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

1 Answer
Narad T.
Aug 12, 2018

The dimensions of the rectangle is sqrt2r√2r and r/sqrt2r√2

Explanation:

The equation of the semicircle is

x^2+y^2=r^2x2+y2=r2.......................(1)(1)

The area of the rectangle is

A=2xyA=2xy....................(2)(2)

From equation (1)(1), we get

y^2=r^2-x^2y2=r2−x2

y=sqrt(r^2-x^2)y=√r2−x2

Plugging this value in equation (2)(2)

A=2xsqrt(r^2-x^2)A=2x√r2−x2

Differentiating wrt xx using the product rule

(dA)/dx=2sqrt(r^2-x^2)-2x^2/sqrt(r^2-x^2)dAdx=2√r2−x2−2x2√r2−x2

=(2r^2-2x^2-2x^2)/(sqrt(r^2-x^2))=2r2−2x2−2x2√r2−x2

=(2r^2-4x^2)/(sqrt(r^2-x^2))=2r2−4x2√r2−x2

The critical points are when

(dA)/dx=0dAdx=0

That is

(2r^2-4x^2)/(sqrt(r^2-x^2))=02r2−4x2√r2−x2=0

r^2=2x^2r2=2x2

x=r/sqrt2x=r√2

Then,

y=sqrt(r^2-x^2)=sqrt(r^2-r^2/2)=r/sqrt2y=√r2−x2=√r2−r22=r√2

The maximum area is

A=2*r/sqrt2*r/sqrt2=r^2A=2⋅r√2⋅r√2=r2

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