Find the Maximum/Minimum Value H(t)=40t-16t^2

$H (t) = 40 t - 16 t^{2}$

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 $b$ .

$t = - \frac{40}{2 (- 16)}$

Remove parentheses.

$t = - \frac{40}{2 (- 16)}$

Simplify $- \frac{40}{2 (- 16)}$ .

Cancel the common factor of $40$ and $2$ .

Factor $2$ out of $40$ .

$t = - \frac{2 \cdot 20}{2 \cdot - 16}$

Cancel the common factors.

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

$t = - \frac{2 \cdot 20}{2 (- 16)}$

Cancel the common factor.

$t = - \frac{2 \cdot 20}{2 \cdot - 16}$

Rewrite the expression.

$t = - \frac{20}{- 16}$

Cancel the common factor of $20$ and $- 16$ .

Factor $4$ out of $20$ .

$t = - \frac{4 (5)}{- 16}$

Cancel the common factors.

Factor $4$ out of $- 16$ .

$t = - \frac{4 \cdot 5}{4 \cdot - 4}$

Cancel the common factor.

$t = - \frac{4 \cdot 5}{4 \cdot - 4}$

Rewrite the expression.

$t = - \frac{5}{- 4}$

Move the negative in front of the fraction.

$t = - - \frac{5}{4}$

Multiply $- - \frac{5}{4}$ .

Multiply $- 1$ by $- 1$ .

$t = 1 (\frac{5}{4})$

Multiply $\frac{5}{4}$ by $1$ .

$t = \frac{5}{4}$

Evaluate $f (\frac{5}{4})$ .

Replace the variable $t$ with $\frac{5}{4}$ in the expression.

$H (\frac{5}{4}) = 40 (\frac{5}{4}) - 16 {(\frac{5}{4})}^{2}$

Simplify the result.

Simplify each term.

Cancel the common factor of $4$ .

Factor $4$ out of $40$ .

$H (\frac{5}{4}) = 4 (10) (\frac{5}{4}) - 16 {(\frac{5}{4})}^{2}$

Cancel the common factor.

$H (\frac{5}{4}) = 4 \cdot (10 (\frac{5}{4})) - 16 {(\frac{5}{4})}^{2}$

Rewrite the expression.

$H (\frac{5}{4}) = 10 \cdot 5 - 16 {(\frac{5}{4})}^{2}$

Multiply $10$ by $5$ .

$H (\frac{5}{4}) = 50 - 16 {(\frac{5}{4})}^{2}$

Apply the product rule to $\frac{5}{4}$ .

$H (\frac{5}{4}) = 50 - 16 \frac{5^{2}}{4^{2}}$

Raise $5$ to the power of $2$ .

$H (\frac{5}{4}) = 50 - 16 \frac{25}{4^{2}}$

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

$H (\frac{5}{4}) = 50 - 16 (\frac{25}{16})$

Cancel the common factor of $16$ .

Factor $16$ out of $- 16$ .

$H (\frac{5}{4}) = 50 + 16 (- 1) (\frac{25}{16})$

Cancel the common factor.

$H (\frac{5}{4}) = 50 + 16 \cdot (- 1 (\frac{25}{16}))$

Rewrite the expression.

$H (\frac{5}{4}) = 50 - 1 \cdot 25$

Multiply $- 1$ by $25$ .

$H (\frac{5}{4}) = 50 - 25$

Subtract $25$ from $50$ .

$H (\frac{5}{4}) = 25$

The final answer is $25$ .

$25$

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

$(\frac{5}{4}, 25)$

Find the Maximum/Minimum Value H(t)=40t-16t^2

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