Find the Maximum/Minimum Value g(x)=1/20x(x-100)

$g (x) = \frac{1}{20} x (x - 100)$

Apply the distributive property.

$g (x) = \frac{1}{20} \cdot (x \cdot x + x \cdot - 100)$

Simplify the expression.

Multiply $x$ by $x$ .

$g (x) = \frac{1}{20} \cdot (x^{2} + x \cdot - 100)$

Move $- 100$ to the left of $x$ .

$g (x) = \frac{1}{20} \cdot (x^{2} - 100 \cdot x)$

Apply the distributive property.

$g (x) = \frac{1}{20} x^{2} + \frac{1}{20} \cdot (- 100 x)$

Combine $\frac{1}{20}$ and $x^{2}$ .

$g (x) = \frac{x^{2}}{20} + \frac{1}{20} \cdot (- 100 x)$

Cancel the common factor of $20$ .

Factor $20$ out of $- 100 x$ .

$g (x) = \frac{x^{2}}{20} + \frac{1}{20} \cdot (20 (- 5 x))$

Cancel the common factor.

$g (x) = \frac{x^{2}}{20} + \frac{1}{20} \cdot (20 (- 5 x))$

Rewrite the expression.

$g (x) = \frac{x^{2}}{20} - 5 x$

The minimum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is positive, the minimum value of the function is $f (- \frac{b}{2 a})$ .

$f_{min}$ $x = a x^{2} + b x + c$ occurs at $x = - \frac{b}{2 a}$

Find the value of $x = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

$x = - \frac{- 5}{2 (0.05)}$

Remove parentheses.

$x = - \frac{- 5}{2 (0.05)}$

Simplify $- \frac{- 5}{2 (0.05)}$ .

Multiply $2$ by $0.05$ .

$x = - \frac{- 5}{0.1}$

Divide $- 5$ by $0.1$ .

$x = - - 50$

Multiply $- 1$ by $- 50$ .

$x = 50$

Evaluate $f (50)$ .

Replace the variable $x$ with $50$ in the expression.

$g (50) = \frac{{(50)}^{2}}{20} - 5 \cdot 50$

Simplify the result.

Simplify each term.

Raise $50$ to the power of $2$ .

$g (50) = \frac{2500}{20} - 5 \cdot 50$

Divide $2500$ by $20$ .

$g (50) = 125 - 5 \cdot 50$

Multiply $- 5$ by $50$ .

$g (50) = 125 - 250$

Subtract $250$ from $125$ .

$g (50) = - 125$

The final answer is $- 125$ .

$- 125$

Use the $x$ and $y$ values to find where the minimum occurs.

$(50, - 125)$

Find the Maximum/Minimum Value g(x)=1/20x(x-100)

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