Area of physical fitness room

[abby_07]

i am sorry that i do not have any work, but i do not know what to do.
i am suppose to find the dimensions that will make the area of the rectangular region as large as possible. Thus, i think i am suppose to find the maximum.
the problem reads

A physical fitness room consists of a rectangular region with a semicircle on each end. If the perimeter of the room is to be a 200-meter running track, find the dimensions that will make the area of the rectangular region as large as possible.

can you please help?

 

[daon]

The first thing you need to do is write down the formulas. IN ANY OF THESE KIND OF PROBLEMS, the first thing you ALWAYS should do is write down what your objective is and what you know.

What you need to do: Maximize the area of the rectangular region of the room.

What you know:
1) You know the perimiter is 200 meters. You also know the perimiter is going to be P=2L + C where L is the length of the rectangular region and C is the circumference of the circle created by addition of the semi-circles.

..___L__
.(...........)
(.............) 1/2C
.( _____)

So, 200 = 2L+C.

2) You also know that the AREA of the rectangular region will be the Diameter of the (semi)Circle (D) times the Length of the side(L). So, A=L*D. But we have too many vairiables. Note: for these kinds of questions you will usually be given 1 more variable than you have equations... You have FOUR variables and TWO equations. Your variables are C,L,A,D. Note that P is not a variable (we know it must be 200 meters). Your equations are P=2L+C and A=L*D. SO we need to somehow get rid of one of these extra variables. Note that you want to get rid of one of these variables by finding it in terms of another variable you arleady are using.

So, with a little inevstigation, it seems easy enough to eliminate the D in the Area equation. The Circumference of the (semi)circle is C=Pi*D, so D=C/Pi. Now rewrite your equations:

200 = 2L+C
A=L*(C/Pi)

Now we have Two equations and three variables. With these two equations we can get a SINGLE equation with the variable to maximize (A) and another variable (you choose which). I will choose L.

200=2L+C means that C = 200-2L

Now, substitute 200-2L in for C in the Area equation:

A=L*(C/Pi) means that A = L*(200-2L)/Pi.

Simplify: A = (1/Pi)(-2L² + 200L)

Now take derivative (of A with respect to L): dA/dL = (1/Pi)(-2L + 200)

Set derivative equal to zero to determine the maximum: (1/Pi)(-2L+200)=0

Solve for L: L = 50.

So, when L is 50 we have a MAXIMUM area. Note that this is a negative quadratic so it MUST be a maximum and not a minimum.

Now, Plug in L=50 into the formula for A: A = 50(200-2(50))/Pi = 5000/Pi.

I did this in a hurry so check my work, but get the idea?
-Daon

 

[soroban]

Hello, Abby!

A physical fitness room consists of a rectangular region with a semicircle on each end.
If the perimeter of the room is to be a 200-meter running track,
find the dimensions that will make the area of the rectangular region as large as possible.

First, make a sketch!

                         x
              * - - - - - - - - - - *
          *   :                     :   *
        *     :                     :     *
       *      :y                   y:      *
              :                     :
      *       :                     :       *
      *       +                     +       *
      *       :                     :       *
              :                     :
       *      :y                   y:      *
        *     :                     :     *
          *   :                     :   *
              * - - - - - - - - - - *
                         x

Let \(\displaystyle x\) = length of the rectangular region.
Let \(\displaystyle y\) = radius of the semicircles.

The perimeter of the track is: \(\displaystyle \,P \:=\:2x + 2\pi y \:=\:200\;\;\Rightarrow\;\;y\:=\:\frac{1}{\pi}(100\,-\,x)\;\) [1]

The area of the rectangular region is: \(\displaystyle \,A\:=\:x(2y) \:=\:2xy\;\) [2]


Substitute [1] into [2]: \(\displaystyle \;A \:=\:2x\,\cdot\,\frac{1}{\pi}(100\,-\,x)\:=\:\frac{2}{\pi}(100x\,-\,x^2)\)

Differentiate and equate to zero:
. . \(\displaystyle A'\:=\:\frac{2}{\pi}(100\,-\,2x)\:=\:0\;\;\Rightarrow\;\;\fbox{x\,=\,50}\)

Substitute into [1]: \(\displaystyle \,y\:=\:\frac{1}{\pi}(100\,-\,50)\;\;\Rightarrow\;\;\fbox{y \:=\:\frac{50}{\pi}}\)


Therefore, the length of the rectangular region should be \(\displaystyle 50\) meters
. . and its width (diameter of the semicircles) should be \(\displaystyle \frac{100}{\pi}\) meters.