# A rectangle is to be placed in a semicircle of radius 2. What can be the largest area of the rectangle and what are its dimensions.

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### 2 Answers

Let the length of the rectangle be 2x. When the rectangle is placed in the semicircle we see that:

length is 2x

width is sqrt [4- x^2]

Therefore the area is 2x*sqrt [4- x^2]

Now we have a function for the area = f(x) = 2x*sqrt [4- x^2]

To find the maximum value for area we need to find the derivative of f(x) and equate it to 0.

f’(x) = 2x*(-2x) * (4- x^2) ^ (-1/2) + 2* sqrt (4-x^2) =0

=> -2*x^2/ sqrt (4- x^2) + 2 sqrt (4- x^2) =0

=> -2*x^2/ sqrt (4- x^2) = -2 sqrt (4- x^2) =0

=>2*x^2 = 2*(4-x^2)

=> 4*x^2 = 8

=> x^2 = 2

=> x = 2 or x= -2

For 0<= x <= 2,

the critical values are f (sqrt 2) = 2sqrt 2* sqrt (4-2) = 4

f (2) =0

the endpoint values are f (0) = 0

f (2) = 0

Therefore the maximum f(x) is 4.

**So, the maximum area is 4. The dimensions are sqrt 2 and 2* sqrt 2.**

Let AB be the diameter, and O be the centre of the circle of radius diameter. To find the rectangle of maximum area.

Let C and D be the points on either side of the centere O from which CE and DF are drawn perpendicular to the diameter AB to meet the circumference at E and F on one side and E' and F' on the other side of the circle.

Therefore , the rectangle CDFE has the area = CD*CE.

But OCE is a right angled triangle with OC = x say. And CE^2 = OE^2-x^2 = r^2-x^2 = 2^2 = x^2 = 4-x^2 as r = 2 is given.

So CD = 2x and CE = sqrt(4-x^2).

Therefore Area of the rectangle A(x) = 2x*sqrt(4-x^2).

A(x) is therefore maximum for x = c, where c is a solution of A '(x) = 0 and A"(c) < 0.

A'(x) = {xsqr(4-x^2)}' = (x)'sqr(4-x^2) +x{sqrt(4-x^2)}'

A'(x)=1*sqrt(4-x^2) + {x/2sqr(4-x^2)}(4-x^2)'

A'(x)=sqrt(4-x^2)+x(-2x)/2sqrt(4-x^2).

A'(x)= {2(4-x^2) -2x^2)/2sqrt(4-x^2) = 4(2-x^2)/2sqrt(4-x^2)

A'(x) = (2-x^2)/sqr(4-x^2).

A'(x) = 0 gives: Numerator 2-x^2 = 0. Or x = sqrt 2.

A"(x) = {2-x^2)/sqrt(4-x^2)}' = {-2x sqrt(4-x^2) -(2-x^2) (-2x)/2sqrt(4-x^2)}/(4-x^2).

= -2x{sqr(4-x^2)-(2-x^2)/(4-x^2)}/(4-x^2)

= -2x{(4-x^2 -2+x^2}/(4-x^2)^(3/2)

= -2x*2/(4-x^2)^(3/2)

A"(sqrt2) : -2*2/(4-2)^3/2 <0.

So for x = sqrt2, the area of the recrangle A(x) = 2x^2 = 2(sqrt2)^2 = 4sq units is the maximum area.