A window frame - maximization problem

[Guest]

A window frame is in the shape of a semi-circle jointed to a rectangle. Find the maximum area of a window using 300cm of the framework.


My working:

Circ'ference = 2 x pi x r
perimeter = pi x r + w + 2L = 300cm
ie. pi * r + 2r + 2 L = 300cm
ie. L = 300 - pi * r - 2 * r/2

Area of rectangle will be = 2 * r * L
Area of semicircle = 1/2 pi r^2
total area = 2 * r * L + 1/2 pi r^2 ie. r( pi * r /2 + 2L)
therefore
L = 300r - 1/2 pi * r^2 - 2 * r ^2
finding its first derivative we get... 300 - pi * r - 4r
if we let that = 0 then
0 = 300 - pi * r - 4r
thus r = 300/pi + 4

We then use this in the original formula and that should answer it. Correct?

 

[galactus]

Since r is the radius of the semicircle(You could use r=x/2 also).

The perimeter of the framework is:

\(\displaystyle 4r+2y+{\pi}r=300\)

\(\displaystyle A=2ry+\frac{{\pi}r^{2}}{2}\)

Solving for y, we get:

\(\displaystyle y=\frac{300-({\pi}+4)r}{2}\)

Inserting this into the area equation, we arrive at:

\(\displaystyle \frac{600r-({\pi}+8)r^{2}}{2}\)

Take derivative:

\(\displaystyle 300-({\pi}+8)r\)

Set to 0 and solve for r, we get:

\(\displaystyle r=\frac{300}{{\pi}+8}\)

Therefore, \(\displaystyle x=y=\frac{600}{{\pi}+8}\)

 

[Guest]

i was close enough...just out by a factor of 1/2. thank you

thank you...