Find the Maximum/Minimum Value -2.5x^2+500x

$- 2.5 x^{2} + 500 x$

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 $b$ .

$x = - \frac{500}{2 (- 2.5)}$

Remove parentheses.

$x = - \frac{500}{2 (- 2.5)}$

Simplify $- \frac{500}{2 (- 2.5)}$ .

Cancel the common factor of $500$ and $2$ .

Factor $2$ out of $500$ .

$x = - \frac{2 \cdot 250}{2 \cdot - 2.5}$

Cancel the common factors.

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

$x = - \frac{2 \cdot 250}{2 (- 2.5)}$

Cancel the common factor.

$x = - \frac{2 \cdot 250}{2 \cdot - 2.5}$

Rewrite the expression.

$x = - \frac{250}{- 2.5}$

Simplify the expression.

Divide $250$ by $- 2.5$ .

$x = - - 100$

Multiply $- 1$ by $- 100$ .

$x = 100$

Step 3

Evaluate $f (100)$ .

Replace the variable $x$ with $100$ in the expression.

$f (100) = - 2.5 {(100)}^{2} + 500 (100)$

Simplify the result.

Simplify each term.

Raise $100$ to the power of $2$ .

$f (100) = - 2.5 \cdot 10000 + 500 (100)$

Multiply $- 2.5$ by $10000$ .

$f (100) = - 25000 + 500 (100)$

Multiply $500$ by $100$ .

$f (100) = - 25000 + 50000$

Add $- 25000$ and $50000$ .

$f (100) = 25000$

The final answer is $25000$ .

$25000$

Step 4

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

$(100, 25000)$

Step 5

Find the Maximum/Minimum Value -2.5x^2+500x

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