First slide

Application of diffrentiation

Question

If Q = 4v³ + 3v² , then the value of 'v' such that, there exist maxima of 'Q'

Difficult

A

0
B

$- \frac{1}{2}$
C

$\frac{1}{2}$
D

none

Solution

$\begin{array}{l} Q = 4 v^{3} + 3 v^{2} \\ \Rightarrow \frac{d Q}{d v} = 12 V^{2} + 6 V = 0 \\ \Rightarrow v = 0, - \frac{1}{2} \\ N o w, \frac{d^{2} Q}{d v^{2}} = 24 v + 6 \\ F o r v = 0, (\frac{d^{2} Q}{d v^{2}}) = 6 (+ v e) \\ F o r v = - \frac{1}{2}, (\frac{d^{2} Q}{d v^{2}}) = - 12 + 6 = - 6 (- v e) \\ T h u s, v = - 1 / 2 for maximum Q \end{array}$

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