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 )
Solution:
To find the equation of the regression line for the given set of points, I will be calculating the slope (m) and the y-intercept (b) of the line using the least squares method.
x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
data <- data.frame(x, y)
# linear regression
model <- lm(y ~ x, data = data)
summary(model)##
## Call:
## lm(formula = y ~ x, data = data)
##
## Residuals:
## 1 2 3 4 5
## -0.24 0.38 -0.20 0.22 -0.16
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -14.8000 1.0365 -14.28 0.000744 ***
## x 4.2571 0.1466 29.04 8.97e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.3246 on 3 degrees of freedom
## Multiple R-squared: 0.9965, Adjusted R-squared: 0.9953
## F-statistic: 843.1 on 1 and 3 DF, p-value: 8.971e-05So:
y=4.26x−14.80
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
Solution:
Finding Critical Points and Classifying Them: Given the function: \[ f(x, y) = 24x - 6xy^2 - 8y^3 \]
Let’s first compute the partial derivatives First partial derivatives of \(f\) are: \[ \frac{\partial f}{\partial x} = 24 - 6y^2 \] \[ \frac{\partial f}{\partial y} = -12xy - 24y^2 \]
Than, let’s solve for critical points Set the partial derivatives equal to zero to find the critical points: \[ 24 - 6y^2 = 0 \] \[ -12xy - 24y^2 = 0 \]
From \(24 - 6y^2 = 0\), we find: \[ y^2 = 4 \implies y = \pm 2 \]
Substitute \(y = 2\) into the second equation: \[ -12x(2) - 24(2)^2 = 0 \] \[ -24x - 96 = 0 \] \[ x = -4 \]
Substitute \(y = -2\) into the second equation: \[ -12x(-2) - 24(-2)^2 = 0 \] \[ 24x - 96 = 0 \] \[ x = 4 \]
Then, let’s do the classification using second derivative test second derivatives are: \[ \frac{\partial^2 f}{\partial x^2} = 0, \quad \frac{\partial^2 f}{\partial y^2} = -12x - 48y, \quad \frac{\partial^2 f}{\partial x \partial y} = -12y \]
The Hessian matrix \(H\) is: \[ H = \begin{pmatrix} 0 & -12y \\ -12y & -12x - 48y \end{pmatrix} \]
The determinant of the Hessian matrix is: \[ \text{det}(H) = 144y^2 \]
The critical points and their classification are: \[ (x, y, z) = (-4, 2, f(-4, 2)), (4, -2, f(4, -2)) \] Since \(\text{det}(H) > 0\) and \(\frac{\partial^2 f}{\partial y^2} < 0\) for both points, both are saddle points.
# function
f <- function(x, y) {24 * x - 6 * x * y^2 - 8 * y^3}
# partial derivatives
f_x <- function(x, y) {24 - 6 * y^2}
f_y <- function(x, y) {-12 * x * y - 24 * y^2}
# solving the system of equations f_x = 0 and f_y = 0
library(rootSolve)
solution <- multiroot(f = function(z) c(f_x(z[1], z[2]), f_y(z[1], z[2])), start = c(0, 0))## diagonal element is zero
## [1] 1x_solution <- solution$root[1]
y_solution <- solution$root[2]
# second derivatives
f_xx <- function(x, y) {0}
f_yy <- function(x, y) {-12 * x - 48 * y}
f_xy <- function(x, y) {-12 * y}
# determinant of the Hessian
hessian_det <- function(x, y) {-144 * y^2}
hessian_determination <- hessian_det(x_solution, y_solution)
# results
list(solution = solution, hessian_determination = hessian_determination)## $solution
