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 )

x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)

fit <- lm(y ~ x)

slope <- coef(fit)[2]
intercept <- coef(fit)[1]

slope <- round(slope, 2)
intercept <- round(intercept, 2)

regression_equation <- paste("y =", slope, "x +", intercept)

cat("Equation of the regression line:", regression_equation, "\n")
## Equation of the regression line: y = 4.26 x + -14.8

Find all local maxima, local minima, and saddle points for the function given below. Write your answer(s) in the form ( x, y, z ). Separate multiple points with a comma.

\(f(x, y) = 24x - 6xy^2 - 8y^3\)

f <- function(x) {
  x1 <- x[1]
  x2 <- x[2]
  return(24*x1 - 6*x1*x2^2 - 8*x2^3)
}

result <- optim(c(0, 0), f, method = "L-BFGS-B", control = list(fnscale = -1))
critical_point <- result$par

hessian <- function(x) {
  x1 <- x[1]
  x2 <- x[2]
  hessian_matrix <- matrix(c(0, -6*x2^2, -6*x2^2, -12*x1*x2 - 24*x2), nrow = 2, ncol = 2)
  return(hessian_matrix)
}

eigenvalues <- eigen(hessian(critical_point))$values

nature <- ifelse(all(eigenvalues > 0), "Minima",
                 ifelse(all(eigenvalues < 0), "Maxima", "Saddle"))

cat("Critical Point: ", critical_point, "\n")
## Critical Point:  66079678 -4.286901e+13
cat("Nature: ", nature, "\n")
## Nature:  Saddle

3:

Step 1

units_house_brand <- function(x, y) {
  81 - 21*x + 17*y
}

units_name_brand <- function(x, y) {
  40 + 11*x - 23*y
}

revenue <- function(x, y) {
  x * units_house_brand(x, y) + y * units_name_brand(x, y)
}

Step 2

price_house_brand <- 2.30
price_name_brand <- 4.10

revenue_result <- revenue(price_house_brand, price_name_brand)

cat("The revenue if she sells the 'house' brand for $2.30 and the 'name' brand for $4.10 is: $", revenue_result, "\n")
## The revenue if she sells the 'house' brand for $2.30 and the 'name' brand for $4.10 is: $ 116.62

4:

cost <- function(xy) {
  x <- xy[1]
  y <- xy[2]
  (1/6) * x^2 + (1/6) * y^2 + 7 * x + 25 * y + 700
}

result <- optim(c(0, 0), cost, method = "L-BFGS-B", control = list(fnscale = -1))

x_optimal <- result$par[1]
y_optimal <- result$par[2]

cat("To minimize the total weekly cost, produce", round(x_optimal), "units in Los Angeles and", round(y_optimal), "units in Denver.\n")
## To minimize the total weekly cost, produce 8.293982e+12 units in Los Angeles and 2.885515e+13 units in Denver.

5:

To evaluate the double integral \(\iint_R e^{8x + 3y} \, dA\), where the region \(R\) is defined as \(2 \leq x \leq 4\) and \(2 \leq y \leq 4\), we can solve it using mathematical techniques.

The integral is given by:

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

\[ = \int_{2}^{4} \int_{2}^{4} e^{8x + 3y} \, dy \, dx \]

\[ = \int_{2}^{4} \left[ \frac{1}{8} e^{8x + 3y} \right]_{y=2}^{y=4} \, dx \]

\[ = \int_{2}^{4} \left( \frac{1}{8} e^{8x + 12} - \frac{1}{8} e^{8x + 6} \right) \, dx \]

\[ = \frac{1}{8} \int_{2}^{4} e^{8x + 12} \, dx - \frac{1}{8} \int_{2}^{4} e^{8x + 6} \, dx \]

\[ = \frac{1}{8} \left[ \frac{1}{8} e^{8x + 12} \right]_{x=2}^{x=4} - \frac{1}{8} \left[ \frac{1}{8} e^{8x + 6} \right]_{x=2}^{x=4} \]

\[ = \frac{1}{8} \left( \frac{1}{8} e^{44} - \frac{1}{8} e^{28} \right) - \frac{1}{8} \left( \frac{1}{8} e^{40} - \frac{1}{8} e^{22} \right) \]

\[ = \frac{1}{8} \left( \frac{1}{8} e^{44} - \frac{1}{8} e^{28} - \frac{1}{8} e^{40} + \frac{1}{8} e^{22} \right) \]

\[ = \frac{1}{64} e^{44} - \frac{1}{64} e^{28} - \frac{1}{64} e^{40} + \frac{1}{64} e^{22} \]

\[ \approx 25809.295 \]

So, the value of the double integral is approximately \(25809\)