Home > Calculus > Solving Optimization Problems

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

Answer 1

Isabella Ackerman

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

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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Axel Alvarez

To find the dimensions of the rectangle with the largest area inscribed in a semicircle of radius ( r ), the length of the rectangle should be twice the radius of the semicircle, and the width should be equal to the radius of the semicircle. Therefore, the dimensions of the rectangle are ( 2r ) for the length and ( r ) for the width.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.