Find the Maximum/Minimum Value 18t-3t^2

$18 t - 3 t^{2}$

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

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

Step 2

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

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

$t = - \frac{18}{2 (- 3)}$

Remove parentheses.

$t = - \frac{18}{2 (- 3)}$

Simplify $- \frac{18}{2 (- 3)}$ .

Cancel the common factor of $18$ and $2$ .

Factor $2$ out of $18$ .

$t = - \frac{2 \cdot 9}{2 \cdot - 3}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 3$ .

$t = - \frac{2 \cdot 9}{2 (- 3)}$

Cancel the common factor.

$t = - \frac{2 \cdot 9}{2 \cdot - 3}$

Rewrite the expression.

$t = - \frac{9}{- 3}$

Cancel the common factor of $9$ and $- 3$ .

Factor $3$ out of $9$ .

$t = - \frac{3 (3)}{- 3}$

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

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

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

Multiply $- 1$ by $3$ .

$t = - - 3$

Multiply $- 1$ by $- 3$ .

$t = 3$

Step 3

Evaluate $f (3)$ .

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

$f (3) = 18 (3) - 3 {(3)}^{2}$

Simplify the result.

Simplify each term.

Multiply $18$ by $3$ .

$f (3) = 54 - 3 {(3)}^{2}$

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

$f (3) = 54 - 3 \cdot 9$

Multiply $- 3$ by $9$ .

$f (3) = 54 - 27$

Subtract $27$ from $54$ .

$f (3) = 27$

The final answer is $27$ .

$27$

Step 4

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

$(3, 27)$

Step 5

Find the Maximum/Minimum Value 18t-3t^2

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