Question

Find the dimensions of the rectangle of greatest area that can be inscribed in a semi-circle of radius r.

Solution

Let ABCD be the rectangle inscribed in the semi-circle of radius r such that OC = r.
Let $angle BOC space equals space straight theta comma space space space 0 less than straight theta less than straight pi over 2 space so space that space OB space equals space straight r space cos space straight theta comma space space space BC space equals space straight r space sinθ$
$therefore space space space AB space equals space 2 comma space space space OB space equals space 2 straight r space cos space straight theta comma space space space space BC space equals space straight r space sin space straight theta$
$Let space straight P space be space area space of space rectangle space ABCD space space therefore space space space space space straight P space equals space AB. space BC space equals space 2 space straight r space cos space straight theta. space space straight r space sin space straight theta space space space space space space space space space space space space space space space equals space 2 straight r squared space sin space space straight theta space space cos space straight theta space equals space straight r squared space sin space 2 straight theta$
$therefore space space space space dP over dθ space equals space 2 straight r squared space cos space space 2 straight theta space space space space space space space space space dP over dθ space equals space 0 space space space space space space space space space space space space space space space space space rightwards double arrow space space 2 straight r squared space cos space 2 straight theta space equals space 0 rightwards double arrow space space space space space cos space 2 straight theta space equals space 0 space space space space space space space space space rightwards double arrow space space space space 2 space straight theta space equals space straight pi over 2 space space space space rightwards double arrow space space space space straight theta space equals space straight pi over 4 space straight epsilon space open parentheses 0 comma space straight pi over 2 close parentheses space space space space space fraction numerator straight d squared straight P over denominator dθ end fraction space equals space minus 4 straight r squared space sin space 2 straight theta$

$At space straight theta space equals straight pi over 4. space fraction numerator straight d squared straight P over denominator dθ squared end fraction space equals space minus 4 straight r squared space sin space straight pi over 2 space equals space minus 4 straight r squared space.1 space equals space minus 4 straight r squared less than 0 therefore space space space space straight P space is space maximum space when space straight theta space equals space straight pi over 4 therefore space space space length space of space rectangle space space equals space AB space equals space 2 space straight r space cos space straight pi over 4 space equals space 2 straight r space fraction numerator 1 over denominator square root of 2 end fraction space equals space square root of 2 space straight r$
and width of rectangle = BC = $straight r space sin straight pi over 4 space equals space straight r space sin space straight pi over 4 space equals space straight r space cross times space fraction numerator 1 over denominator square root of 2 end fraction space equals space fraction numerator straight r over denominator square root of 2 end fraction$

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Application Of Derivatives

Find the dimensions of the rectangle of greatest area that can be inscribed in a semi-circle of radius r.

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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