Find the Maximum/Minimum Value -5t^2+40t

$- 5 t^{2} + 40 t$

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

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

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

$t = - \frac{40}{2 (- 5)}$

Remove parentheses.

$t = - \frac{40}{2 (- 5)}$

Simplify $- \frac{40}{2 (- 5)}$ .

Cancel the common factor of $40$ and $2$ .

Factor $2$ out of $40$ .

$t = - \frac{2 \cdot 20}{2 \cdot - 5}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 5$ .

$t = - \frac{2 \cdot 20}{2 (- 5)}$

Cancel the common factor.

$t = - \frac{2 \cdot 20}{2 \cdot - 5}$

Rewrite the expression.

$t = - \frac{20}{- 5}$

Cancel the common factor of $20$ and $- 5$ .

Factor $5$ out of $20$ .

$t = - \frac{5 (4)}{- 5}$

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

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

Multiply.

Multiply $- 1$ by $4$ .

$t = - - 4$

Multiply $- 1$ by $- 4$ .

$t = 4$

Evaluate $f (4)$ .

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

$f (4) = - 5 {(4)}^{2} + 40 (4)$

Simplify the result.

Simplify each term.

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

$f (4) = - 5 \cdot 16 + 40 (4)$

Multiply $- 5$ by $16$ .

$f (4) = - 80 + 40 (4)$

Multiply $40$ by $4$ .

$f (4) = - 80 + 160$

Add $- 80$ and $160$ .

$f (4) = 80$

The final answer is $80$ .

$80$

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

$(4, 80)$

Find the Maximum/Minimum Value -5t^2+40t

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