Find the Maximum/Minimum Value 100x-0.4x^2

$100 x - 0.4 x^{2}$

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 $b$ .

$x = - \frac{100}{2 (- 0.4)}$

Remove parentheses.

$x = - \frac{100}{2 (- 0.4)}$

Simplify $- \frac{100}{2 (- 0.4)}$ .

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

Factor $2$ out of $100$ .

$x = - \frac{2 \cdot 50}{2 \cdot - 0.4}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 0.4$ .

$x = - \frac{2 \cdot 50}{2 (- 0.4)}$

Cancel the common factor.

$x = - \frac{2 \cdot 50}{2 \cdot - 0.4}$

Rewrite the expression.

$x = - \frac{50}{- 0.4}$

Simplify the expression.

Divide $50$ by $- 0.4$ .

$x = - - 125$

Multiply $- 1$ by $- 125$ .

$x = 125$

Step 3

Evaluate $f (125)$ .

Replace the variable $x$ with $125$ in the expression.

$f (125) = 100 (125) - 0.4 {(125)}^{2}$

Simplify the result.

Simplify each term.

Multiply $100$ by $125$ .

$f (125) = 12500 - 0.4 {(125)}^{2}$

Raise $125$ to the power of $2$ .

$f (125) = 12500 - 0.4 \cdot 15625$

Multiply $- 0.4$ by $15625$ .

$f (125) = 12500 - 6250$

Subtract $6250$ from $12500$ .

$f (125) = 6250$

The final answer is $6250$ .

$6250$

Step 4

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

$(125, 6250)$

Step 5

Find the Maximum/Minimum Value 100x-0.4x^2

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