A window frame - maximization problem

G

Guest

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?
 
window8pq.gif


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

The perimeter of the framework is:

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

A=2ry+πr22\displaystyle A=2ry+\frac{{\pi}r^{2}}{2}

Solving for y, we get:

y=300(π+4)r2\displaystyle y=\frac{300-({\pi}+4)r}{2}

Inserting this into the area equation, we arrive at:

600r(π+8)r22\displaystyle \frac{600r-({\pi}+8)r^{2}}{2}

Take derivative:

300(π+8)r\displaystyle 300-({\pi}+8)r

Set to 0 and solve for r, we get:

r=300π+8\displaystyle r=\frac{300}{{\pi}+8}

Therefore, x=y=600π+8\displaystyle x=y=\frac{600}{{\pi}+8}
 
i was close enough...just out by a factor of 1/2. thank you

thank you...
 
Top