Find the Maximum/Minimum Value -2.5x^2+550x

$- 2.5 x^{2} + 550 x$

Step 1

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 $x = - \frac{b}{2 a}$

Step 2

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

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

$x = - \frac{550}{2 (- 2.5)}$

Remove parentheses.

$x = - \frac{550}{2 (- 2.5)}$

Simplify $- \frac{550}{2 (- 2.5)}$ .

Cancel the common factor of $550$ and $2$ .

Factor $2$ out of $550$ .

$x = - \frac{2 \cdot 275}{2 \cdot - 2.5}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 2.5$ .

$x = - \frac{2 \cdot 275}{2 (- 2.5)}$

Cancel the common factor.

$x = - \frac{2 \cdot 275}{2 \cdot - 2.5}$

Rewrite the expression.

$x = - \frac{275}{- 2.5}$

Simplify the expression.

Divide $275$ by $- 2.5$ .

$x = - - 110$

Multiply $- 1$ by $- 110$ .

$x = 110$

Step 3

Evaluate $f (110)$ .

Replace the variable $x$ with $110$ in the expression.

$f (110) = - 2.5 {(110)}^{2} + 550 (110)$

Simplify the result.

Simplify each term.

Raise $110$ to the power of $2$ .

$f (110) = - 2.5 \cdot 12100 + 550 (110)$

Multiply $- 2.5$ by $12100$ .

$f (110) = - 30250 + 550 (110)$

Multiply $550$ by $110$ .

$f (110) = - 30250 + 60500$

Add $- 30250$ and $60500$ .

$f (110) = 30250$

The final answer is $30250$ .

$30250$

Step 4

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

$(110, 30250)$

Step 5

Find the Maximum/Minimum Value -2.5x^2+550x

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