LArgest area?

Question

A playing field is to be built in the shape of a rectangle plus a semicircular region at each end. a 400-m racetrack is to form the perimeter of the field. what dimensions will give the rectangle part the largest area?

I dont' get this, can you please help me ? I dont' even noe how to set this up..

Answer 1

let the length of the rectangle field be 'a'. let the length of the

width of rectangle field be '2r'. we give 2r because there are

semicular field at each side. since these two semicular field

forms the one circle, their perimeter is 2 pi r.


by the problem, 2 pi r + 2a = 400 , since 400 is perimeter of the whole field.

2a = 400 - 2 pi r

a = 200 - pi r

area of rectangle is A= 2ar

A = 2 ( 200 - pi r) r

A= 400 r -2 pi r^2

to get the maximum area, we have to differentiate the area.

so dA/dr= 400 - 4 pi r , dA/dr is stationare when dA/dr is equal to '0'.

so, 400 - 4 pi r = 0

we get, r = 100/ pi

to know area is maximum or minimum, we have to differentiate again dA/dr.

d^2A/dr^2 = - 4 pi < 0,
since -4 pi is less than 0, it is maximum.if d^2A/dr^2 > 0,

area is minimum.

we got r = 100/ pi,substitute this value in 2 pi r + 2a =400

2 pi x 100 / pi + 2a = 400

2a = 200

a = 100

so a = 100 and r = 100/pi aare able to form the largest area of rectangle field.

'pi' in this case means 22/7 or 3.14.

Answer 2

You have a rectangle with half-circles on opposite sides. The area is the area of the rectangle a*b where a and b are the length of the sides. Say the circular parts are attached to the "a" side. The diameter of the circle is then a. The area of the clicular parts (both together) is pi*a^2/4. The total area of the field is a*b+pi*a^2/4. The track forms the perimeter. The lenght of the perimeter is the combination of the two "b" sides of the rectangle, plus the circumference of the circular ends 2*b + pi*a. Since you were given the value of the perimeter, this latter must = 400m. This allow you to get b in terms of a: 2*b+pi*a = 400, b = (400-pi*a)/2.

Plug this value of b into the area formula to get

A(a) = a*(400-pi*a)/2 + pi*a^2/4 = 200a - pi*a^2/2 *pi*a^2/4

A(a) = 200a -pi*a^2/4;

To find a maximum, differential\te A(a) with resp to a to get

200 - 2*pi*a/4 = 200 - pi*a/2

set this = 0 to get pi*a = 400, or a = 400/pi

Answer 3

yes, i think it is worded slightly wrong there
have you ever been to a high school football field? usually they have a 400meter track around the outside of it ... basically between it and the bleachers? picture that

the 400 meter track won't necessarily be the entire perimeter ..it will be the boundary of the field though is what i ascertain here (why i stated the problem isnt worded very well) : the playing field must be entirely inside of the 400 meter racetrack

sooo now you want to maximize the area of the playing field
area = area of rectangle + 2 * area of semicircular region at each end [i'm assuming here that the racetrack will be an oval , so it follows that each semicircular region is going to be the same size and a true-semicircle ;especially since it adjoins the rectangle at opposite ends and assuming that the problem statement intended ONLY 2 semicircles NOT 4]

now draw yourself a diagram and you'll get what the diameter of each semicircle must be based on the rectangle ....
set up your equation then .... and i assume you know how to differentiate so just take the 1st derivative and find the values
that will give you your max
(if you dont know how to differentiate ... then you are probably in some adv algebra class and know how to find the max of a 2nd degree parabolic equation .... anyhow you have enough info to have at it)

Answer 4

Let w = the width of the rectangle and h = its height
Then h is also the diameter of each semicircular region.
pi X h = the length of both semicircular ends of the track together, and 2w is the total length of the straight sides of the track, so
(pi X h ) + 2w = 400 m. Then w = 1/2 ( 400 - (pi)(h)).
the area A of the rectangle = hw = h/2 ( 400 - (pi)(h))
= 200h - (pi/2) h^2
A' = 200 - (pi)(h). If A' = 0 then h = 200/pi and A has either a maximum or minimum when h = 200/pi
A'' = -pi which is < 0, so it is a maximum, which is what you're looking for.
when h = 200/pi then w = 1/2 ( 400 - (pi)(h)) = 1/2 ( 400 - 200)
= 100
The rectangular field has largest area when it is 100 meters wide and 200/pi (approx. 63 2/3) meters high

Answer 5

Draw a rectangle. Label the length and width x and y. Put a half circle on the side of the rectangle that you labeled x. This is the shape of the track.

Go once around the track to get the perimeter. In other words add up all the lengths. You already have 2 parts of the perimiter labeled y. You need to figure out the perimeter of the half circles. (hint: the radii of the half circles are x/2)
Add up the permiter and set it equal to 400m because the problem said the track length is 400m.
Find an equation for the area of the rectangle. A = ?
Solve the perimeter equation for either x or y then substitute it in the equation for A
Then use calculus to find the maximum of A.

Answer 6

Is the racetrack a rectangle around the field or like a circle?

Answer 7

Draw a picture of the situation first.
____
(____)

The width of the field is also the diameter of the circle.
Thus we have pi *W + 2L = 400

We also have that A=W*L

We want to maximize the area, so solve for L or W in the first equation and substitute it into the second equation

Then take the derivative of the result and set it equal to zero

The rest you should be able to figure out.