1.a Norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular window. fIND THE DIMENSION OF A nORMAN WINDOW of maximum area if the total perimeter is 16 feet.
2. Find the dimension of the trapezoid of greatest area that can be inscribed in a semicircle of radius r.
3. Find the dimension of the largest isosceles triangle that can be inscribed in a circle of radius of 4.
Can someone help me solve these problems? I'm really stuck right now. Thanx.
The first one isn't bad.
P = ((pi r)/2 + 2h + w) ; w = 2r ; P = (pi r + 4h + 4r) / 2
A = (pi r^2)/2 + hw ; A = (pi r^2 + 4rh) / 2
Solve P for h and place in area function
h = 2P - 4r - pi r
A = (pi r^2 + 8Pr - 16 r^2 - 4pi r^2) / 2 = (-3 pi r^2 - 16 r^2 + 8Pr) / 2
dA/dr = (-6 pi r -32r + 8P) / 2
dA/dr = 0 when 8P = r(32 + 6 pi), r = 8P / (32 + 6 pi) ≈2.157
r = 2.157, w = 5.034, h = 16.595 (total height is 16.595 + 2.157)
I figured out number three for you.
I am sorry that I can't draw this for you, but maybe you can draw it yourself and follow along. Any triangle will meet the circle in three places. Using the top vertice as a bisector of the circle you notice that it also bisects the base of the triangle also. Now if you draw a line from the center of the circle to one of the triangle's lower vertices you will see another right triangle appear with the hypotenuse being the radius of the circle and the base being that of half of the triangle's base. The other side will be the amount of distance greater than the radius that the height of the triangle will occupy. I hope that you can see this.
So, h = (r^2 - b^2/4)^(1/2) + 4
A = bh/2
A = b/2 [r + (r^2 - b^2/4)^(1/2)
A = (b^2 r^2/ 4 - b^4/ 16)^(1/2) + br/ 2
Substituting in our value of r = 4 and taking the derivative gives:
dA/db = [(32b - b^3) / 8 (4b^2 - .25b^4)^(1/2)] + 2
This derivative equals 0 when b = 48^(1/2) = 6.9282
h = 6
Maximum area is 3 (48)^(1/2) = 20.78
I hope that helped, I will try and work on the trapazoid later.