If `Q=4v^(3)+3v^(2)`, then the value of `'v'` such that,there exist maxima of `'Q'`-

Options

A. `0`
B. `-(1)/(2)`
C. `(1)/(2)`
D. none

Explanation

To find the value of \( v \) such that there exists a maximum of \( Q \), we follow these steps:

1. **Define the function**: We have \( Q = 4v^3 + 3v^2 \).

2. **Differentiate \( Q \)**: To find the critical points, we need to differentiate \( Q \) with respect to \( v \).
\[ \frac{dQ}{dv} = \frac{d}{dv}(4v^3 + 3v^2) = 12v^2 + 6v \]

3. **Set the derivative to zero**: We set the first derivative equal to zero to find the critical points.
\[ 12v^2 + 6v = 0 \]

4. **Factor the equation**: We can factor out \( 6v \) from the equation.
\[ 6v(2v + 1) = 0 \]

5. **Solve for \( v \)**: Setting each factor equal to zero gives us:
\[ 6v = 0 \quad \Rightarrow \quad v = 0 \]
\[ 2v + 1 = 0 \quad \Rightarrow \quad v = -\frac{1}{2} \]

6. **Determine the nature of the critical points**: To check whether these points are maxima or minima, we need to find the second derivative of \( Q \).
\[ \frac{d^2Q}{dv^2} = \frac{d}{dv}(12v^2 + 6v) = 24v + 6 \]

7. **Evaluate the second derivative at the critical points**:
- For \( v = 0 \):
\[ \frac{d^2Q}{dv^2} \bigg|_{v=0} = 24(0) + 6 = 6 \quad (\text{which is } > 0, \text{ indicating a minima}) \]
- For \( v = -\frac{1}{2} \):
\[ \frac{d^2Q}{dv^2} \bigg|_{v=-\frac{1}{2}} = 24\left(-\frac{1}{2}\right) + 6 = -12 + 6 = -6 \quad (\text{which is } < 0, \text{ indicating a maxima}) \]

8. **Conclusion**: The value of \( v \) such that there exists a maximum of \( Q \) is:
\[ v = -\frac{1}{2} \]