Find the Maximum/Minimum Value h=-16t^2+48t+448

$h = - 16 t^{2} + 48 t + 448$

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 $b$ .

$t = - \frac{48}{2 (- 16)}$

Remove parentheses.

$t = - \frac{48}{2 (- 16)}$

Simplify $- \frac{48}{2 (- 16)}$ .

Cancel the common factor of $48$ and $2$ .

Factor $2$ out of $48$ .

$t = - \frac{2 \cdot 24}{2 \cdot - 16}$

Cancel the common factors.

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

$t = - \frac{2 \cdot 24}{2 (- 16)}$

Cancel the common factor.

$t = - \frac{2 \cdot 24}{2 \cdot - 16}$

Rewrite the expression.

$t = - \frac{24}{- 16}$

Cancel the common factor of $24$ and $- 16$ .

Factor $8$ out of $24$ .

$t = - \frac{8 (3)}{- 16}$

Cancel the common factors.

Factor $8$ out of $- 16$ .

$t = - \frac{8 \cdot 3}{8 \cdot - 2}$

Cancel the common factor.

$t = - \frac{8 \cdot 3}{8 \cdot - 2}$

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Multiply $- - \frac{3}{2}$ .

Multiply $- 1$ by $- 1$ .

$t = 1 (\frac{3}{2})$

Multiply $\frac{3}{2}$ by $1$ .

$t = \frac{3}{2}$

Step 3

Evaluate $f (\frac{3}{2})$ .

Replace the variable $t$ with $\frac{3}{2}$ in the expression.

$f (\frac{3}{2}) = - 16 {(\frac{3}{2})}^{2} + 48 (\frac{3}{2}) + 448$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{3}{2}$ .

$f (\frac{3}{2}) = - 16 \frac{3^{2}}{2^{2}} + 48 (\frac{3}{2}) + 448$

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

$f (\frac{3}{2}) = - 16 \frac{9}{2^{2}} + 48 (\frac{3}{2}) + 448$

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

$f (\frac{3}{2}) = - 16 (\frac{9}{4}) + 48 (\frac{3}{2}) + 448$

Cancel the common factor of $4$ .

Factor $4$ out of $- 16$ .

$f (\frac{3}{2}) = 4 (- 4) (\frac{9}{4}) + 48 (\frac{3}{2}) + 448$

Cancel the common factor.

$f (\frac{3}{2}) = 4 \cdot (- 4 (\frac{9}{4})) + 48 (\frac{3}{2}) + 448$

Rewrite the expression.

$f (\frac{3}{2}) = - 4 \cdot 9 + 48 (\frac{3}{2}) + 448$

Multiply $- 4$ by $9$ .

$f (\frac{3}{2}) = - 36 + 48 (\frac{3}{2}) + 448$

Cancel the common factor of $2$ .

Factor $2$ out of $48$ .

$f (\frac{3}{2}) = - 36 + 2 (24) (\frac{3}{2}) + 448$

Cancel the common factor.

$f (\frac{3}{2}) = - 36 + 2 \cdot (24 (\frac{3}{2})) + 448$

Rewrite the expression.

$f (\frac{3}{2}) = - 36 + 24 \cdot 3 + 448$

Multiply $24$ by $3$ .

$f (\frac{3}{2}) = - 36 + 72 + 448$

Simplify by adding numbers.

Add $- 36$ and $72$ .

$f (\frac{3}{2}) = 36 + 448$

Add $36$ and $448$ .

$f (\frac{3}{2}) = 484$

The final answer is $484$ .

$484$

Step 4

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

$(\frac{3}{2}, 484)$

Step 5

Find the Maximum/Minimum Value h=-16t^2+48t+448

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