The cost of producing x units of a certain commodity is given by P(x)=1000+ int(MC(s))ds from 0 to x, where P is in dollars and M(x) is marginal cost in dollars per unit.

A. Suppose the marginal cost at a production level of 500 units is $10 per unit, and the cost of producing 500 units is $12,000 (that is, M(500)=10 and P(500)= 12,000). Use tangent line approximation to estimate the cost of producing only 497 units.
B. Suppose the production schedule is such that the company produces 5 units each day. That is, the number of units produced in x=5t, where t is in days, and t=0 corresponds to the beginning of production. Write an equation for the cost of production P as a function of time t.
C. Use your equation for P(t) from part B to find dP/dt. Be sure to indicate units and describe what dP/dt means, practically speaking.

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A. To estimate the cost of producing only 497 units using the tangent line approximation, we can use the concept of the derivative. The derivative of the cost function represents the rate of change of cost with respect to the number of units produced.

Given that the marginal cost at a production level of 500 units is $10 per unit, we can assume that the marginal cost function, M(x), is constant or approximately constant for small changes in x around 500 units.

Since M(x) is constant at $10, the tangent line approximation states that a small change in the number of units produced (Δx) will result in a proportional change in the cost (ΔP), where the proportionality constant is the value of the derivative M'(x) at x=500.

Therefore, we can calculate the slope of the tangent line as:
M'(500) = 10

Using the formula for a line, y = mx + b, where m is the slope and b is the y-intercept, we can find the equation of the tangent line:
ΔP = M'(500) * Δx

To estimate the cost of producing only 497 units, we consider Δx = -3 units (since we are decreasing from 500 to 497). Plugging in the values, we get:
ΔP ≈ 10 * (-3) = -30

To find the cost at x=500, we know that P(500) = $12,000. So by subtracting the change in cost (ΔP) from the cost at x=500, we can estimate the cost of producing only 497 units:
P(497) ≈ P(500) + ΔP = $12,000 - $30 = $11,970

Therefore, the estimated cost of producing only 497 units is approximately $11,970.

B. The equation for the production schedule states that the number of units produced, x, is equal to 5t, where t is in days. To find the equation for the cost of production, P, as a function of time, we need to substitute the given expression into the cost function P(x).

Since x = 5t, we have:
P(5t) = 1000 + int(MC(s))ds from 0 to 5t

C. To find dP/dt, we differentiate the equation for P(t) with respect to time t. We need to apply the chain rule since the upper limit of the integral is a function of t.

dP/dt = d/dt (1000 + int(MC(s))ds from 0 to 5t)

Applying the chain rule to the integral:
dP/dt = d/dt (1000 + MC(5t)*d(5t)/dt)

Since x = 5t, we have dx/dt = 5 (using the chain rule).

Simplifying further:
dP/dt = 0 + MC(5t) * 5

Therefore, dP/dt = 5 * MC(5t).
The units of dP/dt will be dollars per day, as MC(5t) represents the marginal cost of producing 5t units in dollars per unit. Practically speaking, dP/dt represents the rate of change of the cost of production with respect to time, or how quickly the cost is increasing or decreasing per day as production continues.

A. To estimate the cost of producing only 497 units, we can use the tangent line approximation. The formula for the cost of producing x units is given by P(x) = 1000 + ∫(MC(s))ds from 0 to x.

We are given that the marginal cost at a production level of 500 units is $10 per unit, and the cost of producing 500 units is $12,000 (M(500)=10 and P(500)=12,000).

First, let's find the equation of the tangent line at x = 500. We can use the point-slope form of a linear equation:

P(x) - P(500) = (MC(500))(x - 500)

Substituting the given values:

P(x) - 12,000 = 10(x - 500)
P(x) = 10x - 5,000 + 12,000
P(x) = 10x + 7,000

Now, we can use this equation to estimate the cost of producing 497 units:

P(497) ≈ 10(497) + 7,000
P(497) ≈ 4,970 + 7,000
P(497) ≈ $11,970

Therefore, the estimated cost of producing 497 units is $11,970.

B. The production schedule is given by x = 5t, where t represents time in days. To write the equation for the cost of production P as a function of time t, we can substitute x = 5t into the original formula:

P(5t) = 1000 + ∫(MC(s))ds from 0 to 5t

C. To find dP/dt, we need to differentiate P(5t) with respect to t. The derivative of the integral is given by the Fundamental Theorem of Calculus:

dP/dt = (d/dt) ∫(MC(s))ds from 0 to 5t

Since the integral is evaluated with respect to s, we can consider 5t as a constant and take its derivative:

dP/dt = 5(MC(5t))

The units for dP/dt would be dollars per day, as it represents the rate of change of cost with respect to time.

Practically speaking, dP/dt represents the rate at which the cost of production is changing with respect to time. It tells us how much the cost is increasing or decreasing per day.