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# Semicircle + Rectangle Running Track?

A track and field playing area is in the shape of a rectangle with semicircles at each end. The inside perimeter of the track is to be 1500 meters. What should

the dimensions of the rectangle be so that the area of the rectangle is a maximum?

### 2 Answers

- D gLv 72 years agoFavorite Answer
if the track and field hs half a circle at BOTH ends you have a WHOLE circle if you include both ends

rectangle has dimensions L and W

say the W is the short side and is the side where the circles attach

TOTAL Perimeter = perimeter of the circle and 2L

SINCE the semi circles attach to short sides or W THE diameter is W

the r = W/2

the perimeter of the circle = 2pi (W/2)= piW

THAT means the total length of the field = piW + 2L = 1500

area of rectangle = LW

we need either L or W replaced since we dont use W lets replace it

piW + 2L = 1500

solve for W

piW = (1500 - 2L)

W = (1500 - 2L)/pi

put that in the area formula

Area = LW = L*(1500 -2L)/pi

Area = (1500L - 2L^2)/ pi

dA/dL = (1/pi)* d(1500L - 2L^2)/dL

= (1/pi) * (1500 - 4L)

a minimum max happens when dA/dlL is 0

0 = (1500 - 4fL)/pi

0 = 1500 - 4L

4L = 1500

L = 1500/4 = 350 m

we can find if its a max or min by taking the double derivative

dA/dL = (1/pi)*(1500 - 4L)

d^2L/dL^2 = -4/pi

this value means the tangent line of the area is LEVELING OUT concave down

and then that means our value is a MAXIMUM

SO from the work we found that

area is max when L = 350 .. THAT IS area of the rectangle only

if 1500 = W pi + 2L

1500 = W pi + 700

800 = W pi

W = 800/pi meters

- geezerLv 72 years ago
To achieve the maximum area inside the track the ''rectangle'' would need to be a square.

So ..

You now know that the circumfrence of the circle (semicircle at each end) + half the perimeter of the square (because the other half in ''inside the circular ends) .. added together .. make 1500 meters.

Over to you !!