Find the Maximum/Minimum Value r(x)=1600x-2x^2

$r (x) = 1600 x - 2 x^{2}$

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

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

Step 2

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

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

$x = - \frac{1600}{2 (- 2)}$

Remove parentheses.

$x = - \frac{1600}{2 (- 2)}$

Simplify $- \frac{1600}{2 (- 2)}$ .

Cancel the common factor of $1600$ and $2$ .

Factor $2$ out of $1600$ .

$x = - \frac{2 \cdot 800}{2 \cdot - 2}$

Cancel the common factors.

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

$x = - \frac{2 \cdot 800}{2 (- 2)}$

Cancel the common factor.

$x = - \frac{2 \cdot 800}{2 \cdot - 2}$

Rewrite the expression.

$x = - \frac{800}{- 2}$

Cancel the common factor of $800$ and $- 2$ .

Factor $2$ out of $800$ .

$x = - \frac{2 (400)}{- 2}$

Move the negative one from the denominator of $\frac{400}{- 1}$ .

$x = - (- 1 \cdot 400)$

Multiply $- (- 1 \cdot 400)$ .

Multiply $- 1$ by $400$ .

$x = - - 400$

Multiply $- 1$ by $- 400$ .

$x = 400$

Step 3

Evaluate $f (400)$ .

Replace the variable $x$ with $400$ in the expression.

$r (400) = 1600 (400) - 2 {(400)}^{2}$

Simplify the result.

Simplify each term.

Multiply $1600$ by $400$ .

$r (400) = 640000 - 2 {(400)}^{2}$

Raise $400$ to the power of $2$ .

$r (400) = 640000 - 2 \cdot 160000$

Multiply $- 2$ by $160000$ .

$r (400) = 640000 - 320000$

Subtract $320000$ from $640000$ .

$r (400) = 320000$

The final answer is $320000$ .

$320000$

Step 4

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

$(400, 320000)$

Step 5

Find the Maximum/Minimum Value r(x)=1600x-2x^2

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