Find the Maximum/Minimum Value -2.5p^2+650p

$- 2.5 p^{2} + 650 p$

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 $f (- \frac{b}{2 a})$ .

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

Step 2

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

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

$p = - \frac{650}{2 (- 2.5)}$

Remove parentheses.

$p = - \frac{650}{2 (- 2.5)}$

Simplify $- \frac{650}{2 (- 2.5)}$ .

Cancel the common factor of $650$ and $2$ .

Factor $2$ out of $650$ .

$p = - \frac{2 \cdot 325}{2 \cdot - 2.5}$

Cancel the common factors.

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

$p = - \frac{2 \cdot 325}{2 (- 2.5)}$

Cancel the common factor.

$p = - \frac{2 \cdot 325}{2 \cdot - 2.5}$

Rewrite the expression.

$p = - \frac{325}{- 2.5}$

Simplify the expression.

Divide $325$ by $- 2.5$ .

$p = - - 130$

Multiply $- 1$ by $- 130$ .

$p = 130$

Step 3

Evaluate $f (130)$ .

Replace the variable $p$ with $130$ in the expression.

$f (130) = - 2.5 {(130)}^{2} + 650 (130)$

Simplify the result.

Simplify each term.

Raise $130$ to the power of $2$ .

$f (130) = - 2.5 \cdot 16900 + 650 (130)$

Multiply $- 2.5$ by $16900$ .

$f (130) = - 42250 + 650 (130)$

Multiply $650$ by $130$ .

$f (130) = - 42250 + 84500$

Add $- 42250$ and $84500$ .

$f (130) = 42250$

The final answer is $42250$ .

$42250$

Step 4

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

$(130, 42250)$

Step 5

Find the Maximum/Minimum Value -2.5p^2+650p

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