• Radius of the semicircle is ‘r’.
• Area of the rectangle is maximum.
Let us consider,
• The base of the rectangle be ‘x’ and the height be ‘y’.
Consider the ΔCEB,
CE2 = EB2 + BC2
As CE = r, and CB = y
Now the area of the rectangle is
A = x × y
Squaring on both sides
A2 = x2 y2
Substituting (1) in the above Area equation
For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function Z(x) has a maximum/minimum at a point c then Z’(c) = 0.
Differentiating the equation (2) with respect to x:
To find the critical point, we need to equate equation (3) to zero.
x(2r2 – x2) = 0
x = 0 (or) x2 = 2r2
x = 0 (or)
[as x cannot be zero]
Now to check if this critical point will determine the maximum area, we need to check with second differential which needs to be negative.
Consider differentiating the equation (3) with x:
Now let us find the value of
As , so the function Z is maximum at
Substituting x in equation (1)
As the area of the rectangle is maximum, and and
So area of the rectangle is
A = r2
Hence the maximum area of the rectangle inscribed inside a semicircle is r2 square units.
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