Data 605 homework 15

Exercise 1

Find the equation of the regression line for the given points. Round any final values to the nearest hundredth, if necessary. ( 5.6, 8.8 ), ( 6.3, 12.4 ), ( 7, 14.8 ), ( 7.7, 18.2 ), ( 8.4, 20.8 )

To find the equation of the regression line for the given points, we’ll use linear regression. Let’s denote the given points as \((x_i, y_i)\) for \(i = 1, 2, \ldots, n\). The equation of the regression line is given by:

\[ y = a + bx \]

where \(a\) is the intercept and \(b\) is the slope. We can find these coefficients by solving the following equations:

\[ \begin{aligned} b &= \frac{n \sum(x_i y_i) - \sum x_i \sum y_i}{n \sum(x_i^2) - (\sum x_i)^2} \\ a &= \frac{\sum y_i - b \sum x_i}{n} \end{aligned} \]

Now, let’s calculate \(a\) and \(b\) using the given points: \[ (5.6, 8.8), (6.3, 12.4), (7, 14.8), (7.7, 18.2), (8.4, 20.8) \]

After calculating \(a\) and \(b\), we can substitute these values into the equation \(y = a + bx\) to find the regression line.

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Exercise 2

To find all local maxima, local minima, and saddle points for the function \(f(x, y) = 24x - 6xy^2 - 8y^3\), we need to compute its partial derivatives with respect to \(x\) and \(y\) and set them equal to zero.

The partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\) are: \[ \begin{aligned} \frac{\partial f}{\partial x} &= 24 - 6y^2 \\ \frac{\partial f}{\partial y} &= -12xy - 24y^2 \end{aligned} \]

Setting these partial derivatives equal to zero and solving for \(x\) and \(y\), we can find the critical points. After finding the critical points, we can classify them as local maxima, local minima, or saddle points using the second derivative test or by analyzing the behavior of the function around each critical point.

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Exercise 3

A grocery store sells two brands of a product, the “house” brand and a “name” brand. The manager estimates that if she sells the “house” brand for x dollars and the “name” brand for y dollars, she will be able to sell 81  21x + 17y units of the “house” brand and 40 + 11x  23y units of the “name” brand.

Step 1. Find the revenue function R ( x, y ).

To find the revenue function \(R(x, y)\), we need to multiply the number of units sold for each brand by their respective prices and sum them together.

Let \(x\) be the price of the “house” brand and \(y\) be the price of the “name” brand. The number of units sold for the “house” brand is \(81 - 21x + 17y\), and for the “name” brand is \(40 + 11x - 23y\).

Therefore, the revenue function \(R(x, y)\) is given by: \[ R(x, y) = (81 - 21x + 17y) \cdot x + (40 + 11x - 23y) \cdot y \]

Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?

To calculate the revenue when the “house” brand is sold for $2.30 and the “name” brand is sold for $4.10, we substitute \(x = 2.30\) and \(y = 4.10\) into the revenue function \(R(x, y)\).

Using the revenue function \(R(x, y) = (81 - 21x + 17y) \cdot x + (40 + 11x - 23y) \cdot y\), we have: \[ R(2.30, 4.10) = (81 - 21 \times 2.30 + 17 \times 4.10) \times 2.30 + (40 + 11 \times 2.30 - 23 \times 4.10) \times 4.10 \]

After calculating this expression, we will obtain the revenue.

Exercise 4

To minimize the total weekly cost, we need to find the number of units produced in Los Angeles (denoted as \(x\)) and Denver (denoted as \(y\)).

The total weekly cost function is given by: \[ C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700 \]

We can minimize the total weekly cost by minimizing the function \(C(x, y)\). This can be done by finding the critical points of \(C(x, y)\) and determining whether they correspond to a minimum, maximum, or saddle point.

After finding the critical points, we’ll use the second derivative test or evaluate the function around each critical point to determine which one minimizes the total weekly cost.

Exercise 5

To evaluate the double integral over the region \(R\), given by \(2 \leq x \leq 4\) and \(2 \leq y \leq 4\), of the function \(e^{8x + 3y}\), we can use the following double integral expression:

\[ \iint_R e^{8x + 3y} \,dA \]

We’ll integrate this function over the given region \(R\) with respect to both \(x\) and \(y\) to find the result. Since the limits of integration are constants, we can directly evaluate the integral.

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