word problem

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ca.chick

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May 7, 2007
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the question is:

a norman window has a shape of a rectangle surmonted by a semicircle of diameter equal to the width of the rectangle. if the perimeter of the winder is 20 feet, what dimensions will admit the most light(maximize area)?

i have no idea what it means and also says to express the function of the width, x, fo the rectangle and to find the dimensions that maximize the area.

please help :(
 
The area of the window is given by \(\displaystyle \L\\A=\frac{{\pi}(\frac{x}{2})^{2}}{2}+xy\).......[1]- i dont understand what the one is doing on the very end of the equation



\(\displaystyle \L\\2y+x+\frac{{\pi}x}{2}=20\)........[2]

Solve [2] for x and sub into [1]. -what do u mean here

It'll then be entirely in terms of y.

Differentiate, set to 0 and solve for y. x will follow and you have your area.
 
galactus, since circle has area > square (same perimeter),
can we not make y = 0?
Then we have a simpler pi(x)/2 + x = 20

No?
 
I'm sorry, I must have misundertood the question. Looks like a max/min calc to me.
 
ca.chick said:
u said this was a calculas problem and im in trig
Click to expand...
If you aren't sure if this is a calculus problem or not, why are you posting a topic in "Calculus"?

This problem can be solved using calculus.
 
It was moved to calculus. It was originally in Intermediate Algebra.
 
ca.chick said:
u said this was a calculas problem and im in trig
Click to expand...
This is a stereotypical calculus problem, which could explain the general confusion here. But since this was assigned in your trigonometry class, then that class must have covered some method for doing this trigonometrically.

Please reply with the method and/or formulas you are supposed to be using. Thank you! :D

Eliz.
 
We can do this without calculus by using the vertex of a parabola form the resulting quadratic.

\(\displaystyle \L\\A=\frac{{\pi}(\frac{x}{2})^{2}}{2}+xy\)

Perimeter: \(\displaystyle \L\\2y+x+\frac{{\pi}x}{2}=20\)

Now, solve the perimeter equation for y and sub into the area equation.

The area equation will be entirely in terms of x. It will be a quadratic.

Use \(\displaystyle \frac{-b}{2a}\) to find the x-coordinate of the quadratic.

Once you have x, y will follow by substitution.

The result will be the same as the calc method.
 
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