Find the Maximum/Minimum Value P(c)=-12c^2+120c

$P (c) = - 12 c^{2} + 120 c$

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

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

Step 2

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

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

$c = - \frac{120}{2 (- 12)}$

Remove parentheses.

$c = - \frac{120}{2 (- 12)}$

Simplify $- \frac{120}{2 (- 12)}$ .

Cancel the common factor of $120$ and $2$ .

Factor $2$ out of $120$ .

$c = - \frac{2 \cdot 60}{2 \cdot - 12}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 12$ .

$c = - \frac{2 \cdot 60}{2 (- 12)}$

Cancel the common factor.

$c = - \frac{2 \cdot 60}{2 \cdot - 12}$

Rewrite the expression.

$c = - \frac{60}{- 12}$

Cancel the common factor of $60$ and $- 12$ .

Factor $12$ out of $60$ .

$c = - \frac{12 (5)}{- 12}$

Move the negative one from the denominator of $\frac{5}{- 1}$ .

$c = - (- 1 \cdot 5)$

Multiply $- (- 1 \cdot 5)$ .

Multiply $- 1$ by $5$ .

$c = - - 5$

Multiply $- 1$ by $- 5$ .

$c = 5$

Step 3

Evaluate $f (5)$ .

Replace the variable $c$ with $5$ in the expression.

$P (5) = - 12 {(5)}^{2} + 120 (5)$

Simplify the result.

Simplify each term.

Raise $5$ to the power of $2$ .

$P (5) = - 12 \cdot 25 + 120 (5)$

Multiply $- 12$ by $25$ .

$P (5) = - 300 + 120 (5)$

Multiply $120$ by $5$ .

$P (5) = - 300 + 600$

Add $- 300$ and $600$ .

$P (5) = 300$

The final answer is $300$ .

$300$

Step 4

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

$(5, 300)$

Step 5

Find the Maximum/Minimum Value P(c)=-12c^2+120c

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