Hello? A little help? Please?
Imagine if you will a fenced-in area composed of a rectangle with one side being a semicircle. The perimeter of this fence is 498 feet. I need to maximize the area of this pen. How do I do this? (I already know the major steps: set up your two equations, relate variables, take the first derivative and set it to zero, etc. I need specifics.)
call one side of the rectangle x and the other y. let the semicircle be attached to one of the x sides.
can you continue? (by the way, is this for a calculus class?)
the point of my question was that i wanted to know if this was calculus or precalculus, since we will use different methods depending on which class this is for. obviously you don't do this for funand yes, this is a homework assignment I'm trying to complete for a class. What? Do you think I just ask this stuff for the fun of it?
Imagine a fenced-in area composed of a rectangle with one side being a semicircle.
The perimeter of this fence is 498 feet.
Maximize the area of this pen.Code:* * * * * * * * * * - - - - * - - - - * | r r | | | x | | x | | | | *-------------------* 2r
The radius of the semicircle is
Hence, the length of the rectangle is
Let the width of the rectangle be
The semicircle has perimeter
The rectangular portion of the pen has perimeter
We have: .
The area of the semicircle is: .
The area of the rectangle is: .
The area of the pen is: .
Substitute  into : .
. . which simplifies to: .
And that is the function we must maximize . . .
I'm...not seeing how. In fact, I'm not even sure how he simplified it like that in the first place. When I simplify, I get .5pi(r^2) + 498r + (pi + 2)r^2. If I add the two pi(r^2) terms together, I get 1.5pi(r^2) + 498r + 2r^2.
Anyway, it's almost midnight here. I'm hittin' the hay If I haven't figured this out by tomorrow I'll be back. In the mean time, if you'd like, you can give me the next step and I may get a chance to check this in the morning before I leave. Anyway, good night.