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 profit function $$P(x)$$?

Accepted Answer

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

Question 2

To maximize the weekly profit, which equation must be solved for $$x$$?

Accepted Answer

$$\frac{d}{dx}P(x) = 0$$

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

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

Question 4

If you set $$\frac{dP}{dx} = 0 to $$find the optimal $$x$$, which equation do you solve?

Accepted Answer

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

Question 5

What is the economic interpretation of the solution to $$3 + \frac{2}{x} - 2.5 = 0$$?

Accepted Answer

It is the number of kilograms that maximizes weekly profit.

Question 6

If the initial guess for $$x is$$ $$50$$, and you use the secant method to solve $$f(x) = 3 + \frac{2}{x} - 2.5 = 0$$, what is the formula for the next approximation $$x_{k+1}$$?

Accepted Answer

$$x_{k+1} = x_k - f(x_k) \frac{x_k - x_{k-1}}{f(x_k) - f(x_{k-1})}$$

Question 7

Suppose after two iterations of the secant method you have $$x_0 = 50$, x_1$ = 81.81$$, and $$f(x_0$) = -17.17$$, $$f(x_1$) = 0.28. $$What is the next value $$x_2$$?

Accepted Answer

$$x_2 = 81.81 - 0.28$$ \frac{81.81 - 50}{0.28 - (-17.17)}$$

Question 8

If the cost per kilogram increases from $$2.5 to$$ $$3$$, how does the profit-maximizing quantity change?

Accepted Answer

It decreases, because the marginal cost increases.

Question 9

Which of the following best describes the secant method used in this context?

Accepted Answer

An iterative numerical method to find roots of nonlinear equations.

Question 10

If the weekly fixed cost increases from $$50 to$$ $$100$$, what is the effect on the profit-maximizing quantity?

Accepted Answer

There is no effect, because the fixed cost does not affect the marginal profit.

Question 11

If the income function changes to $$I(x) = 4x + 2 \ln(x)$$, what is the new equation to maximize profit?

Accepted Answer

$$4 + \frac{2}{x} - 2.5 = 0$$

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 with respect to the quantity sold.

Question 13

If the solution to the profit-maximizing equation is $$x^* = 82.35$$, what does this value represent?

Accepted Answer

The optimal number of kilograms to sell per week to maximize profit.

Question 14

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

Accepted Answer

The value of $$x$$ that solves $$3x + 2 \ln(x) = 50 + 2.5x$$

Optimization Methods in Business: Solving for Maximum Weekly Profit

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