Math  /  Calculus

QuestionFor the given cost function C(x)=48400+200x+x2C(x)=48400+200 x+x^{2}, which gives the total cost (\)for) for xitems.Findtheaveragecostfunction: items. Find the average cost function: \bar{C}(x)= \squareFindtheproductionlevelthatwillminimizetheaveragecost: Find the production level that will minimize the average cost: x= \square$ items
Find the minimal average cost: \square Question Help: \square Video Submit Question

Studdy Solution

STEP 1

What is this asking? We're given a fancy cost function, and we need to find the average cost function, the number of items that makes the average cost as low as possible, and what that lowest average cost actually is! Watch out! Don't mix up the total cost and the average cost!
The average cost is the total cost divided by the number of items.

STEP 2

1. Define the average cost function.
2. Find the derivative of the average cost function.
3. Find the critical points.
4. Determine the minimum average cost.

STEP 3

Alright, so we're given the **total cost** function C(x)=48400+200x+x2C(x) = 48400 + 200x + x^2.
The **average cost**, Cˉ(x)\bar{C}(x), is just the total cost divided by the number of items, xx.

STEP 4

So, we **divide** the total cost function by xx: Cˉ(x)=C(x)x=48400+200x+x2x \bar{C}(x) = \frac{C(x)}{x} = \frac{48400 + 200x + x^2}{x}

STEP 5

Let's **simplify** this by dividing each term in the numerator by xx: Cˉ(x)=48400x+200xx+x2x=48400x+200+x \bar{C}(x) = \frac{48400}{x} + \frac{200x}{x} + \frac{x^2}{x} = \frac{48400}{x} + 200 + x So, our average cost function is Cˉ(x)=48400x+200+x\bar{C}(x) = \frac{48400}{x} + 200 + x.
Awesome!

STEP 6

To minimize the average cost, we need to find where its derivative is zero.
Let's **find** that derivative!
Remember, 48400x\frac{48400}{x} is the same as 48400x148400 \cdot x^{-1}.

STEP 7

Using the power rule, the **derivative** of Cˉ(x)\bar{C}(x) is: Cˉ(x)=48400x2+0+1=48400x2+1 \bar{C}'(x) = -48400x^{-2} + 0 + 1 = -\frac{48400}{x^2} + 1

STEP 8

Now, we **set** the derivative equal to zero and **solve** for xx: 48400x2+1=0 -\frac{48400}{x^2} + 1 = 0

STEP 9

**Subtract** 1 from both sides: 48400x2=1 -\frac{48400}{x^2} = -1

STEP 10

**Multiply** both sides by 1-1: 48400x2=1 \frac{48400}{x^2} = 1

STEP 11

**Multiply** both sides by x2x^2: 48400=x2 48400 = x^2

STEP 12

**Take** the square root of both sides: x=±48400=±220 x = \pm \sqrt{48400} = \pm 220

STEP 13

Since we can't produce a negative number of items, we only care about the **positive** solution: x=220x = 220.

STEP 14

Now, we **plug** x=220x = \mathbf{220} back into our average cost function to find the minimum average cost: Cˉ(220)=48400220+200+220 \bar{C}(220) = \frac{48400}{220} + 200 + 220

STEP 15

Cˉ(220)=220+200+220=640 \bar{C}(220) = 220 + 200 + 220 = 640 So, the minimum average cost is $640\$\mathbf{640}!

STEP 16

The average cost function is Cˉ(x)=48400x+200+x\bar{C}(x) = \frac{48400}{x} + 200 + x.
The production level that minimizes the average cost is x=220x = 220 items, and the minimal average cost is $640\$640.

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