Find the Maximum/Minimum Value h=320t-32t^2

$h = 320 t - 32 t^{2}$

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 $b$ .

$t = - \frac{320}{2 (- 32)}$

Remove parentheses.

$t = - \frac{320}{2 (- 32)}$

Simplify $- \frac{320}{2 (- 32)}$ .

Cancel the common factor of $320$ and $2$ .

Factor $2$ out of $320$ .

$t = - \frac{2 \cdot 160}{2 \cdot - 32}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 32$ .

$t = - \frac{2 \cdot 160}{2 (- 32)}$

Cancel the common factor.

$t = - \frac{2 \cdot 160}{2 \cdot - 32}$

Rewrite the expression.

$t = - \frac{160}{- 32}$

Cancel the common factor of $160$ and $- 32$ .

Factor $32$ out of $160$ .

$t = - \frac{32 (5)}{- 32}$

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

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

Multiply $- (- 1 \cdot 5)$ .

Multiply $- 1$ by $5$ .

$t = - - 5$

Multiply $- 1$ by $- 5$ .

$t = 5$

Step 3

Evaluate $f (5)$ .

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

$f (5) = 320 (5) - 32 {(5)}^{2}$

Simplify the result.

Simplify each term.

Multiply $320$ by $5$ .

$f (5) = 1600 - 32 {(5)}^{2}$

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

$f (5) = 1600 - 32 \cdot 25$

Multiply $- 32$ by $25$ .

$f (5) = 1600 - 800$

Subtract $800$ from $1600$ .

$f (5) = 800$

The final answer is $800$ .

$800$

Step 4

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

$(5, 800)$

Step 5

Find the Maximum/Minimum Value h=320t-32t^2

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