Optimization- Calculus?

Question

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 regions as large as possible.



can u work this problem out for me..its an optiziation problem from Calculus and i suck at them...it will be very helpful THANK U!

Answer 1

Let's call the width of the room W (in between the semicircles) and the length L (the sides which have the semicircles). Each semicircle has radius L/2. (A circle of radius L/2 has perimeter 2*pi*L/2, but this is a semicircle so each has half of that, an outside perimeter pi*L/2.

The area of the rectangular portion is:

a = W * L

The perimeter of the room is two widths plus two semicircles:

p = 2*pi*L/2 + 2*W
p = pi*L + 2W

Solve the perimeter for w, and substitute the known value:

p = pi*L + 2W
200 = pi*L + 2W
W = (200 - pi*L)/2

And substitute that equation into the area equation to get area as a function of L only:

a = W * L
a = (200 - pi*L)/2 * L
a = 100*L - pi*L^2/2

Take the derivative of area with respect to L and set it equal to zero:

da/dL = 0
d(100*L - pi*L^2/2)/dL =0
100 - pi*L = 0
L = 100/pi meters

Knowing that the perimeter is 200m, substitute back into the perimeter equation to find W:

p = pi*L + 2*W
200 = pi*(100/pi) + 2*W
100 = 2W
W = 50m

The length of the semicircular sides will be 100/pi meters (about 32m), and the perimeter of each semicircle will be 50m. The length of the sides in between the semicircles will be 50m. The perimeter will be two semicircles (2*50m = 100m) plus two straight sides (2*50m = 100m) which is 200m as desired.

The area of the rectangular region is:

a = W * L
a = 50 * (100/pi)
a = 5,000/pi square meters
a = about 1,592 m^2

(Note that a previous respondent got a square 38.9 meters on a side, which is only 1,513 m^2 of area. That's not the maximum, because it's less than my own solution.)

Answer 2

Ok, we want to maximize the area of the rectangular region. Using x as the long side of the rectangle and d as the short side, we should get the equation:

area = xd

Also, we know that the equation for the perimeter is as follows:

P = 200 = 2x +2(pi*d / 2)

= 200 = 2x + pi*d

solving for d here:

d = (200 - 2x) / pi


Now, we can plug this into the area equation:

x ( 200 - 2x ) / pi = A

(200x / pi) - (2x^2 / pi) = A

Now, we need to take the deriv. with respect to x and set it equal to zero to find the maximum area:

200 / pi - 4x / pi = 0

Solving for x gives us:

x = 50m

and plugging back into the perimeter equation gives us:

d = 31.83m


Dimensions of the rectangle:

50m x 31.83m

Answer 3

Usually the area of a rectangle is maximized when it's a square.
Let's suppose we don't know that.
y = one side
x = other side which is also diamter of semicircle

so P = 2y + πx = 200
y = 100 - π/2 * x

A = xy = x* (100 - π/2 * x)
This is maximum when
x = 100 - π/2 * x
or x = 200 / (2 + π)
= 38.9.
y also turns out to be the same thing.

So the rectangular region should be 38.9 x 38.9 m