Given the data points: (5.6, 8.8), (6.3, 12.4), (7, 14.8), (7.7, 18.2), and (8.4, 20.8).
The equation of the regression line is expressed as: \[ y = mx + b \] To generate the equation, we take the following steps:
The slope (\(m\)) of the regression line \(y = mx + b\) is calculated using the formula: \[ m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2} \] Where \(n\) is the number of data points, \(\sum xy\) is the sum of the product of \(x\) and \(y\) for all points, \(\sum x\) and \(\sum y\) are the sums of \(x\) and \(y\) values, and \(\sum x^2\) is the sum of squared \(x\) values.
Once the slope (\(m\)) is determined, the intercept (\(b\)) is computed as: \[ b = \frac{\sum y - m(\sum x)}{n} \]
# Data points
points <- data.frame(x = c(5.6, 6.3, 7, 7.7, 8.4),
y = c(8.8, 12.4, 14.8, 18.2, 20.8))
# Calculate Means of x and y
mean_x <- mean(points$x)
mean_y <- mean(points$y)
# Compute the Slope (m)
sum_x <- sum(points$x)
sum_y <- sum(points$y)
sum_xy <- sum(points$x * points$y)
sum_x2 <- sum(points$x^2)
n <- nrow(points)
slope <- (n * sum_xy - sum_x * sum_y) / (n * sum_x2 - sum_x^2)
# Calculate the Intercept (b)
intercept <- (sum_y - slope * sum_x) / n
# Display the equation
equation <- paste("The equation of the regression line for the given points is",
paste("y =", format(round(slope, 2)), "x +", format(round(intercept, 2))))
print(equation)## [1] "The equation of the regression line for the given points is y = 4.26 x + -14.8"The first partial derivatives of \(f\) with respect to \(x\) and \(y\) are:
\[ f_x = \frac{\partial f}{\partial x} = 24 - 6y^2 \]
\[ f_y = \frac{\partial f}{\partial y} = -12xy - 24y^2 \]
To find the critical points, we set \(f_x\) and \(f_y\) equal to zero:
\[ 24 - 6y^2 = 0 \quad \Rightarrow \quad y^2 = 4 \quad \Rightarrow \quad y = \pm 2 \]
For \(y = 2\):
\[ -12x(2) - 24(2)^2 = 0 \quad \Rightarrow \quad -24x - 96 = 0 \quad \Rightarrow \quad x = -4 \]
For \(y = -2\):
\[ -12x(-2) - 24(-2)^2 = 0 \quad \Rightarrow \quad 24x - 96 = 0 \quad \Rightarrow \quad x = 4 \]
The critical points are \((-4, 2)\) and \((4, -2)\).
The second partial derivatives of \(f\) are:
\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 0 \]
\[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = -12x - 48y \]
\[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -12y \]
The Hessian determinant \(H\) is given by:
\[ H = f_{xx} f_{yy} - (f_{xy})^2 \]
For the critical point \((-4, 2)\):
\[ f_{xx} = 0, \quad f_{yy} = -12(-4) - 48(2) = 48 - 96 = -48, \quad f_{xy} = -12(2) = -24 \]
\[ H = (0)(-48) - (-24)^2 = -576 \]
For the critical point \((4, -2)\):
\[ f_{xx} = 0, \quad f_{yy} = -12(4) - 48(-2) = -48 + 96 = 48, \quad f_{xy} = -12(-2) = 24 \]
\[ H = (0)(48) - (24)^2 = -576 \]
Since \(H < 0\) at both critical points, \((-4, 2)\) and \((4, -2)\) are saddle points.
The function has no local maxima or minima.
# Define the function
f <- function(x, y) {
24 * x - 6 * x * y^2 - 8 * y^3
}
# Define the partial derivatives
fx <- function(y) {
24 - 6 * y^2
}
fy <- function(x, y) {
-12 * x * y - 24 * y^2
}
# Find critical points
critical_points <- function() {
y <- c(2, -2)
x <- c(-4, 4)
data.frame(x = x, y = y)
}
# Evaluate the Hessian determinant at critical points
hessian_determinant <- function(points) {
fxx <- 0
fyy <- -12 * points$x - 48 * points$y
fxy <- -12 * points$y
H <- fxx * fyy - (fxy)^2
H
}
# Determine saddle points and local extrema
local_extrema <- function() {
points <- critical_points()
H <- hessian_determinant(points)
extrema <- ifelse(H < 0, "saddle_point", ifelse(H > 0 & 0 > 0, "local_minimum", "local_maximum"))
result <- data.frame(points, Hessian = H, Type = extrema)
if (all(result$Type == "saddle_point")) {
cat("No local maxima or minima found. Only saddle points are present.\n")
}
for (i in 1:nrow(result)) {
cat(sprintf("Point (%.2f, %.2f): %s\n", result$x[i], result$y[i], result$Type[i]))
}
}
# Calculate the local extrema and saddle points
local_extrema()## No local maxima or minima found. Only saddle points are present.
