Find the Maximum/Minimum Value P(c)=-15c(c-8)

$P (c) = - 15 c (c - 8)$

Apply the distributive property.

$P (c) = - 15 c \cdot c - 15 c \cdot - 8$

Multiply $c$ by $c$ by adding the exponents.

Move $c$ .

$P (c) = - 15 (c \cdot c) - 15 c \cdot - 8$

Multiply $c$ by $c$ .

$P (c) = - 15 c^{2} - 15 c \cdot - 8$

Multiply $- 8$ by $- 15$ .

$P (c) = - 15 c^{2} + 120 c$

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 $c = - \frac{b}{2 a}$

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

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

$c = - \frac{120}{2 (- 15)}$

Remove parentheses.

$c = - \frac{120}{2 (- 15)}$

Simplify $- \frac{120}{2 (- 15)}$ .

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

Factor $2$ out of $120$ .

$c = - \frac{2 \cdot 60}{2 \cdot - 15}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 15$ .

$c = - \frac{2 \cdot 60}{2 (- 15)}$

Cancel the common factor.

$c = - \frac{2 \cdot 60}{2 \cdot - 15}$

Rewrite the expression.

$c = - \frac{60}{- 15}$

Cancel the common factor of $60$ and $- 15$ .

Factor $15$ out of $60$ .

$c = - \frac{15 (4)}{- 15}$

Move the negative one from the denominator of $\frac{4}{- 1}$ .

$c = - (- 1 \cdot 4)$

Multiply.

Multiply $- 1$ by $4$ .

$c = - - 4$

Multiply $- 1$ by $- 4$ .

$c = 4$

Evaluate $f (4)$ .

Replace the variable $c$ with $4$ in the expression.

$P (4) = - 15 {(4)}^{2} + 120 (4)$

Simplify the result.

Simplify each term.

Raise $4$ to the power of $2$ .

$P (4) = - 15 \cdot 16 + 120 (4)$

Multiply $- 15$ by $16$ .

$P (4) = - 240 + 120 (4)$

Multiply $120$ by $4$ .

$P (4) = - 240 + 480$

Add $- 240$ and $480$ .

$P (4) = 240$

The final answer is $240$ .

$240$

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

$(4, 240)$

Find the Maximum/Minimum Value P(c)=-15c(c-8)

Related Questions

Support LCMGCF.com by disabling your adblocker.