Question

If y = 3t^2- 4t ; then minima of y will be at ?

Answer 1

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Answer 2

Answer:

minimum at t=2/3

Explanation:

Given:

$y = 3 {t}^{2} - 4t$

For y to be minimum:

$\frac{dy}{dt} = 0 \: and \: \frac{ {d}^{2}y }{d {t}^{2} } > 0$

Now,

Differentiating of y with respect to t,

$\frac{dy}{dt} = 6t - 4 \\ \frac{dy}{dt} = 0 \\ 6t- 4 = 0 \\ 6t = 4 \\ t = \frac{4}{6} \\ t = \frac{2}{3}$

$\frac{ {d}^{2}y }{d {t}^{2} } = 6 \: > 0$

Hence, there will be minimum at t=2/3.