An outdoor track consists of a rectangular region with a semi-circle on each end. if the perimeter of the track must be 200 meters, find the dimensions that will make the area of the rectangular region as large as possible.

Length = 50m

Width = 31.84m

Step-by-step explanation:

Perimeter of a track = 200m

The perimeter of a track with two semi-circular end and two rectangular regions = P

Perimeter of a semi-circle = ½(2πr)

Length of a rectangle = x

Perimeter of the rectangle = x+x

P = ½(2πr) + ½(2πr)+x + x

P = 2πr + 2x

200 =2πr + 2x

2πr = 200 – 2x

2πr = 2(100 – x)

r = 2(100 – x) / 2π

r = (100-x)/π

total area of the rectangular region =A

A= x(2r)

= (x)2[(100-x)/π]

= 2x[(100-x)/π]

A = (200x – 2x^2)/π

Differentiate A with respect to x

dA/dx = (200 -4x)/π

at critical point, first derivative vanishes(dA/dx = 0)

(200 -4x)/π = 0

-4x =-200

x =-200/-4

x = 50

this means length (x) = 50m

put x=50 into r = (100-x)/π

r= (100-50)/π

r =50/π

width = 2r

width = 2(50/π)

= 2(19.62)

= 31.84m

the cat is blue because the probability of extension towards the prefix of the diameter of the circle could only mathematically equal to the sum of the roots of the carrots from   your garden.

step-by-step explanation:

nope.   happens to me as well.

-aparri