## Point (-4.00, 2.00): saddle_point
## Point (4.00, -2.00): saddle_pointThe revenue function \(R(x, y)\) given by
\[ R(x, y) = x(81 - 21x + 17y) + y(40 + 11x - 23y) \]
can be simplified by expanding and combining like terms.
Expand the expressions:
The simplified revenue function is:
\[ R(x, y) = -21x^2 + 81x + 28xy + 40y - 23y^2 \]
# Simplified revenue function formula
revenue_function <- function(x, y) {
total_revenue <- -21 * x^2 + 81 * x + 28 * x * y + 40 * y - 23 * y^2
return(total_revenue)
}
# Prices given in the problem
price_house = 2.30
price_name = 4.10
# Calculate the total revenue for given prices
total_revenue_calculated = revenue_function(price_house, price_name)
# Result
print(paste("The total revenue for the house brand priced at $", price_house,
" and the name brand priced at $", price_name, " is $",
sprintf("%.2f", total_revenue_calculated), sep=""))## [1] "The total revenue for the house brand priced at $2.3 and the name brand priced at $4.1 is $116.62"The total weekly cost is given by: \[ C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700 \]
The constraint on production is: \[ x + y = 96 \]
\[ C(x, 96 - x) = \frac{1}{6}x^2 + \frac{1}{6}(96 - x)^2 + 7x + 25(96 - x) + 700 \]
\[ C(x) = \frac{1}{6}x^2 + \frac{1}{6}(96^2 - 192x + x^2) + 7x + 2400 - 25x + 700 \] \[ C(x) = \frac{1}{3}x^2 - 18x + 700 + 2400 + \frac{1}{6} \cdot 96^2 \]
\[ \frac{dC}{dx} = \frac{2}{3}x - 18 = 0 \] \[ \frac{2}{3}x = 18 \] \[ x = 27 \]
Using the constraint \(x + y = 96\): \[ y = 96 - x = 96 - 27 = 69 \]
The number of units that should be produced in Los Angeles is \(x = 27\) and in Denver is \(y = 69\) to minimize the total weekly cost.
# Define the cost function
cost_function <- function(x) {
y <- 96 - x
(1/6)*x^2 + (1/6)*y^2 + 7*x + 25*y + 700
}
# Define the derivative of the cost function with respect to x
cost_derivative_x <- function(x) {
(2/3)*x - 18
}
# Solve for the critical point
solve_critical_point <- function() {
x <- 18 / (2/3)
y <- 96 - x
cat(sprintf("Number of units produced in Los Angeles (x): %.2f\n", x))
cat(sprintf("Number of units produced in Denver (y): %.2f\n", y))
# Return the solution
list(x = x, y = y)
}
# Calculate the optimal production to minimize cost
solve_critical_point()## Number of units produced in Los Angeles (x): 27.00
## Number of units produced in Denver (y): 69.00## $x
## [1] 27
##
## $y
## [1] 69\[ \int_{2}^{4} \int_{2}^{4} e^{8x + 3y} \, dx \, dy \]
\[ \int_{2}^{4} e^{8x + 3y} \, dx \]
\(y\) acts as a constant during the integration over \(x\), we can treat \(e^{3y}\) as a constant multiplier.
\[ \int_{2}^{4} e^{8x + 3y} \, dx = e^{3y} \int_{2}^{4} e^{8x} \, dx \]
\[ \int e^{8x} \, dx = \frac{1}{8} e^{8x} \]
\[ \left[ \frac{1}{8} e^{8x} \right]_{2}^{4} = \frac{1}{8} \left( e^{32} - e^{16} \right) e^{3y} \]
\[ \int_{2}^{4} \frac{1}{8} \left( e^{32} - e^{16} \right) e^{3y} \, dy \]
\[ \frac{1}{8} \left( e^{32} - e^{16} \right) \int_{2}^{4} e^{3y} \, dy \]
\[ \int e^{3y} \, dy = \frac{1}{3} e^{3y} \]
\[ \frac{1}{8} \left( e^{32} - e^{16} \right) \left[ \frac{1}{3} e^{3y} \right]_{2}^{4} = \frac{1}{8} \left( e^{32} - e^{16} \right) \left( \frac{1}{3} \left( e^{12} - e^{6} \right) \right) \]
\[ \frac{1}{8} \left( e^{32} - e^{16} \right) \left( \frac{1}{3} \left( e^{12} - e^{6} \right) \right) \]
\[ \frac{1}{24} \left( e^{32} - e^{16} \right) \left( e^{12} - e^{6} \right) \]
# Define the inner integral function
inner_integral <- function(y) {
# Integrate e^(8x + 3y) with respect to x from 2 to 4
integrand <- Vectorize(function(x) {
exp(8 * x + 3 * y)
})
result <- integrate(integrand, lower = 2, upper = 4)
return(result$value)
}
# Define the outer integral function
outer_integral <- function() {
# Integrate the result of the inner integral with respect to y from 2 to 4
integrand <- Vectorize(function(y) {
inner_integral(y)
})
result <- integrate(integrand, lower = 2, upper = 4)
return(result$value)
}
# Perform the double integral calculation
result <- outer_integral()
cat("The value of the double integral is:", result, "\n")## The value of the double integral is: 5.341559e+17