Find the Maximum/Minimum Value A(w)=-w^2+100w

$A (w) = - w^{2} + 100 w$

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 $f (- \frac{b}{2 a})$ .

$f_{max}$ $w = a w^{2} + b w + c$ occurs at $w = - \frac{b}{2 a}$

Step 2

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

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

$w = - \frac{100}{2 (- 1)}$

Remove parentheses.

$w = - \frac{100}{2 (- 1)}$

Simplify $- \frac{100}{2 (- 1)}$ .

Cancel the common factor of $100$ and $2$ .

Factor $2$ out of $100$ .

$w = - \frac{2 \cdot 50}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{50}{- 1}$ .

$w = - (- 1 \cdot 50)$

Multiply $- (- 1 \cdot 50)$ .

Multiply $- 1$ by $50$ .

$w = - - 50$

Multiply $- 1$ by $- 50$ .

$w = 50$

Step 3

Evaluate $f (50)$ .

Replace the variable $w$ with $50$ in the expression.

$A (50) = - {(50)}^{2} + 100 (50)$

Simplify the result.

Simplify each term.

Raise $50$ to the power of $2$ .

$A (50) = - 1 \cdot 2500 + 100 (50)$

Multiply $- 1$ by $2500$ .

$A (50) = - 2500 + 100 (50)$

Multiply $100$ by $50$ .

$A (50) = - 2500 + 5000$

Add $- 2500$ and $5000$ .

$A (50) = 2500$

The final answer is $2500$ .

$2500$

Step 4

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

$(50, 2500)$

Step 5

Find the Maximum/Minimum Value A(w)=-w^2+100w

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