Find the rectangle of largest area that can be inscribed in a semicircle of diameter 33.?

Question

i tried doing it by using teh area of circle formula and tehn finding derivative but i know i am making a mistake some where please help

The Question is

Find the rectangle of largest area that can be inscribed in a semicircle of diameter 33, assuming that one side of the rectangle lies on the diameter of the semicircle.
The largest possible area is ?

Answer 1

It's the square that has a diagonal of 33. The area of the circle is irrelevant you need to write an expression for the area of the rectangle then take the derivative of that.

Answer 2

Let the center of the circle be the origin of a coordinate system with the diameter along the x-axis. We can call the two vertices of the rectangle on the diameter (-a,0) and (a,0). By symmetry they should have opposite x-coordinates. We can call the other two vertices (which lie on the semicircle) (-a, b) and (a,b). The area of our rectangle will be A = (2a)b. Now we need to relate a to b before we can differentiate. To do that use the eqn for the semicircle: y = sqrt(-x^2 + 17.5^2). Now just let y=b and x=a and substitute into your eqn. for area above. Then differentiate and go from there. See if you can finish.

Answer 3

x^2 + y^2 = 4: equation of circle, consider y positive, the semi-circle The points of the rectangle that are inscribed are found by drawing a triangle in the first and third quadrant that intersects the semicircle at the point (sqrt(2),sqrt(2)), (sqrt(2),0), (-sqrt(2),0), (-sqrt(2),sqrt(2)) Why? because the hypotenuse of the triangle from (0,0) to (sqrt(2),2) is the radius of length 2. Its respective legs are sqrt(2) and sqrt(2) in length 2^2 = sqrt(2)^2 + sqrt(2)^2 The area of the rectangle is 2sqrt(2)*sqrt(2) = 4 cm^2