Find the Maximum/Minimum Value h(t)=44t-16t^2

$h (t) = 44 t - 16 t^{2}$

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 $b$ .

$t = - \frac{44}{2 (- 16)}$

Remove parentheses.

$t = - \frac{44}{2 (- 16)}$

Simplify $- \frac{44}{2 (- 16)}$ .

Cancel the common factor of $44$ and $2$ .

Factor $2$ out of $44$ .

$t = - \frac{2 \cdot 22}{2 \cdot - 16}$

Cancel the common factors.

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

$t = - \frac{2 \cdot 22}{2 (- 16)}$

Cancel the common factor.

$t = - \frac{2 \cdot 22}{2 \cdot - 16}$

Rewrite the expression.

$t = - \frac{22}{- 16}$

Cancel the common factor of $22$ and $- 16$ .

Factor $2$ out of $22$ .

$t = - \frac{2 (11)}{- 16}$

Cancel the common factors.

Factor $2$ out of $- 16$ .

$t = - \frac{2 \cdot 11}{2 \cdot - 8}$

Cancel the common factor.

$t = - \frac{2 \cdot 11}{2 \cdot - 8}$

Rewrite the expression.

$t = - \frac{11}{- 8}$

Move the negative in front of the fraction.

$t = - - \frac{11}{8}$

Multiply $- - \frac{11}{8}$ .

Multiply $- 1$ by $- 1$ .

$t = 1 (\frac{11}{8})$

Multiply $\frac{11}{8}$ by $1$ .

$t = \frac{11}{8}$

Step 3

Evaluate $f (\frac{11}{8})$ .

Replace the variable $t$ with $\frac{11}{8}$ in the expression.

$h (\frac{11}{8}) = 44 (\frac{11}{8}) - 16 {(\frac{11}{8})}^{2}$

Simplify the result.

Simplify each term.

Cancel the common factor of $4$ .

Factor $4$ out of $44$ .

$h (\frac{11}{8}) = 4 (11) (\frac{11}{8}) - 16 {(\frac{11}{8})}^{2}$

Factor $4$ out of $8$ .

$h (\frac{11}{8}) = 4 \cdot (11 (\frac{11}{4 \cdot 2})) - 16 {(\frac{11}{8})}^{2}$

Cancel the common factor.

$h (\frac{11}{8}) = 4 \cdot (11 (\frac{11}{4 \cdot 2})) - 16 {(\frac{11}{8})}^{2}$

Rewrite the expression.

$h (\frac{11}{8}) = 11 (\frac{11}{2}) - 16 {(\frac{11}{8})}^{2}$

Combine $11$ and $\frac{11}{2}$ .

$h (\frac{11}{8}) = \frac{11 \cdot 11}{2} - 16 {(\frac{11}{8})}^{2}$

Multiply $11$ by $11$ .

$h (\frac{11}{8}) = \frac{121}{2} - 16 {(\frac{11}{8})}^{2}$

Apply the product rule to $\frac{11}{8}$ .

$h (\frac{11}{8}) = \frac{121}{2} - 16 \frac{11^{2}}{8^{2}}$

Raise $11$ to the power of $2$ .

$h (\frac{11}{8}) = \frac{121}{2} - 16 \frac{121}{8^{2}}$

Raise $8$ to the power of $2$ .

$h (\frac{11}{8}) = \frac{121}{2} - 16 (\frac{121}{64})$

Cancel the common factor of $16$ .

Factor $16$ out of $- 16$ .

$h (\frac{11}{8}) = \frac{121}{2} + 16 (- 1) (\frac{121}{64})$

Factor $16$ out of $64$ .

$h (\frac{11}{8}) = \frac{121}{2} + 16 \cdot (- 1 \frac{121}{16 \cdot 4})$

Cancel the common factor.

$h (\frac{11}{8}) = \frac{121}{2} + 16 \cdot (- 1 \frac{121}{16 \cdot 4})$

Rewrite the expression.

$h (\frac{11}{8}) = \frac{121}{2} - 1 (\frac{121}{4})$

Rewrite $- 1 (\frac{121}{4})$ as $- (\frac{121}{4})$ .

$h (\frac{11}{8}) = \frac{121}{2} - \frac{121}{4}$

To write $\frac{121}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$h (\frac{11}{8}) = \frac{121}{2} \cdot \frac{2}{2} - \frac{121}{4}$

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{121}{2}$ by $\frac{2}{2}$ .

$h (\frac{11}{8}) = \frac{121 \cdot 2}{2 \cdot 2} - \frac{121}{4}$

Multiply $2$ by $2$ .

$h (\frac{11}{8}) = \frac{121 \cdot 2}{4} - \frac{121}{4}$

Combine the numerators over the common denominator.

$h (\frac{11}{8}) = \frac{121 \cdot 2 - 121}{4}$

Simplify the numerator.

Multiply $121$ by $2$ .

$h (\frac{11}{8}) = \frac{242 - 121}{4}$

Subtract $121$ from $242$ .

$h (\frac{11}{8}) = \frac{121}{4}$

The final answer is $\frac{121}{4}$ .

$\frac{121}{4}$

Step 4

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

$(\frac{11}{8}, \frac{121}{4})$

Step 5

Find the Maximum/Minimum Value h(t)=44t-16t^2

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