Given points: (5.6, 8.8), (6.3, 12.4), (7, 14.8), (7.7, 18.2), (8.4, 20.8)
We use the method of least squares to find the best-fitting line for the given data points.
The linear regression equation is given by:
\[ y = mx + b \]
where: - \(y\) is the dependent variable, - \(x\) is the independent variable, - \(m\) is the slope of the line, - \(b\) is the y-intercept.
The slope \(m\) and intercept \(b\) are calculated using:
\[ m = \frac{N(\sum xy) - (\sum x)(\sum y)}{N(\sum x^2) - (\sum x)^2} \] \[ b = \frac{(\sum y) - m(\sum x)}{N} \]
where: - \(N\) is the number of data points, - \(\sum xy\) is the sum of the product of each pair of \(x\) and \(y\), - \(\sum x\), \(\sum y\) are the sums of \(x\) and \(y\) values respectively, - \(\sum x^2\) is the sum of the squares of \(x\) values.
# Defining the data points
x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
# Calculating for slope (m) and intercept (b)
N <- length(x)
sum_x <- sum(x)
sum_y <- sum(y)
sum_xy <- sum(x*y)
sum_x2 <- sum(x^2)
# Slope
m <- (N * sum_xy - sum_x * sum_y) / (N * sum_x2 - sum_x^2)
# Intercept
b <- (sum_y - m * sum_x) / N
# Rounding to the nearest hundredth
m <- round(m, 2)
b <- round(b, 2)
# Output the slope and intercept
c(m = m, b = b)## m b
## 4.26 -14.80We have:
Given these estimates, the revenue function, representing the total sales from each product, is:
\[ R(x, y) = x(81 - 21x + 17y) + y(40 + 11x - 23y) \]
Expanding and simplifying, we have:
\[ R(x, y) = -21x^2 + 28xy - 23y^2 + 81x + 40y \]
This function calculates the total revenue based on prices \(x\) and \(y\) for the house and name brands, respectively.
For prices: - House brand \(x = \$2.30\) - Name brand \(y = \$4.10\)
The revenue can be calculated by substituting these prices into the revenue function:
\[ R(2.30, 4.10) = -21(2.30)^2 + 28(2.30)(4.10) - 23(4.10)^2 + 81(2.30) + 40(4.10) \]
Then, this will yield the total revenue from selling the products at these prices, which is:
x <- 2.30
y <- 4.10
R <- -21*x^2 + 28*x*y - 23*y^2 + 81*x + 40*y
R_rounded <- round(R, 2)
R_rounded## [1] 116.62A company must produce 96 units weekly between two plants. The cost function is:
\[ C(x, y) = 16x^2 + 16y^2 + 7x + 25y + 700 \]
where \(x\) and \(y\) are the units produced in Los Angeles and Denver, respectively.
Our objective is to determine how many units should be produced at each location to minimize the total cost, given the constraint \(x + y = 96\).
Using the method of Lagrange multipliers, we set up the Lagrangian as follows:
\[ L(x, y, \lambda) = 16x^2 + 16y^2 + 7x + 25y + 700 + \lambda (96 - x - y) \]
Derivatives of the Lagrangian are set to zero to solve for \(x\), \(y\), and \(\lambda\):
\[ \frac{\partial L}{\partial x} = 32x + 7 - \lambda = 0 \] \[ \frac{\partial L}{\partial y} = 32y + 25 - \lambda = 0 \] \[ \frac{\partial L}{\partial \lambda} = 96 - x - y = 0 \]
library(nleqslv)
# System of equations based on the Lagrangian derivatives
system_eq <- function(z) {
x <- z[1]
y <- z[2]
lambda <- z[3]
# System of equations
eq1 <- 32*x + 7 - lambda
eq2 <- 32*y + 25 - lambda
eq3 <- 96 - x - y
c(eq1, eq2, eq3)
}
# Initial guesses for x, y, and lambda
z_start <- c(x = 48, y = 48, lambda = 100)
# Solving the system
solution <- nleqslv(x = z_start, fn = system_eq)
# Extracting solutions
x_opt <- solution$x[1]
y_opt <- solution$x[2]
# Output the results
c("Units to produce in Los Angeles" = x_opt, "Units to produce in Denver" = y_opt)## Units to produce in Los Angeles.x Units to produce in Denver.y
## 48.28125 47.71875We are given the double integral over the region \(R\) defined by \(2 \leq x \leq 4\) and \(2 \leq y \leq 4\) for the function \(e^{8x+3y}\).
The function to integrate is \(e^{8x + 3y}\), and the region of integration is defined by the ranges of \(x\) and \(y\).
\[ R: 2 \leq x \leq 4, 2 \leq y \leq 4 \]
We start by setting up the double integral:
\[ \int_{2}^{4} \int_{2}^{4} e^{8x+3y} \, dy \, dx \]
First, integrate with respect to \(y\), treating \(x\) as a constant:
\[ \int_{2}^{4} e^{8x+3y} \, dy = e^{8x} \int_{2}^{4} e^{3y} \, dy \]
The integration with respect to \(y\) yields:
\[ = e^{8x} \left[ \frac{e^{3y}}{3} \right]_{y=2}^{y=4} \] \[ = e^{8x} \left( \frac{e^{12}}{3} - \frac{e^6}{3} \right) \] \[ = \frac{e^{8x}}{3} (e^{12} - e^6) \]
Now, integrating the result with respect to \(x\):
\[ \int_{2}^{4} \frac{e^{8x}}{3} (e^{12} - e^6) \, dx \] \[ = \frac{e^{12} - e^6}{3} \int_{2}^{4} e^{8x} \, dx \]
The integration with respect to \(x\) is:
\[ = \frac{e^{12} - e^6}{3} \left[ \frac{e^{8x}}{8} \right]_{x=2}^{x=4} \] \[ = \frac{e^{12} - e^6}{3} \left( \frac{e^{32}}{8} - \frac{e^{16}}{8} \right) \] \[ = \frac{(e^{12} - e^6) (e^{32} - e^{16})}{24} \]
Therefore, the total value of the double integral over the specified region \(R\) is:
\[ \frac{(e^{12} - e^6) (e^{32} - e^{16})}{24} \]
# Evaluate the final expression
e <- exp(1) # Base of the natural logarithm
result <- (e^12 - e^6) * (e^32 - e^16) / 24
result## [1] 5.341559e+17