Answer :
Answer:
r = 13.33 m = (40/3) mm is the size of object for which the maximum velocity is needed to remove the object.
Step-by-step explanation:
v(r) = k(20r² - r³)
The function is at a maximum when (dv/dr) = 0 and (d²v/dr²) < 0
v(r) = (20kr² - kr³)
(dv/dr) = 40kr - 3kr²
At maximum v, (dv/dr) = 40kr - 3kr² = 0
40kr - 3kr² = 0
r (40k - 3kr) = 0
r = 0 or 40k - 3kr = 0
3kr = 40k
r = 40/3 = 13.333 mm
r = 0 mm or 13.33 mm
To investigate which point truly corresponds to the maximum value for the velocity,
(dv/dr) = 40kr - 3kr²
(d²v/dr²) = 40k - 6kr
At r = 0 mm
(d²v/dr²) = 40k - 6kr = 40k - 6k(0) = 40k
Since k is described as a positive constant, 40k > 0 and is positive,
Hence, (d²v/dr²) = 40k > 0,
r = 0 mm does not corresponds to a maximum point for the given function, rather, the conditions obtained show that it corresponds to a minimum point.
At r = 13.333 mm = (40/3) mm
(d²v/dr²) = 40k - 6kr = 40k - 6k(40/3) = 40k - 80k = - 40k.
Since k is described as a positive constant, - 40k < 0
Hence, (d²v/dr²) = - 40k < 0,
r = (40/3) mm corresponds to a maximum point for the given function, the conditions obtained show that it is indeed the value of r, when the velocity is at maximum value.
The size of the object must be 13.3 m to attain a maximum velocity
The velocity function is given as:
[tex]\mathbf{V(r) = k(20r^2 - r^3)}[/tex]
Differentiate with respect to r
[tex]\mathbf{V'(r) = k(40r - 3r^2)}[/tex]
Set to 0
[tex]\mathbf{k(40r - 3r^2) = 0}[/tex]
Divide both sides by k
[tex]\mathbf{40r - 3r^2 = 0}[/tex]
Factorize
[tex]\mathbf{r(40 - 3r) = 0}[/tex]
Split
[tex]\mathbf{r = 0\ or\ 40 - 3r = 0}[/tex]
Solve for r
[tex]\mathbf{r = 0\ or\ 3r = 40}[/tex]
Divide through by 3
[tex]\mathbf{r = 0\ or\ r = 13.3}[/tex]
The interval is given as [0,20].
So, the critical points are 0, 13.3 and 20
Substitute these values for r in V(r)
[tex]\mathbf{V(r) = k(20r^2 - r^3)}[/tex]
[tex]\mathbf{V(0) = k(20 \times 0^2 - 0^3)}[/tex]
[tex]\mathbf{V(0) =0}[/tex]
[tex]\mathbf{V(13.3) = k(20 \times 13.3^2 - 13.3^3)}[/tex]
[tex]\mathbf{V(13.3) = 1185.163k}[/tex]
[tex]\mathbf{V(20) = k(20 \times 20^2 - 20^3)}[/tex]
[tex]\mathbf{V(20) =0}[/tex]
The value of V(r) is maximum at r = 13.3.
Hence, the size of the object must be 13.3 m to attain a maximum velocity
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