Find the Maximum/Minimum Value h(t)=112t-16t^2

$h (t) = 112 t - 16 t^{2}$

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 $t = - \frac{b}{2 a}$

Find the value of $t = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

$t = - \frac{112}{2 (- 16)}$

Remove parentheses.

$t = - \frac{112}{2 (- 16)}$

Simplify $- \frac{112}{2 (- 16)}$ .

Cancel the common factor of $112$ and $2$ .

Factor $2$ out of $112$ .

$t = - \frac{2 \cdot 56}{2 \cdot - 16}$

Cancel the common factors.

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

$t = - \frac{2 \cdot 56}{2 (- 16)}$

Cancel the common factor.

$t = - \frac{2 \cdot 56}{2 \cdot - 16}$

Rewrite the expression.

$t = - \frac{56}{- 16}$

Cancel the common factor of $56$ and $- 16$ .

Factor $8$ out of $56$ .

$t = - \frac{8 (7)}{- 16}$

Cancel the common factors.

Factor $8$ out of $- 16$ .

$t = - \frac{8 \cdot 7}{8 \cdot - 2}$

Cancel the common factor.

$t = - \frac{8 \cdot 7}{8 \cdot - 2}$

Rewrite the expression.

$t = - \frac{7}{- 2}$

Move the negative in front of the fraction.

$t = - - \frac{7}{2}$

Multiply $- - \frac{7}{2}$ .

Multiply $- 1$ by $- 1$ .

$t = 1 (\frac{7}{2})$

Multiply $\frac{7}{2}$ by $1$ .

$t = \frac{7}{2}$

Evaluate $f (\frac{7}{2})$ .

Replace the variable $t$ with $\frac{7}{2}$ in the expression.

$h (\frac{7}{2}) = 112 (\frac{7}{2}) - 16 {(\frac{7}{2})}^{2}$

Simplify the result.

Simplify each term.

Cancel the common factor of $2$ .

Factor $2$ out of $112$ .

$h (\frac{7}{2}) = 2 (56) (\frac{7}{2}) - 16 {(\frac{7}{2})}^{2}$

Cancel the common factor.

$h (\frac{7}{2}) = 2 \cdot (56 (\frac{7}{2})) - 16 {(\frac{7}{2})}^{2}$

Rewrite the expression.

$h (\frac{7}{2}) = 56 \cdot 7 - 16 {(\frac{7}{2})}^{2}$

Multiply $56$ by $7$ .

$h (\frac{7}{2}) = 392 - 16 {(\frac{7}{2})}^{2}$

Apply the product rule to $\frac{7}{2}$ .

$h (\frac{7}{2}) = 392 - 16 \frac{7^{2}}{2^{2}}$

Raise $7$ to the power of $2$ .

$h (\frac{7}{2}) = 392 - 16 \frac{49}{2^{2}}$

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

$h (\frac{7}{2}) = 392 - 16 (\frac{49}{4})$

Cancel the common factor of $4$ .

Factor $4$ out of $- 16$ .

$h (\frac{7}{2}) = 392 + 4 (- 4) (\frac{49}{4})$

Cancel the common factor.

$h (\frac{7}{2}) = 392 + 4 \cdot (- 4 (\frac{49}{4}))$

Rewrite the expression.

$h (\frac{7}{2}) = 392 - 4 \cdot 49$

Multiply $- 4$ by $49$ .

$h (\frac{7}{2}) = 392 - 196$

Subtract $196$ from $392$ .

$h (\frac{7}{2}) = 196$

The final answer is $196$ .

$196$

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

$(\frac{7}{2}, 196)$

Find the Maximum/Minimum Value h(t)=112t-16t^2

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