Find the equation of the regression line for the given points. Round any final values to the nearest hundredth, if necessary. \[ (5.6,8.8),(6.3,12.4),(7,14.8),(7.7,18.2),(8.4,20.8) \] The equation for a linear regression line can be represented as: \[ y=m x+c \] - \(m\) is the slope of the line - \(c\) is the \(y\)-intercept of the line
The slope \(m\) can be calculated as: \[ m=\frac{n(\Sigma x y)-(\Sigma x)(\Sigma y)}{n\left(\Sigma x^2\right)-(\Sigma x)^2} \]
The \(y\)-intercept \(c\) can be calculated as: \[ c=\frac{\sum y-m(\Sigma x)}{n} \] - \(\Sigma x\) is sum of all \(\mathrm{x}\) values - \(\Sigma y\) is the sum of all \(y\) values - \(\quad \sum x y\) is the sum of the products of corresponding \(\mathrm{x}\) and \(\mathrm{y}\) values - \(\Sigma x^2\) is the sum of the squares of all \(\mathrm{x}\) values - \(n\) is the number of data points
First, compute the value of sums: \[ \begin{aligned} & \sum x=5.6+6.3+7+7.7+8.4=35 \\ & \sum y=8.8+12.4+14.8+18.2+20.8=75 \\ & \sum x y=(5.6 * 8.8)+(6.3 * 12.4)+(7 * 14.8)+(7.7 * 18.2)+(8.4 * \\ & 20.8)=545.86 \\ & \sum x^2=5.6^2+6.3^2+7^2+7.7^2+8.4^2=249.9 \end{aligned} \]
Find the slope \((m)\) : \[ \begin{aligned} m & =\frac{n(\Sigma x y)-(\Sigma x)(\Sigma y)}{n\left(\Sigma x^2\right)-(\Sigma x)^2} \\ m & =\frac{5(545.86)-(35)(75)}{5(249.9)-(35)^2} \\ m & =4.257 \end{aligned} \] calculate the \(y\)-intercept \((c)\) \[ \begin{aligned} & c=\frac{\sum y-m\left(\sum x\right)}{n} \\ & c=\frac{75-(4.257 * 35)}{5} \\ & c=-14.799 \end{aligned} \]
Therefore, the equation of the regression line is approximately: \[ y \approx 4.257 x-14.799 \]
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.
\[ f(x, y)=24 x-6 x y^2-8 y^3 \]
To find local maxima, minima, or saddle points: calculate the partial derivatives: \[ \begin{gathered} f_x=\frac{\partial f}{\partial x}=24-6 y^2 \\ f_y=\frac{\partial f}{\partial y}=-12 x y-24 y^2 \end{gathered} \]
Set both partial derivatives to zero: \[ \begin{aligned} & f_x=0 \Rightarrow 24-6 y^2=0 \\ & f_y=0 \Rightarrow-12 x y-24 y^2=0 \end{aligned} \]
From \(f_x=0\), we get \(y= \pm 2\) When \(y=2, x=-4\). When \(y=-2, x=4\) These critical points are \((-4,2)\) and \((4,-2)\) \[ \begin{gathered} z=24(-4)-6(-4) * 2^2-8(2)^3=-64 \\ z=24(4)-6(4) * (-2)^2-8(-2)^3=64 \end{gathered} \]
Hence, both critical points \((-4,2,-64)\) and \((4,-2,64)\) are saddle points.
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).
Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10 ?
Answer:
Step 1. The revenue function is \[ \begin{aligned} R(x, y) & =(81-21 x+17 y) x+(40+11 x-23 y) y \\ & =81 x-21 x^2+17 x y+40 y+11 x y-23 y^2 \\ & =81 x+40 y+28 x y-21 x^2-23 y^2 \end{aligned} \]
Step 2. The revenue when the prices are \(\$ 2.30\) for the “house” brand and \(\$ 4.10\) for the “name” brand is \[ R(2.3,4.1)=81 * 2.3+40 * 4.1+28 * 2.3 * 4.1-21 *(2.3)^{\wedge} 2-23 *(4.1)^2=116.62 \]
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 \(C(x, y)=\frac{1}{6} x^2+\frac{1}{6} y^2+7 x+25 y+700\), 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?
Given that \[ \begin{gathered} C(x, y)=\frac{1}{6} x^2+\frac{1}{6} y^2+7 x+25 y+700 \\ x+y=96 \end{gathered} \]
So, \(y=96-x\) Substitute \(y\) in terms of \(x\) into \(C(x, y)\) \[ C(x)=\frac{1}{6} x^2+\frac{1}{6}(96-x)^2+7 x+25(96-x)+700 \]
Simplify this equation \[ \begin{gathered} C(x)=\frac{1}{6} x^2+\frac{1}{6}\left(9216-192 x+x^2\right)+7 x+2400-25 x+700 \\ C(x)=\frac{1}{6} x^2+1536-32 x+\frac{1}{6} x^2+7 x-25 x+3100 \\ C(x)=\frac{1}{3} x^2-50 x+4636 \end{gathered} \]
Find \(C^{\prime}(x)\) and \(x\) \[ C^{\prime}(x)=\frac{2}{3} x-50 \] \(\operatorname{Set} C^{\prime}(x)=0\) \[ \begin{gathered} \frac{2}{3} x-50=0 \\ x=\frac{3}{2} \times 50=75 \end{gathered} \]
Substitute \(x=75\) into \(y=96-x\) \[ y=96-75=21 \]
Therefore, Los Angeles plant should produce 75 units, and Denver plant should produce 21 units to minimize the total weekly cost.
Evaluate the double integral on the given region. \[ \iint_R\left(e^{8 x+3 y}\right) d A ; R: 2 \leq x \leq 4 \text { and } 2 \leq y \leq 4 \]
Write your answer in exact form without decimals.
library(pracma) # Load the pracma package
# Define the integrand
Integral_function <- function(x, y) {
exp(8*x + 3*y)
}
# Calculate the double integral
result <- integral2(Integral_function, 2, 4, 2, 4)
# Display the result in exact form without decimals
cat("The exact form without decimals is", format(result$Q, scientific = FALSE), "\n")## The exact form without decimals is 534155947871806976