Find the Maximum/Minimum Value h(t)=-16t^2+108t+28

$h (t) = - 16 t^{2} + 108 t + 28$

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 $b$ .

$t = - \frac{108}{2 (- 16)}$

Remove parentheses.

$t = - \frac{108}{2 (- 16)}$

Simplify $- \frac{108}{2 (- 16)}$ .

Cancel the common factor of $108$ and $2$ .

Factor $2$ out of $108$ .

$t = - \frac{2 \cdot 54}{2 \cdot - 16}$

Cancel the common factors.

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

$t = - \frac{2 \cdot 54}{2 (- 16)}$

Cancel the common factor.

$t = - \frac{2 \cdot 54}{2 \cdot - 16}$

Rewrite the expression.

$t = - \frac{54}{- 16}$

Cancel the common factor of $54$ and $- 16$ .

Factor $2$ out of $54$ .

$t = - \frac{2 (27)}{- 16}$

Cancel the common factors.

Factor $2$ out of $- 16$ .

$t = - \frac{2 \cdot 27}{2 \cdot - 8}$

Cancel the common factor.

$t = - \frac{2 \cdot 27}{2 \cdot - 8}$

Rewrite the expression.

$t = - \frac{27}{- 8}$

Move the negative in front of the fraction.

$t = - - \frac{27}{8}$

Multiply $- - \frac{27}{8}$ .

Multiply $- 1$ by $- 1$ .

$t = 1 (\frac{27}{8})$

Multiply $\frac{27}{8}$ by $1$ .

$t = \frac{27}{8}$

Step 3

Evaluate $f (\frac{27}{8})$ .

Replace the variable $t$ with $\frac{27}{8}$ in the expression.

$h (\frac{27}{8}) = - 16 {(\frac{27}{8})}^{2} + 108 (\frac{27}{8}) + 28$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{27}{8}$ .

$h (\frac{27}{8}) = - 16 \frac{27^{2}}{8^{2}} + 108 (\frac{27}{8}) + 28$

Raise $27$ to the power of $2$ .

$h (\frac{27}{8}) = - 16 \frac{729}{8^{2}} + 108 (\frac{27}{8}) + 28$

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

$h (\frac{27}{8}) = - 16 (\frac{729}{64}) + 108 (\frac{27}{8}) + 28$

Cancel the common factor of $16$ .

Factor $16$ out of $- 16$ .

$h (\frac{27}{8}) = 16 (- 1) (\frac{729}{64}) + 108 (\frac{27}{8}) + 28$

Factor $16$ out of $64$ .

$h (\frac{27}{8}) = 16 \cdot (- 1 \frac{729}{16 \cdot 4}) + 108 (\frac{27}{8}) + 28$

Cancel the common factor.

$h (\frac{27}{8}) = 16 \cdot (- 1 \frac{729}{16 \cdot 4}) + 108 (\frac{27}{8}) + 28$

Rewrite the expression.

$h (\frac{27}{8}) = - 1 (\frac{729}{4}) + 108 (\frac{27}{8}) + 28$

Rewrite $- 1 (\frac{729}{4})$ as $- (\frac{729}{4})$ .

$h (\frac{27}{8}) = - (\frac{729}{4}) + 108 (\frac{27}{8}) + 28$

Cancel the common factor of $4$ .

Factor $4$ out of $108$ .

$h (\frac{27}{8}) = - \frac{729}{4} + 4 (27) (\frac{27}{8}) + 28$

Factor $4$ out of $8$ .

$h (\frac{27}{8}) = - \frac{729}{4} + 4 \cdot (27 (\frac{27}{4 \cdot 2})) + 28$

Cancel the common factor.

$h (\frac{27}{8}) = - \frac{729}{4} + 4 \cdot (27 (\frac{27}{4 \cdot 2})) + 28$

Rewrite the expression.

$h (\frac{27}{8}) = - \frac{729}{4} + 27 (\frac{27}{2}) + 28$

Combine $27$ and $\frac{27}{2}$ .

$h (\frac{27}{8}) = - \frac{729}{4} + \frac{27 \cdot 27}{2} + 28$

Multiply $27$ by $27$ .

$h (\frac{27}{8}) = - \frac{729}{4} + \frac{729}{2} + 28$

Find the common denominator.

Multiply $\frac{729}{2}$ by $\frac{2}{2}$ .

$h (\frac{27}{8}) = - \frac{729}{4} + \frac{729}{2} \cdot \frac{2}{2} + 28$

Multiply $\frac{729}{2}$ by $\frac{2}{2}$ .

$h (\frac{27}{8}) = - \frac{729}{4} + \frac{729 \cdot 2}{2 \cdot 2} + 28$

Write $28$ as a fraction with denominator $1$ .

$h (\frac{27}{8}) = - \frac{729}{4} + \frac{729 \cdot 2}{2 \cdot 2} + \frac{28}{1}$

Multiply $\frac{28}{1}$ by $\frac{4}{4}$ .

$h (\frac{27}{8}) = - \frac{729}{4} + \frac{729 \cdot 2}{2 \cdot 2} + \frac{28}{1} \cdot \frac{4}{4}$

Multiply $\frac{28}{1}$ by $\frac{4}{4}$ .

$h (\frac{27}{8}) = - \frac{729}{4} + \frac{729 \cdot 2}{2 \cdot 2} + \frac{28 \cdot 4}{4}$

Multiply $2$ by $2$ .

$h (\frac{27}{8}) = - \frac{729}{4} + \frac{729 \cdot 2}{4} + \frac{28 \cdot 4}{4}$

Combine the numerators over the common denominator.

$h (\frac{27}{8}) = \frac{- 729 + 729 \cdot 2 + 28 \cdot 4}{4}$

Simplify each term.

Multiply $729$ by $2$ .

$h (\frac{27}{8}) = \frac{- 729 + 1458 + 28 \cdot 4}{4}$

Multiply $28$ by $4$ .

$h (\frac{27}{8}) = \frac{- 729 + 1458 + 112}{4}$

Simplify by adding numbers.

Add $- 729$ and $1458$ .

$h (\frac{27}{8}) = \frac{729 + 112}{4}$

Add $729$ and $112$ .

$h (\frac{27}{8}) = \frac{841}{4}$

The final answer is $\frac{841}{4}$ .

$\frac{841}{4}$

Step 4

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

$(\frac{27}{8}, \frac{841}{4})$

Step 5

Find the Maximum/Minimum Value h(t)=-16t^2+108t+28

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