find the perimeter of the rectangle with maximum area that can be inscribed in a semicircle of radius 2 ft. if two of its vertices are on the diameter of the semicircle and the other two are on the semicircle.
calculas - Reiny, Tuesday, December 7, 2010 at 10:06pm
Did you make a sketch?
draw a line from the centre to a vertex of the rectangle lying on the semicircle.
Let the length of the rectangle be 2x and its height y
I see a right-angled triangle with sides x and y and hypotenuse 2
x^2 + y^2 = 4
let the area be A
A = 2xy
A^2 = 4x^2y^2 = 4x^2(4-x^2)
= 16x^2 - 4x^4
2A dA/dx = 32x - 16x^3 = 0 for max of A
16x(2 - x^2) = 0
sub back into equation above
y = √2
so length = 2√2 and width = √2
Perimeter = 6√2
calculas - Anonymous, Tuesday, December 7, 2010 at 10:10pm
thank you so muchh :)