Find the Maximum/Minimum Value -16t^2+152t

$- 16 t^{2} + 152 t$

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 $b$ .

$t = - \frac{152}{2 (- 16)}$

Remove parentheses.

$t = - \frac{152}{2 (- 16)}$

Simplify $- \frac{152}{2 (- 16)}$ .

Cancel the common factor of $152$ and $2$ .

Factor $2$ out of $152$ .

$t = - \frac{2 \cdot 76}{2 \cdot - 16}$

Cancel the common factors.

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

$t = - \frac{2 \cdot 76}{2 (- 16)}$

Cancel the common factor.

$t = - \frac{2 \cdot 76}{2 \cdot - 16}$

Rewrite the expression.

$t = - \frac{76}{- 16}$

Cancel the common factor of $76$ and $- 16$ .

Factor $4$ out of $76$ .

$t = - \frac{4 (19)}{- 16}$

Cancel the common factors.

Factor $4$ out of $- 16$ .

$t = - \frac{4 \cdot 19}{4 \cdot - 4}$

Cancel the common factor.

$t = - \frac{4 \cdot 19}{4 \cdot - 4}$

Rewrite the expression.

$t = - \frac{19}{- 4}$

Move the negative in front of the fraction.

$t = - - \frac{19}{4}$

Multiply $- - \frac{19}{4}$ .

Multiply $- 1$ by $- 1$ .

$t = 1 (\frac{19}{4})$

Multiply $\frac{19}{4}$ by $1$ .

$t = \frac{19}{4}$

Step 3

Evaluate $f (\frac{19}{4})$ .

Replace the variable $t$ with $\frac{19}{4}$ in the expression.

$f (\frac{19}{4}) = - 16 {(\frac{19}{4})}^{2} + 152 (\frac{19}{4})$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{19}{4}$ .

$f (\frac{19}{4}) = - 16 \frac{19^{2}}{4^{2}} + 152 (\frac{19}{4})$

Raise $19$ to the power of $2$ .

$f (\frac{19}{4}) = - 16 \frac{361}{4^{2}} + 152 (\frac{19}{4})$

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

$f (\frac{19}{4}) = - 16 (\frac{361}{16}) + 152 (\frac{19}{4})$

Cancel the common factor of $16$ .

Factor $16$ out of $- 16$ .

$f (\frac{19}{4}) = 16 (- 1) (\frac{361}{16}) + 152 (\frac{19}{4})$

Cancel the common factor.

$f (\frac{19}{4}) = 16 \cdot (- 1 (\frac{361}{16})) + 152 (\frac{19}{4})$

Rewrite the expression.

$f (\frac{19}{4}) = - 1 \cdot 361 + 152 (\frac{19}{4})$

Multiply $- 1$ by $361$ .

$f (\frac{19}{4}) = - 361 + 152 (\frac{19}{4})$

Cancel the common factor of $4$ .

Factor $4$ out of $152$ .

$f (\frac{19}{4}) = - 361 + 4 (38) (\frac{19}{4})$

Cancel the common factor.

$f (\frac{19}{4}) = - 361 + 4 \cdot (38 (\frac{19}{4}))$

Rewrite the expression.

$f (\frac{19}{4}) = - 361 + 38 \cdot 19$

Multiply $38$ by $19$ .

$f (\frac{19}{4}) = - 361 + 722$

Add $- 361$ and $722$ .

$f (\frac{19}{4}) = 361$

The final answer is $361$ .

$361$

Step 4

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

$(\frac{19}{4}, 361)$

Step 5

Find the Maximum/Minimum Value -16t^2+152t

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