Math / Calculus

Question(1 point)
The total cost (in hundreds of dollars) of producing $x$ slide rules per day is $C(x)=12+\sqrt{2 x+20}$
Find the marginal cost at each production level $x$ given below: $\begin{array}{l} x=25 \square \\ x=40 \square \end{array}$

Studdy Solution

STEP 1

1. The function $C(x) = 12 + \sqrt{2x + 20}$ represents the total cost in hundreds of dollars.
2. The marginal cost is the derivative of the cost function $C(x)$ with respect to $x$ .
3. We will evaluate the derivative at specific values of $x$ .

STEP 2

1. Find the derivative of the cost function $C(x)$ .
2. Evaluate the derivative at $x = 25$ .
3. Evaluate the derivative at $x = 40$ .

STEP 3

To find the marginal cost, we need to differentiate the cost function $C(x) = 12 + \sqrt{2x + 20}$ .
The derivative of a constant is zero, so we focus on differentiating $\sqrt{2x + 20}$ .
Using the chain rule, let $u = 2x + 20$ , then $\sqrt{u} = u^{1/2}$ .
The derivative of $u^{1/2}$ with respect to $u$ is $\frac{1}{2}u^{-1/2}$ .
The derivative of $u = 2x + 20$ with respect to $x$ is $2$ .
Thus, applying the chain rule, the derivative of $\sqrt{2x + 20}$ with respect to $x$ is:
$\frac{d}{dx} \sqrt{2x + 20} = \frac{1}{2}(2x + 20)^{-1/2} \cdot 2 = (2x + 20)^{-1/2}$
So, the derivative of $C(x)$ is:
$C'(x) = (2x + 20)^{-1/2}$

STEP 4

Evaluate the derivative at $x = 25$ :
$C'(25) = (2 \cdot 25 + 20)^{-1/2} = (50 + 20)^{-1/2} = 70^{-1/2}$
$C'(25) = \frac{1}{\sqrt{70}}$

STEP 5

Evaluate the derivative at $x = 40$ :
$C'(40) = (2 \cdot 40 + 20)^{-1/2} = (80 + 20)^{-1/2} = 100^{-1/2}$
$C'(40) = \frac{1}{\sqrt{100}} = \frac{1}{10}$
The marginal cost at $x = 25$ is $\frac{1}{\sqrt{70}}$ and at $x = 40$ is $\frac{1}{10}$ .

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