Math  /  Calculus

QuestionSection 12.6: Problem 1 Previous Problem Problem List Next Problem (1 point) The average cost function for the weekly manufacture of portable gramophones is given by Cˉ(x)=750,000x1+45+0.045x\bar{C}(x)=750,000 x^{-1}+45+0.045 x dollars per gramophone, where xx is the number of gramophones manufactured that week. Weekly production is currently 280 gramophones and is increasing at a rate of 60 gramophones per week. What is happening to the average cost?

Studdy Solution

STEP 1

1. The average cost function is given by Cˉ(x)=750,000x1+45+0.045x\bar{C}(x) = 750,000 x^{-1} + 45 + 0.045 x.
2. xx represents the number of gramophones manufactured per week.
3. The current production level is 280 gramophones.
4. The production rate is increasing at 60 gramophones per week.
5. We need to determine the rate of change of the average cost with respect to time.

STEP 2

1. Find the derivative of the average cost function Cˉ(x)\bar{C}(x) with respect to xx.
2. Use the chain rule to find the rate of change of the average cost with respect to time.
3. Evaluate the derivative at the current production level and rate.

STEP 3

Differentiate Cˉ(x)=750,000x1+45+0.045x\bar{C}(x) = 750,000 x^{-1} + 45 + 0.045 x with respect to xx.
ddx(750,000x1)=750,000x2\frac{d}{dx}\left(750,000 x^{-1}\right) = -750,000 x^{-2}
ddx(45)=0\frac{d}{dx}(45) = 0
ddx(0.045x)=0.045\frac{d}{dx}(0.045 x) = 0.045
Thus, the derivative of the average cost function is:
Cˉ(x)=750,000x2+0.045\bar{C}'(x) = -750,000 x^{-2} + 0.045

STEP 4

Use the chain rule to find the rate of change of the average cost with respect to time, dCˉdt \frac{d\bar{C}}{dt} .
Given dxdt=60\frac{dx}{dt} = 60, we have:
dCˉdt=Cˉ(x)dxdt\frac{d\bar{C}}{dt} = \bar{C}'(x) \cdot \frac{dx}{dt}
Substitute Cˉ(x)\bar{C}'(x) and dxdt\frac{dx}{dt}:
dCˉdt=(750,000x2+0.045)60\frac{d\bar{C}}{dt} = \left(-750,000 x^{-2} + 0.045\right) \cdot 60

STEP 5

Evaluate dCˉdt\frac{d\bar{C}}{dt} at x=280x = 280.
Cˉ(280)=750,000(280)2+0.045\bar{C}'(280) = -750,000 (280)^{-2} + 0.045
Calculate 750,000(280)2-750,000 (280)^{-2}:
750,000×12802=750,000×1784009.57-750,000 \times \frac{1}{280^2} = -750,000 \times \frac{1}{78400} \approx -9.57
Cˉ(280)=9.57+0.045=9.525\bar{C}'(280) = -9.57 + 0.045 = -9.525
Now, calculate dCˉdt\frac{d\bar{C}}{dt}:
dCˉdt=(9.525)60571.5\frac{d\bar{C}}{dt} = (-9.525) \cdot 60 \approx -571.5
The average cost is decreasing at a rate of approximately 571.5-571.5 dollars per week.

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