Find the Maximum/Minimum Value P(c)=-20(c-3)^2+180

$P (c) = - 20 {(c - 3)}^{2} + 180$

Step 1

Simplify each term.

Rewrite ${(c - 3)}^{2}$ as $(c - 3) (c - 3)$ .

$P (c) = - 20 ((c - 3) (c - 3)) + 180$

Expand $(c - 3) (c - 3)$ using the FOIL Method.

Apply the distributive property.

$P (c) = - 20 (c (c - 3) - 3 (c - 3)) + 180$

Apply the distributive property.

$P (c) = - 20 (c \cdot c + c \cdot - 3 - 3 (c - 3)) + 180$

Apply the distributive property.

$P (c) = - 20 (c \cdot c + c \cdot - 3 - 3 c - 3 \cdot - 3) + 180$

Simplify and combine like terms.

Simplify each term.

Multiply $c$ by $c$ .

$P (c) = - 20 (c^{2} + c \cdot - 3 - 3 c - 3 \cdot - 3) + 180$

Move $- 3$ to the left of $c$ .

$P (c) = - 20 (c^{2} - 3 \cdot c - 3 c - 3 \cdot - 3) + 180$

Multiply $- 3$ by $- 3$ .

$P (c) = - 20 (c^{2} - 3 c - 3 c + 9) + 180$

Subtract $3 c$ from $- 3 c$ .

$P (c) = - 20 (c^{2} - 6 c + 9) + 180$

Apply the distributive property.

$P (c) = - 20 c^{2} - 20 (- 6 c) - 20 \cdot 9 + 180$

Simplify.

Multiply $- 6$ by $- 20$ .

$P (c) = - 20 c^{2} + 120 c - 20 \cdot 9 + 180$

Multiply $- 20$ by $9$ .

$P (c) = - 20 c^{2} + 120 c - 180 + 180$

Step 2

Combine the opposite terms in $- 20 c^{2} + 120 c - 180 + 180$ .

Add $- 180$ and $180$ .

$P (c) = - 20 c^{2} + 120 c + 0$

Add $- 20 c^{2} + 120 c$ and $0$ .

$P (c) = - 20 c^{2} + 120 c$

Step 3

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 4

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

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

$c = - \frac{120}{2 (- 20)}$

Remove parentheses.

$c = - \frac{120}{2 (- 20)}$

Simplify $- \frac{120}{2 (- 20)}$ .

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

Factor $2$ out of $120$ .

$c = - \frac{2 \cdot 60}{2 \cdot - 20}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 20$ .

$c = - \frac{2 \cdot 60}{2 (- 20)}$

Cancel the common factor.

$c = - \frac{2 \cdot 60}{2 \cdot - 20}$

Rewrite the expression.

$c = - \frac{60}{- 20}$

Cancel the common factor of $60$ and $- 20$ .

Factor $20$ out of $60$ .

$c = - \frac{20 (3)}{- 20}$

Move the negative one from the denominator of $\frac{3}{- 1}$ .

$c = - (- 1 \cdot 3)$

Multiply $- (- 1 \cdot 3)$ .

Multiply $- 1$ by $3$ .

$c = - - 3$

Multiply $- 1$ by $- 3$ .

$c = 3$

Step 5

Evaluate $f (3)$ .

Replace the variable $c$ with $3$ in the expression.

$P (3) = - 20 {(3)}^{2} + 120 (3)$

Simplify the result.

Simplify each term.

Raise $3$ to the power of $2$ .

$P (3) = - 20 \cdot 9 + 120 (3)$

Multiply $- 20$ by $9$ .

$P (3) = - 180 + 120 (3)$

Multiply $120$ by $3$ .

$P (3) = - 180 + 360$

Add $- 180$ and $360$ .

$P (3) = 180$

The final answer is $180$ .

$180$

Step 6

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

$(3, 180)$

Step 7

Find the Maximum/Minimum Value P(c)=-20(c-3)^2+180

Related Questions

Support LCMGCF.com by disabling your adblocker.