Question

suppose that for a company manufacturing calculators, the cost and revenue equations (in dollars) are given by
$c = 80000 + 40x$,
$r = 400x - \frac{x^2}{2000}$,
where the production output in one week is $x$ calculators. if the production rate is increasing at a rate of 500 calculators per week when the production output is 5000 calculators, find each of the following with correct units. round your answer to two decimal places if necessary. (note: profit is revenue minus costs.)
the rate of change of the cost with respect to time is
the rate of change of the revenue with respect to time is
the rate of change of the profit with respect to time is
submit answer
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Explanation:

Step 1: Find the rate of change of cost

The cost function is \( C = 80000 + 40x \). To find the rate of change of cost with respect to time \( t \), we use the chain rule. First, find the derivative of \( C \) with respect to \( x \): \( \frac{dC}{dx} = 40 \). We know that \( \frac{dx}{dt} = 500 \) (since production is increasing at 500 calculators per week). Then, by the chain rule, \( \frac{dC}{dt} = \frac{dC}{dx} \cdot \frac{dx}{dt} = 40 \times 500 = 20000 \).

Step 2: Find the rate of change of revenue

The revenue function is \( R = 400x - \frac{x^2}{2000} \). First, find the derivative of \( R \) with respect to \( x \): \( \frac{dR}{dx} = 400 - \frac{2x}{2000} = 400 - \frac{x}{1000} \). When \( x = 5000 \) (current production output), \( \frac{dR}{dx} = 400 - \frac{5000}{1000} = 400 - 5 = 395 \). Then, using the chain rule with \( \frac{dx}{dt} = 500 \), we get \( \frac{dR}{dt} = \frac{dR}{dx} \cdot \frac{dx}{dt} = 395 \times 500 = 197500 \).

Step 3: Find the rate of change of profit

Profit \( P = R - C \). So, the rate of change of profit with respect to time is \( \frac{dP}{dt} = \frac{dR}{dt} - \frac{dC}{dt} \). We already found \( \frac{dR}{dt} = 197500 \) and \( \frac{dC}{dt} = 20000 \), so \( \frac{dP}{dt} = 197500 - 20000 = 177500 \). Wait, no, wait. Wait, in step 2, when \( x = 5000 \), let's recalculate \( \frac{dR}{dx} \): \( R = 400x - \frac{x^2}{2000} \), so derivative is \( 400 - \frac{2x}{2000} = 400 - \frac{x}{1000} \). So when \( x = 5000 \), \( \frac{dR}{dx} = 400 - 5000/1000 = 400 - 5 = 395 \). Then \( \frac{dR}{dt} = 395 * 500 = 197500 \). \( \frac{dC}{dt} = 40 * 500 = 20000 \). So \( \frac{dP}{dt} = 197500 - 20000 = 177500 \). Wait, but let's check again. Wait, maybe I made a mistake in the revenue derivative. Wait, \( R = 400x - \frac{x^2}{2000} \), so derivative is \( 400 - (2x)/2000 = 400 - x/1000 \). Yes, that's correct. And \( x = 5000 \), so \( 400 - 5 = 395 \). Then \( 395 * 500 = 197500 \). Cost derivative is 40, so 40*500=20000. Then profit rate is 197500 - 20000 = 177500. Wait, but let's confirm the steps again.

Wait, the problem says "the production output in one week is \( x \) calculators. If the production rate is increasing at a rate of 500 calculators per week when the production output is 5000 calculators". So \( x = 5000 \), \( dx/dt = 500 \).

So:

1. Rate of change of cost: \( dC/dt = dC/dx * dx/dt \). \( C = 80000 + 40x \), so \( dC/dx = 40 \). Thus, \( dC/dt = 40 * 500 = 20000 \) dollars per week.

2. Rate of change of revenue: \( dR/dt = dR/dx * dx/dt \). \( R = 400x - x²/2000 \), so \( dR/dx = 400 - (2x)/2000 = 400 - x/1000 \). At \( x = 5000 \), \( dR/dx = 400 - 5000/1000 = 400 - 5 = 395 \). Thus, \( dR/dt = 395 * 500 = 197500 \) dollars per week.

3. Rate of change of profit: \( dP/dt = dR/dt - dC/dt = 197500 - 20000 = 177500 \) dollars per week. Wait, but that seems high. Wait, maybe I messed up the revenue function. Wait, the revenue function is \( R = 400x - \frac{x^2}{2000} \)? Let me check the original problem. The user wrote: \( C = 80000 + 40x \), \( R = 400x - \frac{x^2}{2000} \). Yes. So when \( x = 5000 \), \( R \) derivative is 400 - (5000)/1000 = 395. Then 395 * 500 = 197500. Cost derivative is 40 * 500 = 20000. So profit rate is 197500 - 20000 = 177500. But let's check the calculations again.

Wait, 400 - (2x)/2000: the derivative of \( x²/2000 \) is (2x)/2000 = x/1000. So yes, derivative of R is 400 - x/1000. So at x=5000, that's 400 - 5 = 395. Then dR/dt = 395 * 500 = 197500. dC/dt = 40 * 500 = 20000. So dP/dt = 197500 - 20000 = 177500. So:

- Rate of change of cost with respect to time: 20000.00 dollars per week.

- Rate of change of revenue with respect to time: 197500.00 dollars per week.

- Rate of change of profit with respect to time: 177500.00 dollars per week. Wait, but that seems like the profit rate is very high. Maybe I made a mistake in the revenue function's derivative. Wait, let's re-express the revenue function: \( R = 400x - \frac{x^2}{2000} \). So derivative is \( 400 - \frac{2x}{2000} = 400 - \frac{x}{1000} \). Yes, that's correct. So when x=5000, that's 400 - 5 = 395. Then 395 * 500 = 197500. Cost derivative is 40, so 40*500=20000. So profit rate is 197500 - 20000 = 177500. So that's the calculation.

Answers:

- The rate of change of the cost with respect to time is \(\boldsymbol{20000.00}\) dollars per week.
- The rate of change of the revenue with respect to time is \(\boldsymbol{197500.00}\) dollars per week.
- The rate of change of the profit with respect to time is \(\boldsymbol{177500.00}\) dollars per week.

Answer:

s:

• The rate of change of the cost with respect to time is \(\boldsymbol{20000.00}\) dollars per week.
• The rate of change of the revenue with respect to time is \(\boldsymbol{197500.00}\) dollars per week.
• The rate of change of the profit with respect to time is \(\boldsymbol{177500.00}\) dollars per week.