# How do you find the maximum value of f(x) = -3x^2 +30x -50?

Aug 26, 2017

Maximum value of $f \left(x\right)$ is $25$

#### Explanation:

$f \left(x\right) = - 3 {x}^{2} + 30 x - 50$

${f}^{'} \left(x\right) = - 6 x + 30$ , At stationary point ${f}^{'} \left(x\right) = 0$ or

$- 6 x + 30 = 0 \mathmr{and} x = \frac{30}{6} = 5$

At x=5 ; f(x) = - 3(5^2)+30*(5) -50= -75+150-50=25

Stationary point is $\left(5 , 25\right)$

2nd derivative test for maximum or minimum :

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - 6$ (negative)

If $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$ is negative, then the point is a maximum turning point.

So $\left(5 , 25\right)$ is maximum turning point and maximum value of

$f \left(x\right)$ is $25$

graph{-3x^2+30x-50 [-50.64, 50.6, -25.28, 25.37]} [Ans]