Area Of A Norman Window

• #1

Howdy Mathematicians,

I have run into more issues with a dissimilar optimization problem.

A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is 30 feet, notice the dimensions of the window so that the greatest possible amount of calorie-free is admitted.

I fastened a copy of my work.

http://www.flickr.com/photos/77835031@N02/6858707594/

http://world wide web.flickr.com/photos/77835031@N02/6858707940/

Please expect at both if you lot have the chance. I did something slightly different betwixt the ii.

Thanks for any and all input.

• #4

thanks

Thank you both TKHunny and Pappas for your input. It is profoundly appreciated.

@TKHunny. Neither of my pictures are correct. They are close, only not right. The correct answers are \(\displaystyle 60/(4+pi) and 30/(4+pi)\)

I redid my work according to Pappus'south perimeter equation, but I still did not come up up with the correct answer.

Here is my work redone according to Pappus's perimeter equation.

• #5

Let r be the radius of the semicircle and let h exist the meridian of the rectangle.

Then, the area of the window is \(\displaystyle A=2rh+\frac{{\pi}r^{2}}{2}\)

The perimeter is \(\displaystyle P=2r+2h+\pi r\)

\(\displaystyle h=\frac{i}{two}(p-(2+\pi)r)\)

And so, \(\displaystyle A=pr-(two+\frac{\pi}{2})r^{2}\)

\(\displaystyle \frac{dA}{dr}=p-(4+\pi)r=0\)

when \(\displaystyle r=\frac{p}{4+\pi}\)

\(\displaystyle \frac{d^{ii}A}{dr^{ii}}<0\), so A is a max when:

\(\displaystyle r=\frac{p}{4+\pi}\)

• #6

thank you

Cheers Galactus!

I see how you addressed the formulas for the master and secondary equations, which helped tremendously.

Area Of A Norman Window,

Source: https://www.freemathhelp.com/forum/threads/norman-window-area.75105/

Posted by: pedersenworign.blogspot.com

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