An indoor physical fitness room consists of a rectangular region with a semicircle on each end. The perimeter of the room is to be a 200-meter running track.
a) Draw a figure that visually represents the problem. Let x and y represent the length and width of the rectangular region respectively
b) Determine the radius of the semicircular ends of the track. Determine the distance, in terms of y, around the inside edge of each semicircular part of the track.
c) Use the result of part b to write an equation in terms of x and y, for the distance traveled in one lap around the track. Solve for x.
d) Use the result of part c to write the area A of the rectangular region as a function of x. What dimensions will produce a rectangle of maximum area?
c) is (200-2x)/π because of 2x + πy = 200 you are trying to find what y=
In Reiny's answer question b is right but I am not sure if the questions after c is right.
Hello people of 2021 and the future of precal :D
a) I'm great at visual humor, but sadly, I cannot draw pictures!
b) Let's do some math here. The perimeter of the room is the sum of the lengths of all sides. In this case, it is the sum of the lengths of the rectangular sides and the semicircular ends. The semicircular ends have a radius, which we'll call r. So, if we let y be the width of the rectangular region and x be the length, the perimeter equation is:
2x + 2y + 2πr = 200
To find the distance around the inside edge of a semicircle (which is half a circumference), we can use the formula for the circumference of a circle:
C = 2πr
So, the distance around the inside edge of each semicircular part of the track is just πr.
c) Now that we know the distance around the inside edge of each semicircular part of the track is πr, we can write the equation for the distance traveled in one lap around the track:
Distance = 2x + 2y + 2πr
d) To write the area A of the rectangular region as a function of x, we need to multiply the length (x) by the width (y):
A = xy
To find the dimensions that will produce a rectangle of maximum area, we need to find the maximum value of A with respect to x. This can be done by taking the derivative of A with respect to x, setting it equal to zero, and solving for x. However, since we don't have a specific value for y, we cannot find the exact dimensions that will produce a rectangle of maximum area. However, we can determine that the length (x) and width (y) should be equal to maximize the area.
a) To visually represent the problem, we can draw a rectangle with two semicircles attached at each end. The length of the rectangle can be represented as 'x' and the width can be represented as 'y'. The figure will look something like this:
```
+------------------------------------------------------------+
| |
| |
| |
| x |
| |
| |
| |
+--------------------+-----------------+---------------------+
| |
| |
| y |
| |
| |
+--------------------+-----------------+---------------------+
```
b) The perimeter of the room is a 200-meter running track. Since the perimeter is made up of the length of the rectangle and the semicircles on each end, we can set up an equation.
Perimeter = 2(length of rectangle) + circumference of semicircle
200 = 2x + 2πr (where r is the radius of the semicircle)
We know that the circumference of a semicircle is equal to half the circumference of a full circle, which is 2πr. So, the circumference of a semicircle is just πr.
200 = 2x + 2πr
200 = 2x + 2πr
To determine the radius of the semicircles, we need to find the value of r in terms of y. Since the length of the semicircle is equal to the width of the rectangle, it means the radius of the semicircle is equal to half the width of the rectangle.
So, r = y/2
Now, we can substitute the value of r back into the equation.
200 = 2x + 2π(y/2)
200 = 2x + πy
To determine the distance around the inside edge of each semicircular part of the track, we need to find the circumference of a semicircle which is πr.
Circumference of each semicircle = π(y/2)
c) To find the distance traveled in one lap around the track, we need to find the perimeter of the entire track which is the sum of the lengths of the rectangle and the circumferences of the two semicircles.
Distance traveled in one lap = Length of rectangle + 2(circumference of each semicircle)
= x + 2(π(y/2))
= x + πy
Since we are given that one lap around the track is equal to 200 meters, we can set up an equation.
x + πy = 200
Now, we can solve this equation to find the value of x in terms of y.
x = 200 - πy
d) To write the area A of the rectangular region as a function of x, we multiply the length and width of the rectangle.
Area A = Length × Width
= x × y
= (200 - πy)y
= 200y - πy^2
To find the dimensions that will produce a rectangle of maximum area, we need to find the value of y that maximizes the area A. To do this, we can take the derivative of the area function with respect to y and set it equal to zero.
dA/dy = 200 - 2πy
Setting the derivative equal to zero:
200 - 2πy = 0
Solving for y:
2πy = 200
y = 100/π
Substituting the value of y back into the equation for x:
x = 200 - π(100/π)
x = 200 - 100
x = 100
So, the dimensions that will produce a rectangle of maximum area are x = 100 meters and y = 100/π meters.
a) ---- you have done that, I hope
b) radius = y/2
( I would have labeled the radius y, making the width of the rectangle 2y, that way I would have no fractions, but ..... oh well)
So we have two semicircles or one complete circle with radius y/2
distance around = circumference of one circle
= 2π(y/2) = πy
c) so 2x + πy = 200
x = (200 - πy)/2
Area = π(y/2)^2 + xy
= πy^2/4 + y(200-πy)/2
= πy^2 /4 + 100y - (π/2)y^2
d(area)/dy = πy/2 + 100 - πy
= 0 for a max/min of area
πy - πy/2 = 100
πy/2 = 100
y = 200/π
then x = (200 - π(200/π)/2 = 0
ahh, the famously flawed track question.
Of course the answer makes sense. It says to simply build a circular track, since a circle gives us the largest area for a given perimeter.
But in most of the variations of this question, there is some other restriction given on your values of x and y.