Problems involving differentiation applications..headache?

1)An isoceles triangle ABC with AB=AC, is inscribed in a fixed circle, centre O whose radius is 1 unit. Angle BOC=2(tetta) where tetta is acute. Express the area of the triangle ABC in terms of tetta and hence show that as tetta varies, the area if a max when triangle is equilaterial. ( tetta used here = the sign for an angle)

2) A beam is to be cut from a cylindrical log so that its cross-section is a rectangle. The log has diameter d and the beam is to have breadth x cm and depth y. Given that the stiffness of such a beam is proportional to xy^3, find , in terms of d, the values of x and y for the stiffest beam that can be cut from the log.

Problems involving differentiation applications..headache?pacific theater

let denote tetta by x and area by y . draw perpendicular AD on BC

now AD is height of the triangle.

AD = AO + OD = 1 + cosx

BC = 2 BD = 2 sinx

area y = 1/2 * AD * BC = 1/2 * (1+cosx) *2sinx = sinx +sinx * cosx

now dy/dx = cosx + cos2x

for maximum area dy/dx =0

cosx + cos2x = 0

2 cosx/2 * cos3x/2 =0

x = 60 or 180 , since 180 is not acceptable so x =60

so, angle BOC = 2x =120

So, angle BAC = 60

now since AB = AC .so , angle ABC = angle ACB

so, BAC = ABC = ACB = 60

2) since rectangle is inscribed in the circle so,

x^2 + y^2 = d^2 ......equation(1)

now stiffness s = xy^3 = sqrt(d^2 - y^2) * y^3

for maximum stifness ds/dy should be zero

now , ds/dy = 3y^2 * sqrt(d^2 - y^2) + y^3 * (-2y / sqrt(d^2 - y^2)) = (3d^2 * y^2 - y^4)/sqrt(d^2 - y^2) =0

so, 3d^2 * y^2 - 5y^4 = 0

3d^2 - 5y^2 =0

y = sqrt(3/5) * d = 0.775d

put the value of y in equation(1)

x = sqrt(2/5) * d = 0.632d