Calculus Problems.....I can't solve it :(?
A Rectangular race track, with semi-circular ends at the two turns, has a perimeter of 600 yards. If the lengths of the straight tracks have to be at least 200 yards each. Find the max, area possible.
- 1 decade agoFavorite Answer
P = perimeter = 2L + 2πr = 600 ---> L + πr = 300
L = length of straight track > = 200
r = radius of semicircular tracks
A = area = πr² + 2rL
But since L = 300 - πr
A = πr² + 2r(300 - πr)
A = πr² + 600r - 2πr²
A = 600r - πr²
Taking the derivative of A with respect to r,
dA/dr = 600 - 2πr
Set this to zero and solve for r
600 - 2πr = 0
r = 600/(2π) = 300/π
A = maximum area = 600r - πr² = 600 (300/π) - π(300/π)²
A = 180,000/π - 90,000/π
A = 90,000/π square yards << Maximum area (circular track)
Tatiana - you are correct. I did not check my answer and forgot about the constraint of L >= 200 yards or L+x = 200: x>= 0. I tried several methods of calculating this analytically but ultimately it boils down to graphically showing that to meet the constraint L must be = 200 for x > 0 corresponding to the maximum allowable area.
But your calculated Area is also INCORRECT. It must be 50,000/π
The maximum Area of 90,000/π can be achieved when L = 0.
However by setting the constraint of L being at least 200 yds, then you basically reduced the maximum allowable area down from 90,000/π to 50,000/π.
A = area = πr² + 2rL
L + πr = 300 ==> 200 + πr = 300
r = 100/π
*** Here you can observe that as L gets larger than 200, r gets smaller, and thus the Area gets smaller as well ***
A = π(100/π)² + 2 (100/π)(200)
A = 10,000/π + 40,000/π
A = 50,000/π <<< final answer
- 1 decade ago
Note by 200+x the length of the straight part,
and r the radius of the two semi-circle ends.
The perimeter is given by P=2*(200+x)+2Pi * r
The area A=Pi* r^2+(200+x)*2r
You want P=600 so you can solve for r in the expression of P,
Plug into A, you obtain A as a function of x:
You want to find the max of A for non negative values of x.
For x=-200 the derivative A' is zero and you can check that before -200 the function A' is positive and after -200 it's negative.
That means that over |R, A is first increasing, reaches a max at -200 and then is decreasing.
So the max of A over non negative numbers is reached at x=0.
And at x=0 one has A=500/Pi
(In this case the size of the track is 200 for each straight segments and 100/Pi for the radius of the semicircular parts)
@Χέηοτ -- ۩Ю: your answer is wrong because if r=300/Pi then L=0 and you want L>=200.
- modulo_functionLv 71 decade ago
L be the length of the straight part
R be the radius of the curved ends
The perimeter is:
P = 2L+2Rpi
the area is:
A = 2LR + piR^2
Use the given value for P and the first equation to get an expression for L or R
use that expression in the second equation. You now have an equation for area as a function of either L or R. Differentiate and set to zero to get a maximum.