1. ## Optimization

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.)

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

Derive and set to 0:

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

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.)

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

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

Derive and set to 0:

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

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

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