Q: If the cost per kilogram is $$2.5$$ and the fixed weekly cost is $$50$$, what is the total cost for producing $$x = 10$$ kilograms?

$$CC(10) = 50 + 2.5 \times 10 = 75$$

Q: If the weekly income for $$x = 5$$ kilograms is $$I(5) = 3 \times 5 + 2 \ln(5)$$, what is the numerical value of $$I(5) ($$rounded to two decimals)?

$$I(5) \approx 15 + 2 \times 1.61 = 18.22$$

Question 1

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

Accepted Answer

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

Question 2

If the weekly profit function is $$P(x) = 3x + 2 \ln(x) - 50 - 2.5x$$, what value of $$x$$ maximizes the profit? (Hint: Set the derivative to zero and solve for $$x.)$$

Accepted Answer

$$x = \exp\left(\frac{2}{2.5}\right)$$

Question 3

What is the interpretation of the term $$2 \ln(x) in $$the income function $$I(x) = 3x + 2 \ln(x)$$?

Accepted Answer

It represents a variable income that increases logarithmically with the number of units sold.

Question 4

If the cost per kilogram is $$2.5$$ and the fixed weekly cost is $$50$$, what is the total cost for producing $$x = 10$$ kilograms?

Accepted Answer

$$CC(10) = 50 + 2.5 	imes 10 = 75$$

Question 5

Suppose you want to maximize the weekly profit. Which equation would you use to find the optimal $$x$$?

Accepted Answer

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

Question 6

If the weekly profit function is $$P(x) = 3x + 2 \ln(x) - 50 - 2.5x$$, what is the second derivative $$P''(x)$$?

Accepted Answer

$$P''(x) = \frac{2}{x^2$}$$

Question 7

In the context of the provided model, what does the fixed cost $$50$$ represent?

Accepted Answer

The cost incurred regardless of the number of kilograms produced.

Question 8

If the weekly income for $$x = 5$$ kilograms is $$I(5) = 3 	imes 5 + 2 \ln(5)$$, what is the numerical value of $$I(5) ($$rounded to two decimals)?

Accepted Answer

$$I(5) \approx 15 + 2 	imes 1.61 = 18.22$$

Question 9

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

Accepted Answer

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

Question 10

If you use the secant method to solve $$f(x) = 0$$ for the profit maximization problem, which formula would you use for the next iteration?

Accepted Answer

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

Question 11

If the initial guesses for the secant method are $$x_0 = 50$ and x_1$ = 81.81$$, and $$f(x) = 3x + 2 \ln(x) - 50 - 2.5x$$, what is the next value $$x_2$$?

Accepted Answer

$$x_2$$ = $$x_1$$ - \frac{f($$x_1$$)($$x_1$$ - $$x_0$$)}{f($$x_1) - f$$($$x_0$$)}$$

Question 12

What is the main advantage of using the secant method over the Newton-Raphson method in this context?

Accepted Answer

The secant method does not require the calculation of the derivative.

Question 13

If the weekly cost function is $$CC(x) = 50 + 2.5x$$, what is the marginal cost for producing one additional kilogram?

Accepted Answer

$$2.5$$

Question 14

If the weekly income function is $$I(x) = 3x + 2 \ln(x)$$, what is the marginal income for one additional kilogram?

Accepted Answer

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

Question 15

If the secant method produces the sequence $$x_0 = 50$, x_1$ = 81.81$$, $$x_2 = 82.35$, what does this sequence represent?

Accepted Answer

The sequence of approximations to the optimal number of kilograms to maximize profit.

Question 16

If the profit function is $$P(x) = 3x + 2 \ln(x) - 50 - 2.5x$$, what is the value of $$P(82.35) ($$rounded to two decimals)?

Accepted Answer

$$P(82.35) \approx 3 	imes 82.35 + 2 \ln(82.35) - 50 - 2.5 	imes 82.35 \approx 246.99 + 8.38 - 50 - 205.88 = -0.51$$

Método de la Secante: Optimización de Funciones en Estadística

Guida di studio - Domande di pratica

Study guide - Flashcard