
Answer:
Explanation:
P(v) = 16 / v + 10ā»Ā³ v³
differentiating on both sides
dP / dt = - 16 / v² + 3 x 10ā»Ā³ v²
For maxima and minima , the condition is
dP / dt = - 16 / v² + 3 x 10ā»Ā³ v² Ā = 0
v² = 160 / 3 x 10²
v² = 73 m/s
v = 8.54 m /s
To know the condition of minima
again differentiating
d²P / dt² = - 16 x -2 / v² + 6 x 10ā»Ā³ x v
= 32 / v³ + 6 x 10ā»Ā³ x v
= + ve quantity
So at v_p = Ā 8.54 m /s , power consumption will be minimum .