Find the Maximum/Minimum Value r(x)=400x-0.2x^2

$r (x) = 400 x - 0.2 x^{2}$

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 $b$ .

$x = - \frac{400}{2 (- 0.2)}$

Remove parentheses.

$x = - \frac{400}{2 (- 0.2)}$

Simplify $- \frac{400}{2 (- 0.2)}$ .

Cancel the common factor of $400$ and $2$ .

Factor $2$ out of $400$ .

$x = - \frac{2 \cdot 200}{2 \cdot - 0.2}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 0.2$ .

$x = - \frac{2 \cdot 200}{2 (- 0.2)}$

Cancel the common factor.

$x = - \frac{2 \cdot 200}{2 \cdot - 0.2}$

Rewrite the expression.

$x = - \frac{200}{- 0.2}$

Simplify the expression.

Divide $200$ by $- 0.2$ .

$x = - - 1000$

Multiply $- 1$ by $- 1000$ .

$x = 1000$

Step 3

Evaluate $f (1000)$ .

Replace the variable $x$ with $1000$ in the expression.

$r (1000) = 400 (1000) - 0.2 {(1000)}^{2}$

Simplify the result.

Simplify each term.

Multiply $400$ by $1000$ .

$r (1000) = 400000 - 0.2 {(1000)}^{2}$

Raise $1000$ to the power of $2$ .

$r (1000) = 400000 - 0.2 \cdot 1000000$

Multiply $- 0.2$ by $1000000$ .

$r (1000) = 400000 - 200000$

Subtract $200000$ from $400000$ .

$r (1000) = 200000$

The final answer is $200000$ .

$200000$

Step 4

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

$(1000, 200000)$

Step 5

Find the Maximum/Minimum Value r(x)=400x-0.2x^2

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