Find the Maximum/Minimum Value h=256t-32t^2

$h = 256 t - 32 t^{2}$

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 $b$ .

$t = - \frac{256}{2 (- 32)}$

Remove parentheses.

$t = - \frac{256}{2 (- 32)}$

Simplify $- \frac{256}{2 (- 32)}$ .

Cancel the common factor of $256$ and $2$ .

Factor $2$ out of $256$ .

$t = - \frac{2 \cdot 128}{2 \cdot - 32}$

Cancel the common factors.

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

$t = - \frac{2 \cdot 128}{2 (- 32)}$

Cancel the common factor.

$t = - \frac{2 \cdot 128}{2 \cdot - 32}$

Rewrite the expression.

$t = - \frac{128}{- 32}$

Cancel the common factor of $128$ and $- 32$ .

Factor $32$ out of $128$ .

$t = - \frac{32 (4)}{- 32}$

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

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

Multiply $- (- 1 \cdot 4)$ .

Multiply $- 1$ by $4$ .

$t = - - 4$

Multiply $- 1$ by $- 4$ .

$t = 4$

Step 3

Evaluate $f (4)$ .

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

$f (4) = 256 (4) - 32 {(4)}^{2}$

Simplify the result.

Simplify each term.

Multiply $256$ by $4$ .

$f (4) = 1024 - 32 {(4)}^{2}$

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

$f (4) = 1024 - 32 \cdot 16$

Multiply $- 32$ by $16$ .

$f (4) = 1024 - 512$

Subtract $512$ from $1024$ .

$f (4) = 512$

The final answer is $512$ .

$512$

Step 4

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

$(4, 512)$

Step 5

Find the Maximum/Minimum Value h=256t-32t^2

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