Find the Maximum/Minimum Value y=-1/10x^2+10x

$y = - \frac{1}{10} x^{2} + 10 x$

Combine $x^{2}$ and $\frac{1}{10}$ .

$f (x) = - \frac{x^{2}}{10} + 10 x$

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

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

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

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

$x = - \frac{10}{2 (- 0.1)}$

Remove parentheses.

$x = - \frac{10}{2 (- 0.1)}$

Simplify $- \frac{10}{2 (- 0.1)}$ .

Cancel the common factor of $10$ and $2$ .

Factor $2$ out of $10$ .

$x = - \frac{2 \cdot 5}{2 \cdot - 0.1}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 0.1$ .

$x = - \frac{2 \cdot 5}{2 (- 0.1)}$

Cancel the common factor.

$x = - \frac{2 \cdot 5}{2 \cdot - 0.1}$

Rewrite the expression.

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

Simplify the expression.

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.

$f (50) = - \frac{{(50)}^{2}}{10} + 10 (50)$

Simplify the result.

Simplify each term.

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

$f (50) = - \frac{2500}{10} + 10 (50)$

Divide $2500$ by $10$ .

$f (50) = - 1 \cdot 250 + 10 (50)$

Multiply $- 1$ by $250$ .

$f (50) = - 250 + 10 (50)$

Multiply $10$ by $50$ .

$f (50) = - 250 + 500$

Add $- 250$ and $500$ .

$f (50) = 250$

The final answer is $250$ .

$250$

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

$(50, 250)$

Find the Maximum/Minimum Value y=-1/10x^2+10x

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