Question

If $y=3t^2-4t$, then minima of $y$ will be at

• A. $3/2$
• B. $3/4$
• C. $2/3$
• D. $4/3$

Answer 1

The correct option is C $2/3$

$y=3t^2-4t$
For y to be minimum $\frac{dy}{dt}=0\text{and}\frac{d^{2}y}{d{t}^2}>0$
$\frac{dy}{dt}=6t-4=0\Rightarrow t=\frac{2}{3}.$
$\frac{d^{2}y}{d{t}^2}=6>0$
Hence, there will be minima at $t=\frac{2}{3}$