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.
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?
2 answers
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
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