Optimization

**Open that Hampster!** · October 27th 2009, 06:25 PM

1) A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 40 ft, find the value of x so that the greatest possible amount of light is admitted. (Give your answer correct to two decimal places.)

$A_r = x(40-x)$
$A_s = \frac{\pi (\frac{x}{2})^2}{2} = \frac{\pi x^2}{8}$
$A_T = A_r + A_s = x(40-x) + \frac{\pi x^2}{8}$

Derive and set to 0:

$40-2x + \frac{\pi x}{4} = 0$

x = 32.93

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2) A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (Give your answers correct to two decimal places.)

$A_t = x^2 * \frac{\sqrt{3}}{4}$
$A_r = (10-x)^2$

$A_{TOT} = A_r + A_r = (10-x)^2 + x^2 * \frac{\sqrt{3}}{4}$

Derive and set to 0:

$x\sqrt{3} - 20 + 2x => x = 5.36$

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**Pinkk** · October 27th 2009, 06:51 PM

If we let the diameter of the semicircle (which is the bottom, horizontal edge of the rectangle) be $x$ , then the perimeter of the semicircle is $\frac{\pi x}{2}$ . Then if we let each vertical edge of the rectangle be $y$ , then $2y + \frac{\pi x}{2} + x = 40$ , so $y = 20 - \frac{x}{2} - \frac{\pi x}{4}$ . The area of the rectangle, $A_{r} = xy = 20x - \frac{x^{2}}{2} -\frac{\pi x^{2}}{4}$ . The area of the semicircle, $A_{sc}=\frac{\pi x^{2}}{8}$ . So area total is:

$A = A_{r} + A_{sc} = 20x - \frac{x^{2}}{2} -\frac{\pi x^{2}}{4} + \frac{\pi x^{2}}{8}$ .
Set $A'=0$ to find the $x$ value that gives maximum area.

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