## $solution$root
## [1] 0 0
##
## $solution$f.root
## [1] 24 0
##
## $solution$iter
## [1] 1
##
## $solution$estim.precis
## [1] 12
##
##
## $hessian_determination
## [1] 0A 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 ). Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?
Solution:
I will first derive the revenue function:
Given a grocery store sells two brands of a product, “house” and “name” brands, at prices \(x\) dollars for “house” and \(y\) dollars for “name”. The quantity sold of each brand depends on their prices:
The revenue function \(R(x, y)\) is then given by: \[R(x, y) = x(81 - 2x + 17y) + y(40 + 11x - 23y)\]
Expanding and simplifying: \[R(x, y) = 81x - 2x^2 + 28xy + 40y - 23y^2\]
Than I’ll calculate specific revenue:
To find the revenue when the “house” brand sells for $2.30 and the “name” brand for $4.10, substitute \(x = 2.30\) and \(y = 4.10\) into the revenue function: \[R(2.30, 4.10) = 81(2.30) - 2(2.30)^2 + 28(2.30)(4.10) + 40(4.10) - 23(4.10)^2\]
Calculating the above expression gives: \[R(2.30, 4.10) = 81 \times 2.30 - 2 \times (2.30)^2 + 28 \times 2.30 \times 4.10 + 40 \times 4.10 - 23 \times (4.10)^2\]
x <- 2.30
y <- 4.10
R <- 81*x - 2*x^2 + 28*x*y + 40*y - 23*y^2
R## [1] 217.13A 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 the function: \[ C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700, \] 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?
Solution:
Given the cost function: \[ C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700 \] subject to the constraint: \[ x + y = 96 \]
Substituting \(y = 96 - x\) into the cost function: \[ C(x) = \frac{1}{6}x^2 + \frac{1}{6}(96 - x)^2 + 7x + 25(96 - x) + 700 \] Simplifying, we obtain: \[ C(x) = \frac{1}{3}x^2 - \frac{208}{6}x + 4300 \]
Minimizing \(C(x)\) by setting its derivative equal to zero: \[ \frac{dC}{dx} = \frac{2}{3}x - \frac{208}{6} = 0 \] Solving for \(x\): \[ x = 52 \] Then, \(y = 96 - 52 = 44\).
To minimize the total weekly cost, 52 units should be produced in Los Angeles and 44 units in Denver.
Evaluate the double integral on the given region: \[\iint_R e^{8x + 3y} \, dA\] where the region \(R\) is defined by: \[ 2 \leq x \leq 4 \quad \text{and} \quad 2 \leq y \leq 4 \]
Write your answer in exact form without decimals.
Solution:
Considering the double integral over the region \(R\) defined by \(2 \leq x \leq 4\) and \(2 \leq y \leq 4\):
\[\iint_R e^{8x + 3y} \, dA\]
The integral is set up over the rectangular region \(R\): \[\int_{x=2}^{4} \int_{y=2}^{4} e^{8x + 3y} \, dy \, dx\]
Performing the Inner Integral Integrate with respect to \(y\) first, factoring out the \(e^{8x}\) term: \[\int_{y=2}^{4} e^{8x + 3y} \, dy = e^{8x} \int_{y=2}^{4} e^{3y} \, dy\] The integral of \(e^{3y}\) is: \[e^{8x} \left[ \frac{e^{3y}}{3} \right]_{2}^{4} = e^{8x} \left( \frac{e^{12}}{3} - \frac{e^{6}}{3} \right) = \frac{e^{8x} (e^{12} - e^{6})}{3}\]
Performing the Outer Integral Now, integrate with respect to \(x\): \[\int_{x=2}^{4} \frac{e^{8x} (e^{12} - e^{6})}{3} \, dx = \frac{e^{12} - e^{6}}{3} \int_{x=2}^{4} e^{8x} \, dx\] The integral of \(e^{8x}\) is: \[\left[ \frac{e^{8x}}{8} \right]_{2}^{4} = \frac{e^{12} - e^{6}}{3} \left( \frac{e^{32}}{8} - \frac{e^{16}}{8} \right) = \frac{e^{12} - e^{6}}{24} (e^{32} - e^{16})\]
The evaluated double integral in exact form is: \[\frac{e^{12} - e^{6}}{24} (e^{32} - e^{16})\] This expression represents the exact value of the double integral over the given region.