Respuesta :
Answer:
Speed s= 19
Explanation:
Take note of the following parameters:
Speed= s,
Number of cars= N,
Number N of cars that can pass a given spot per minute=
N(s)=88s/16+16(s19)2
The principle of differentiation is here:
We let N(s) = N
N = 86s / (17 + 17((s/19)^2))
17N = 86s /(1 + s²/19²)
17N = 361* 86s /(361 + s²)
17N = 31046s /(361 + s²)
Next step;
Differentiate with respect to s
17N = 31046s /(361 + s²)
Remember the quotient rule [u/v]’ = (vu’ - uv’) / v²
Therefore,
u = 31046s =====> du/ds = u’ = 31046
v = (361 + s²) ====>dv/ds = v’ = 2s
17N = 31046s /(361 + s²)
17 dN/ds = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²
The maximum when dN/ds = 0
17 dN/ds = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²
17 * 0 = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²
( 31046(361 + s²) - 31046s(2s) ) = 0
31046 [ (361 + s²) - 2s² ] = 0
31046 (361 - s²) = 0
(361 - s²) = 0
(19 - s)(19 + s) = 0
either s = -19 or s = 19, but s > 0
s = 19
The speed at which the greatest number of cars travel safely on that road is; s = 19
What is the speed required?
We are given the function to represent Number N of cars that can pass a given spot per minute as;
N(s) = 88s/(16 + 16(s/19)²)
where;
s is Speed
N is number of cars
Differentiating the function gives;
N' = -3971(s² - 361)/(2(s² + 361)²
Now, the speed at the greatest number of cars would be gotten when N' = 0. Thus;
-3971(s² - 361)/(2(s² + 361)² = 0
Cross multiply to get;
-3971(s² - 361) = 0
divide both sides by -3971 to get;
s² - 361 = 0
s = √361
s = 19
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