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Math  /  Calculus

QuestionC(x)=4000+55x+0.1x2C(x) = 4000 + 55x + 0.1x^2 C‾(x)=totalCx=4000x+55+0.1x\overline{C}(x) = \frac{total C}{x} = \frac{4000}{x} + 55 + 0.1x a)a) instantaneous rate of change of C‾(x)\overline{C}(x) C‾(x)′=−4000x2+0.1\overline{C}(x)' = \frac{-4000}{x^2} + 0.1
b)b) level of production C′(x)=0C'(x) = 0 −4000x2+0.1=0\frac{-4000}{x^2} + 0.1 = 0 4000=0.1x24000 = 0.1x^2 x2=40000.1x^2 = \frac{4000}{0.1} x=200x = 200 unit
c)c) C′(x)=55+0.2xC'(x) = 55 + 0.2x C′(200)=95C'(200) = 95 C‾(200)=$95\overline{C}(200) = \$95

Studdy Solution

STEP 1

What is this asking? We're looking at a cost function and figuring out how the *average cost* changes as we produce more stuff, finding where that change is zero, and then checking the actual cost at that production level. Watch out! Don't mix up *total cost* and *average cost*!
Also, remember that the *instantaneous rate of change* is just a fancy way of saying *derivative*.

STEP 2

1. Find the instantaneous rate of change of average cost.
2. Find the production level where the rate of change of average cost is zero.
3. Calculate the marginal cost and average cost at that production level.

STEP 3

The average cost function C‾(x)\overline{C}(x) is given by: C‾(x)=4000x+55+0.1x \overline{C}(x) = \frac{4000}{x} + 55 + 0.1x This tells us the average cost per unit when we produce xx units.

STEP 4

To find the *instantaneous rate of change*, we need to find the *derivative* of the average cost function with respect to xx.
We can differentiate each term separately: C‾′(x)=ddx(4000x+55+0.1x) \overline{C}'(x) = \frac{d}{dx} \left( \frac{4000}{x} + 55 + 0.1x \right) C‾′(x)=ddx(4000x−1)+ddx(55)+ddx(0.1x) \overline{C}'(x) = \frac{d}{dx} \left( 4000x^{-1} \right) + \frac{d}{dx} (55) + \frac{d}{dx} (0.1x) C‾′(x)=−4000x−2+0+0.1 \overline{C}'(x) = -4000x^{-2} + 0 + 0.1 C‾′(x)=−4000x2+0.1 \overline{C}'(x) = \frac{-4000}{x^2} + 0.1 So, the instantaneous rate of change of the average cost is −4000x2+0.1\frac{-4000}{x^2} + 0.1.

STEP 5

We want to find the **production level**, xx, where the *instantaneous rate of change* of average cost is **zero**.
So, we set the derivative we just found equal to zero and solve for xx: −4000x2+0.1=0 \frac{-4000}{x^2} + 0.1 = 0

STEP 6

Now, let's solve for xx: 0.1=4000x2 0.1 = \frac{4000}{x^2} Multiply both sides by x2x^2 to get: 0.1x2=4000 0.1x^2 = 4000 Divide both sides by 0.10.1 to isolate x2x^2: x2=40000.1=40000 x^2 = \frac{4000}{0.1} = 40000 Take the square root of both sides: x=40000=200 x = \sqrt{40000} = 200 Since we're talking about production levels, we only consider the positive value.
So, the production level where the rate of change of average cost is zero is x=200x = \textbf{200} units.

STEP 7

The *marginal cost* is the derivative of the *total cost* function, C(x)=4000+55x+0.1x2C(x) = 4000 + 55x + 0.1x^2: C′(x)=ddx(4000+55x+0.1x2)=0+55+0.2x=55+0.2x C'(x) = \frac{d}{dx} (4000 + 55x + 0.1x^2) = 0 + 55 + 0.2x = 55 + 0.2x At x=200x = \textbf{200}, the marginal cost is: C′(200)=55+0.2⋅200=55+40=95 C'(200) = 55 + 0.2 \cdot 200 = 55 + 40 = \textbf{95}

STEP 8

At x=200x = \textbf{200}, the average cost is: C‾(200)=4000200+55+0.1⋅200=20+55+20=95 \overline{C}(200) = \frac{4000}{200} + 55 + 0.1 \cdot 200 = 20 + 55 + 20 = \textbf{95}

STEP 9

At a production level of 200\textbf{200} units, both the *marginal cost* and the *average cost* are $95\$\textbf{95}.
The instantaneous rate of change of the average cost is C‾′(x)=−4000x2+0.1\overline{C}'(x) = \frac{-4000}{x^2} + 0.1.

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