2#.......................#(1)#

The area of the rectangle is

#A=2xy#....................#(2)#

From equation #(1)#, we get

#y^2=r^2-x^2#

#y=sqrt(r^2-x^2)#

Plugging this value in equation #(2)#

#A=2xsqrt(r^2-x^2)#

Differentiating wrt #x# using the product rule

#(dA)/dx=2sqrt(r^2-x^2)-2x^2/sqrt(r^2-x^2)#

#=(2r^2-2x^2-2x^2)/(sqrt(r^2-x^2))#

#=(2r^2-4x^2)/(sqrt(r^2-x^2))#

The critical points are when

#(dA)/dx=0#

That is

#(2r^2-4x^2)/(sqrt(r^2-x^2))=0#

#r^2=2x^2#

#x=r/sqrt2#

Then,

#y=sqrt(r^2-x^2)=sqrt(r^2-r^2/2)=r/sqrt2#

The maximum area is

#A=2*r/sqrt2*r/sqrt2=r^2#