Find the Maximum/Minimum Value R(p)=-2.5p^2+800p

$R (p) = - 2.5 p^{2} + 800 p$

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 $b$ .

$p = - \frac{800}{2 (- 2.5)}$

Remove parentheses.

$p = - \frac{800}{2 (- 2.5)}$

Simplify $- \frac{800}{2 (- 2.5)}$ .

Cancel the common factor of $800$ and $2$ .

Factor $2$ out of $800$ .

$p = - \frac{2 \cdot 400}{2 \cdot - 2.5}$

Cancel the common factors.

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

$p = - \frac{2 \cdot 400}{2 (- 2.5)}$

Cancel the common factor.

$p = - \frac{2 \cdot 400}{2 \cdot - 2.5}$

Rewrite the expression.

$p = - \frac{400}{- 2.5}$

Simplify the expression.

Divide $400$ by $- 2.5$ .

$p = - - 160$

Multiply $- 1$ by $- 160$ .

$p = 160$

Step 3

Evaluate $f (160)$ .

Replace the variable $p$ with $160$ in the expression.

$R (160) = - 2.5 {(160)}^{2} + 800 (160)$

Simplify the result.

Simplify each term.

Raise $160$ to the power of $2$ .

$R (160) = - 2.5 \cdot 25600 + 800 (160)$

Multiply $- 2.5$ by $25600$ .

$R (160) = - 64000 + 800 (160)$

Multiply $800$ by $160$ .

$R (160) = - 64000 + 128000$

Add $- 64000$ and $128000$ .

$R (160) = 64000$

The final answer is $64000$ .

$64000$

Step 4

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

$(160, 64000)$

Step 5

Find the Maximum/Minimum Value R(p)=-2.5p^2+800p

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