show that the triangle.....
Asked by | 10th Oct, 2009, 05:53: PM
Since the we are talking about the triangle inscribed in a GIVEN CIRCLE, the radius r is constant.
A(x) = (r+x)(r2-x2)1/2 = uv where u = (r+x), v = (r2-x2)1/2
dA(x)/dx = d(uv)/dx = udv/dx + vdu/dx ..................using the product rule.
du/dx = 1, and dv/dx = (1/2)(r2-x2)-1/2(-2x)....................using the differentiation rules..chain rule.
Substituting these, the first derivative is,
dA(x)/dx = (r+x)(1/2)(r2-x2)-1/2(-2x) + (r2-x2)1/2
dA(x)/dx = (r+x)(r2-x2)-1/2(-x) + (r2-x2)1/2 = U/v + v .....where U = -x(r+x)
Similarly the second derivative can be evaluated, using the quotient rule of differentiation,
d2A(x)/dx2 = (vdU/dx - Udv/dx)/v2 + dv/dx
Now to find the value of x that maximizes the area, we set first derivative equal to zero,
(r+x)(r2-x2)-1/2 (-x) + (r2-x2)1/2 = 0
rx + x2 - r2 + x2 = 0,
x = (- r ± 3r)/4, x = r/2,
So the base is r+x =3r/2, so we can similarly draw perpendicular to another side, and following same arguments, we find it's also 3r/2 and again for the third.
It shows that it's an equilateral triangle.
Answered by | 11th Oct, 2009, 11:51: AM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number