Question

Find the minimum value of $y=25x^2+5-10x$.

Answer 1

For maximum and minimum value, we can put $\frac{dy}{dx}=0.$

or $\frac{dy}{dx}=50x-10$

Put, $\frac{dy}{dx}=0,\therefore x=\frac{1}{5}$

Further, $\frac{d^{2}y}{d{x}^2}=50$

or $\frac{d^{2}y}{d{x}^2}$ has positive value at $x=\frac{1}{5}.$

Therefore, $y$ has minimum value at $x=\frac{1}{5}.$

Substituting $x=\frac{1}{5}$ in given equation, we get

$y_{min}=25{\left(\frac{1}{5}\right)}^2+5-10\left(\frac{1}{5}\right)=4$