Find the Maximum/Minimum Value R(p)=-2.5p^2+600p

$R (p) = - 2.5 p^{2} + 600 p$

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 $b$ .

$p = - \frac{600}{2 (- 2.5)}$

Remove parentheses.

$p = - \frac{600}{2 (- 2.5)}$

Simplify $- \frac{600}{2 (- 2.5)}$ .

Cancel the common factor of $600$ and $2$ .

Factor $2$ out of $600$ .

$p = - \frac{2 \cdot 300}{2 \cdot - 2.5}$

Cancel the common factors.

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

$p = - \frac{2 \cdot 300}{2 (- 2.5)}$

Cancel the common factor.

$p = - \frac{2 \cdot 300}{2 \cdot - 2.5}$

Rewrite the expression.

$p = - \frac{300}{- 2.5}$

Simplify the expression.

Divide $300$ by $- 2.5$ .

$p = - - 120$

Multiply $- 1$ by $- 120$ .

$p = 120$

Step 3

Evaluate $f (120)$ .

Replace the variable $p$ with $120$ in the expression.

$R (120) = - 2.5 {(120)}^{2} + 600 (120)$

Simplify the result.

Simplify each term.

Raise $120$ to the power of $2$ .

$R (120) = - 2.5 \cdot 14400 + 600 (120)$

Multiply $- 2.5$ by $14400$ .

$R (120) = - 36000 + 600 (120)$

Multiply $600$ by $120$ .

$R (120) = - 36000 + 72000$

Add $- 36000$ and $72000$ .

$R (120) = 36000$

The final answer is $36000$ .

$36000$

Step 4

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

$(120, 36000)$

Step 5

Find the Maximum/Minimum Value R(p)=-2.5p^2+600p

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