- 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 )
given_points <- c(5.6, 8.8, 6.3, 12.4, 7, 14.8, 7.7, 18.2, 8.4, 20.8)
D <- matrix(given_points, ncol = 2, byrow = TRUE)
df <- data.frame(D)
names(df) <- c("x", "y")
the_model <- lm(y ~ x, df)
b <- the_model$coefficients[1]
m <- the_model$coefficients[2]
paste0("y = ", round(m, 2), "x + ", b)## [1] "y = 4.26x + -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.
\(f ( x, y ) = 24x - 6xy^2 - 8y^3\)
interval <- 10000
fxy <- function(x, y) {
24 * x - 6 * x * y^2 - 8 * y^3
}
(fxy_prime <- Deriv(fxy))## function (x, y)
## c(x = 24 - 6 * y^2, y = -(y * (12 * x + 24 * y)))# f_x critical points
f_x <- function(y) 24 - 6 * y^2
f_x_critical_points <- uniroot.all(f_x, c(-interval, interval))
# f_y critical points
f_y_critical_points <- c(NA, NA)
f_y <- function(x, y) -12 * x * y - 24 * y^2
f_y_critical_points[1] <- uniroot.all(f_y, c(-interval, interval), y = f_x_critical_points[1])
f_y_critical_points[2] <- uniroot.all(f_y, c(-interval, interval), y = f_x_critical_points[2])
(critical_points <- expand.grid(f_x_critical_points, f_y_critical_points))## Var1 Var2
## 1 -1.999975 3.99995
## 2 1.999975 3.99995
## 3 -1.999975 -3.99995
## 4 1.999975 -3.99995# 2nd partial derivatives
(fxy_dblprime <- Deriv(fxy_prime))## function (x, y)
## {
## .e1 <- -(12 * y)
## c(x = c(x = 0, y = .e1), y = c(x = .e1, y = -(12 * x + 48 *
## y)))
## }f_xx <- function(x, y) 0
f_xy <- function(x, y) -(12 * y)
f_yy <- function(x, y) -(12 * x + 48 * y)
f_yx <- function(x, y) -(12 * y)
D <- function(x, y) f_xx(x, y) * f_yy(x, y) - f_xy(x, y) * f_yx(x, y)
assess_points <- function(critical_points) {
r <- nrow(critical_points)
for (i in 1:r) {
D_result <- round(D(critical_points[i, 1], critical_points[i, 2]), 4)
z <- fxy(critical_points[i, 1], critical_points[i, 2])
point <- paste(round(c(critical_points[i, 1], critical_points[i, 2],
z), 4), collapse = " ")
if (D_result == 0) {
conclusion <- paste0(point, "; D = ", D_result, " ;Inconclusive")
} else if (D_result < 0) {
conclusion <- paste0(point, "; D = ", D_result, " ;Saddle Point")
} else {
f_xx_result <- f_xx(critical_points[i, 1], critical_points[i, 2])
if (f_xx_result > 0) {
paste0(point, "; D = ", D_result, " ;f_xx > 0; ", "Local Minimum")
} else if (f_xx_result < 0) {
conclusion <- paste0(point, "; D = ", D_result, " ;f_xx < 0; ",
"Local Maximum")
}
}
print(conclusion)
}
}
assess_points(critical_points)## [1] "-2 3.9999 -367.9873; D = -2303.9419 ;Saddle Point"
## [1] "2 3.9999 -655.974; D = -2303.9419 ;Saddle Point"
## [1] "-2 -3.9999 655.974; D = -2303.9419 ;Saddle Point"
## [1] "2 -3.9999 367.9873; D = -2303.9419 ;Saddle Point"- 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 ).
\(R (x, y) = x(81 - 21x + 17y) + y(40 + 11x - 23y)\)
\(R (x, y) = -21x^2 + 81x + 28xy + 40y - 23y^2\)
Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?
x <- 2.3
y <- 4.1
fxy <- function(x, y) -21 * x^2 + 81 * x + 28 * x * y + 40 * y - 23 * y^2
fxy(x, y)## [1] 116.62- A company has a plant in Los Angeles and a plant in Denver. The firm is committed to produce a total of 96 units of a product each week. The total weekly cost is given by \(C(x, y) = 1/6x^2 + 1/6y^2 + 7x + 25y + 700\), where x is the number of units produced in Los Angeles and y is the number of units produced in Denver. How many units should be produced in each plant to minimize the total weekly cost?
\(96 = x + y \rightarrow y = 96 - x\)
\(1/6x^2 + 1/6y^2 + 7x + 25y + 700 \rightarrow 1/6x^2 + 1/6(96 - x)^2 + 7x + 25(96 - x) + 700\)
Cx <- function(x) 1/6 * x^2 + 1/6 * (96 - x)^2 + 7 * x + 25 * (96 - x) + 700
Cx_prime <- Deriv(Cx)
s <- seq(-100, 100, by = 1)
plot(s, Cx_prime(s), type = "l", col = "red")
abline(0, 0)
(x <- uniroot(Cx_prime, c(-100, 100))$root)## [1] 75(y <- 96 - x)## [1] 21- Evaluate the double integral on the given region. Write your answer in exact form without decimals
\(\iint(e^{8x + 3y }dA)\)
\(R: 2 \leq x \leq 4\)
and
\(2\leq y \leq 4\)
\[\begin{equation} \iint(e^{8x + 3y} dA) = \int_{2}^{4}\int_{2}^{4}e^{8x + 3y} dydx \end{equation}\] \[\begin{equation} = \int_{2}^{4}\left(\left[1/3e^{8x + 3y}|_2^4 \right]\right)dx \end{equation}\] \[\begin{equation} = \int_{2}^{4}\left( 1/3(e^{8x + 12} - e^{8x + 6}) \right)dx \end{equation}\] \[\begin{equation} = \left( 1/24(e^{8x + 12} - e^{8x + 6})\right)|_{2}^{4} \end{equation}\] \[\begin{equation} = 1/24\left(e^{44} - e^{38} - e^{28} + e^{22}\right) \end{equation}\]