2. The cost of producing x units of a certain commodity is given by P�x � 1000 �

0
x
MC�sds,
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 � 12000).
Use a tangent line approximation to estimate the cost of producing only 497 units.

B. Suppose the production schedule is such that the company produces five units each day.
That is, the number of units produced is 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 represents, practically speaking.

What does the question mark within a black square represent? Hard to know what you are asking.

Online, “*” is used to indicate multiplication to avoid confusion with “x” as an unknown.
Online "^" is used to indicate an exponent, e.g., x^2 = x squared.

I got A can someone help me with C and B please?

A. To estimate the cost of producing 497 units using the tangent line approximation, we can use the concept of marginal cost.

Given that at a production level of 500 units, the marginal cost is $10 per unit, and the cost of producing 500 units is $12,000, we can use these values to find the equation of the tangent line.

The equation for the tangent line can be written as:

C(x) = M(500)(x - 500) + C(500)

Where C(x) represents the cost of producing x units, M(500) is the marginal cost at 500 units, and C(500) is the cost of producing 500 units.

Substituting the given values into the equation, we have:

C(x) = 10(x - 500) + 12000

To find the cost of producing 497 units, we substitute x = 497 into the equation:

C(497) = 10(497 - 500) + 12000

Simplifying the equation gives us:

C(497) = -30 + 12000

C(497) = $11,970

Therefore, using the tangent line approximation, the cost of producing 497 units is estimated to be $11,970.

B. The production schedule is given as x = 5t, where x is the number of units produced and t is the time in days.

To write the equation for the cost of production P as a function of time t, we can use the given information that the cost of producing x units is given by P(x) = 1000 + 0.01x^2.

Substituting x = 5t into the equation, we have:

P(t) = 1000 + 0.01(5t)^2

Simplifying the equation gives us:

P(t) = 1000 + 0.01(25t^2)

P(t) = 1000 + 0.25t^2

Therefore, the equation for the cost of production P as a function of time t is P(t) = 1000 + 0.25t^2.

C. To find dP/dt, we differentiate the equation P(t) = 1000 + 0.25t^2 with respect to t.

dP/dt = 0 + 0.5t

The units for dP/dt would be dollars per day.

Practically speaking, dP/dt represents the rate of change of the cost of production with respect to time. It indicates how the cost of production is changing over time, specifically in terms of the increase or decrease in dollars per day.

A. To estimate the cost of producing 497 units using a tangent line approximation, we need to find the slope of the cost function at x = 500.

The given marginal cost function is MC(x) = P'(x) = 1000 - 2x.

To find the slope at x = 500, we substitute x = 500 into the marginal cost function:

MC(500) = 1000 - 2(500) = 1000 - 1000 = 0.

This means that the slope of the cost function at x = 500 is 0.

Using the tangent line approximation, we assume that the cost function is approximately linear near x = 500, and the slope at x = 500 can be used to estimate the cost of producing 497 units.

Since the slope at x = 500 is 0, the cost function is constant in the vicinity of x = 500. Therefore, the cost of producing 497 units would be the same as the cost of producing 500 units, which is $12,000.

Hence, the estimate for the cost of producing 497 units is $12,000.

B. Given the production schedule x = 5t, where t is in days, we need to express the cost of production P as a function of time t.

We know that the cost of producing x units is given by P(x) = 1000x - x^2.

Substituting x = 5t into the cost function, we get:

P(t) = 1000(5t) - (5t)^2
= 5000t - 25t^2.

Therefore, the equation for the cost of production P as a function of time t is P(t) = 5000t - 25t^2.

C. To find dP/dt, we differentiate the cost function P(t) = 5000t - 25t^2 with respect to t.

dP/dt = 5000 - 50t.

The unit of measure for P(t) is dollars, and the unit of measure for t is days. Hence, the unit of measure for dP/dt is dollars per day.

Practically speaking, dP/dt represents the rate at which the cost of production changes with respect to time. In this case, it represents the rate of increase or decrease in cost per day as the production progresses.