Q: If the weekly cost function is $$CC(x) = 50 + 2.5x$$ and the weekly income is $$I(x) = 3x + 2 \ln(x)$$, what is the profit when $$x = 100$$?

$$3 \times 100 + 2 \ln(100) - 50 - 2.5 \times 100 = 300 + 2 \times 4.605 - 50 - 250 = 300 + 9.21 - 50 - 250 = 9.21$$

Question 1

Given the weekly cost function for producing a chemical: $$CC(x) = 50 + 2.5x$$, and the weekly income function: $$I(x) = 3x + 2 \ln(x)$$, where $$x is $$the number of kilograms sold per week, what is the correct equation to maximize weekly profit?

Accepted Answer

$$3x + 2 \ln(x) - (50 + 2.5x)$$

Question 2

To find the value of $$x$$ that maximizes profit, which equation should you solve?

Accepted Answer

$$\frac{d}{dx}[3x + 2 \ln(x) - (50 + 2.5x)] = 0$$

Question 3

What is the derivative of the profit function $$P(x) = 3x + 2 \ln(x) - 50 - 2.5x$$ with respect to $$x$$?

Accepted Answer

$$3 + \frac{2}{x} - 2.5$$

Question 4

If you set the derivative of the profit function to zero, which equation do you get?

Accepted Answer

$$3 + \frac{2}{x} - 2.5 = 0$$

Question 5

What is the correct iterative formula for Newton's method to solve $$3 + \frac{2}{x} - 2.5 = 0$$?

Accepted Answer

$$x_{k+1} = x_k - \frac{3 + \frac{2}{x_k} - 2.5}{-\frac{2}{x_k^2$}}$$

Question 6

If the initial guess for $$x is$$ $$50$$, what is the value of the function $$f(x) = 3 + \frac{2}{x} - 2.5 at$$ $$x = 50$$?

Accepted Answer

$$0.54$$

Question 7

After one iteration of Newton's method starting at $$x_0 = 50$, the next approximation x_1$ is closest to:

Accepted Answer

$$81.81$$

Question 8

Which of the following best describes the purpose of using Newton's method in this context?

Accepted Answer

To find the value of $$x$$ that maximizes the weekly profit by solving a nonlinear equation.

Question 9

If the cost per kilogram increases from $$2.5 to$$ $$3$$, how does this affect the profit-maximizing value of $$x$$?

Accepted Answer

The profit-maximizing $$x$$ decreases.

Question 10

Suppose the weekly income function changes to $$I(x) = 4x + 2 \ln(x). $$What is the new equation to maximize profit?

Accepted Answer

$$4x + 2 \ln(x) - (50 + 2.5x)$$

Question 11

If the fixed cost increases from $$50 to$$ $$100$$, what is the effect on the profit-maximizing value of $$x$$?

Accepted Answer

There is no effect on the profit-maximizing $$x.$$

Question 12

Which of the following is a correct interpretation of the term $$2 \ln(x) in $$the income function?

Accepted Answer

It represents a logarithmic increase in income as more product is sold.

Question 13

If the weekly cost function is $$CC(x) = 50 + 2.5x$$ and the weekly income is $$I(x) = 3x + 2 \ln(x)$$, what is the profit when $$x = 100$$?

Accepted Answer

$$3 	imes 100 + 2 \ln(100) - 50 - 2.5 	imes 100 = 300 + 2 	imes 4.605 - 50 - 250 = 300 + 9.21 - 50 - 250 = 9.21$$

Question 14

Which of the following is a valid reason for using a logarithmic term in the income function?

Accepted Answer

To model diminishing returns as more product is sold.

Question 15

If the weekly cost function is $$CC(x) = 50 + 2.5x$$ and the weekly income is $$I(x) = 3x + 2 \ln(x)$$, what is the break-even point (where profit is zero)?

Accepted Answer

Solve $$3x + 2 \ln(x) = 50 + 2.5x$$

Optimization and Modeling: Weekly Cost and Revenue Analysis

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