Find the Maximum/Minimum Value 160t-16t^2

$160 t - 16 t^{2}$

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 $b$ .

$t = - \frac{160}{2 (- 16)}$

Remove parentheses.

$t = - \frac{160}{2 (- 16)}$

Simplify $- \frac{160}{2 (- 16)}$ .

Cancel the common factor of $160$ and $2$ .

Factor $2$ out of $160$ .

$t = - \frac{2 \cdot 80}{2 \cdot - 16}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 16$ .

$t = - \frac{2 \cdot 80}{2 (- 16)}$

Cancel the common factor.

$t = - \frac{2 \cdot 80}{2 \cdot - 16}$

Rewrite the expression.

$t = - \frac{80}{- 16}$

Cancel the common factor of $80$ and $- 16$ .

Factor $16$ out of $80$ .

$t = - \frac{16 (5)}{- 16}$

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

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

Multiply.

Multiply $- 1$ by $5$ .

$t = - - 5$

Multiply $- 1$ by $- 5$ .

$t = 5$

Evaluate $f (5)$ .

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

$f (5) = 160 (5) - 16 {(5)}^{2}$

Simplify the result.

Simplify each term.

Multiply $160$ by $5$ .

$f (5) = 800 - 16 {(5)}^{2}$

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

$f (5) = 800 - 16 \cdot 25$

Multiply $- 16$ by $25$ .

$f (5) = 800 - 400$

Subtract $400$ from $800$ .

$f (5) = 400$

The final answer is $400$ .

$400$

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

$(5, 400)$

Find the Maximum/Minimum Value 160t-16t^2

Related Questions

Support LCMGCF.com by disabling your adblocker.