Manipal MET2013MathematicsApplication of Derivatives
A B is a diameter of a circle and C is any point on the circumference of the circle. then
Options
- AThe area of A B C is maximum when it is isosceles
- BThe area of A B C is minimum when it is isosceles
- CThe perimeter of A B C is minimum when it is isosceles
- DNone of these
Correct answer
A. The area of A B C is maximum when it is isosceles
Step-by-step solution
Area of A B C, A= 1 2 d^2-x^2 for / d A d x =0 1 2 d^2-x^2 + 1 2 -2 x 2 d^2-x^2 =0 d^2-x^2-x^2 2 d^2-x^2 =0 x= d 2 Also for x d 2 , d A d x 0 and for x d 2 , d A d x 0 . Hence; x= d 2 is the point of maxima. A is max at x= d 2 , the area is maximum y= d 2 when is isosceles,