Find the Maximum/Minimum Value D(x)=1/36(x-60)^2-100

$D (x) = \frac{1}{36} {(x - 60)}^{2} - 100$

Step 1

Simplify each term.

Rewrite ${(x - 60)}^{2}$ as $(x - 60) (x - 60)$ .

$D (x) = \frac{1}{36} \cdot ((x - 60) (x - 60)) - 100$

Expand $(x - 60) (x - 60)$ using the FOIL Method.

Apply the distributive property.

$D (x) = \frac{1}{36} \cdot (x (x - 60) - 60 (x - 60)) - 100$

Apply the distributive property.

$D (x) = \frac{1}{36} \cdot (x \cdot x + x \cdot - 60 - 60 (x - 60)) - 100$

Apply the distributive property.

$D (x) = \frac{1}{36} \cdot (x \cdot x + x \cdot - 60 - 60 x - 60 \cdot - 60) - 100$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$D (x) = \frac{1}{36} \cdot (x^{2} + x \cdot - 60 - 60 x - 60 \cdot - 60) - 100$

Move $- 60$ to the left of $x$ .

$D (x) = \frac{1}{36} \cdot (x^{2} - 60 \cdot x - 60 x - 60 \cdot - 60) - 100$

Multiply $- 60$ by $- 60$ .

$D (x) = \frac{1}{36} \cdot (x^{2} - 60 x - 60 x + 3600) - 100$

Subtract $60 x$ from $- 60 x$ .

$D (x) = \frac{1}{36} \cdot (x^{2} - 120 x + 3600) - 100$

Apply the distributive property.

$D (x) = \frac{1}{36} x^{2} + \frac{1}{36} \cdot (- 120 x) + \frac{1}{36} \cdot 3600 - 100$

Simplify.

Combine $\frac{1}{36}$ and $x^{2}$ .

$D (x) = \frac{x^{2}}{36} + \frac{1}{36} \cdot (- 120 x) + \frac{1}{36} \cdot 3600 - 100$

Cancel the common factor of $12$ .

Factor $12$ out of $36$ .

$D (x) = \frac{x^{2}}{36} + \frac{1}{12 (3)} \cdot (- 120 x) + \frac{1}{36} \cdot 3600 - 100$

Factor $12$ out of $- 120 x$ .

$D (x) = \frac{x^{2}}{36} + \frac{1}{12 (3)} \cdot (12 (- 10 x)) + \frac{1}{36} \cdot 3600 - 100$

Cancel the common factor.

$D (x) = \frac{x^{2}}{36} + \frac{1}{12 \cdot 3} \cdot (12 (- 10 x)) + \frac{1}{36} \cdot 3600 - 100$

Rewrite the expression.

$D (x) = \frac{x^{2}}{36} + \frac{1}{3} \cdot (- 10 x) + \frac{1}{36} \cdot 3600 - 100$

Combine $- 10$ and $\frac{1}{3}$ .

$D (x) = \frac{x^{2}}{36} + \frac{- 10}{3} x + \frac{1}{36} \cdot 3600 - 100$

Combine $\frac{- 10}{3}$ and $x$ .

$D (x) = \frac{x^{2}}{36} + \frac{- 10 x}{3} + \frac{1}{36} \cdot 3600 - 100$

Cancel the common factor of $36$ .

Factor $36$ out of $3600$ .

$D (x) = \frac{x^{2}}{36} + \frac{- 10 x}{3} + \frac{1}{36} \cdot (36 (100)) - 100$

Cancel the common factor.

$D (x) = \frac{x^{2}}{36} + \frac{- 10 x}{3} + \frac{1}{36} \cdot (36 \cdot 100) - 100$

Rewrite the expression.

$D (x) = \frac{x^{2}}{36} + \frac{- 10 x}{3} + 100 - 100$

Move the negative in front of the fraction.

$D (x) = \frac{x^{2}}{36} - \frac{10 x}{3} + 100 - 100$

Step 2

Combine the opposite terms in $\frac{x^{2}}{36} - \frac{10 x}{3} + 100 - 100$ .

Subtract $100$ from $100$ .

$D (x) = \frac{x^{2}}{36} - \frac{10 x}{3} + 0$

Add $\frac{x^{2}}{36} - \frac{10 x}{3}$ and $0$ .

$D (x) = \frac{x^{2}}{36} - \frac{10 x}{3}$

Step 3

The minimum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is positive, the minimum value of the function is $f (- \frac{b}{2 a})$ .

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

Step 4

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

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

$x = - \frac{- 3.\overline{3}}{2 (0.02\overline{7})}$

Remove parentheses.

$x = - \frac{- 3.\overline{3}}{2 (0.02\overline{7})}$

Simplify $- \frac{- 3.\overline{3}}{2 (0.02\overline{7})}$ .

Multiply $2$ by $0.02\overline{7}$ .

$x = - \frac{- 3.\overline{3}}{0.0\overline{5}}$

Divide $- 3.\overline{3}$ by $0.0\overline{5}$ .

$x = - - 60$

Multiply $- 1$ by $- 60$ .

$x = 60$

Step 5

Evaluate $f (60)$ .

Replace the variable $x$ with $60$ in the expression.

$D (60) = \frac{{(60)}^{2}}{36} - \frac{10 (60)}{3}$

Simplify the result.

Simplify each term.

Raise $60$ to the power of $2$ .

$D (60) = \frac{3599.\overline{9}}{36} - \frac{10 (60)}{3}$

Divide $3599.\overline{9}$ by $36$ .

$D (60) = 100 - \frac{10 (60)}{3}$

Multiply $10$ by $60$ .

$D (60) = 100 - \frac{599.99999999}{3}$

Divide $599.99999999$ by $3$ .

$D (60) = 100 - 1 \cdot 200$

Multiply $- 1$ by $200$ .

$D (60) = 100 - 200$

Subtract $200$ from $100$ .

$D (60) = - 100$

The final answer is $- 100$ .

$- 100$

Step 6

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

$(60, - 100)$

Step 7

Find the Maximum/Minimum Value D(x)=1/36(x-60)^2-100

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