CUNY MSDS DATA 605 WEEK 15
Library
library(tidyverse)
library(plotly)
Questions
- 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 )
data <- c(5.6, 8.8, 6.3, 12.4, 7, 14.8, 7.7, 18.2, 8.4, 20.8)
q1 <- matrix(data, ncol=2, byrow = T)
q1
## [,1] [,2]
## [1,] 5.6 8.8
## [2,] 6.3 12.4
## [3,] 7.0 14.8
## [4,] 7.7 18.2
## [5,] 8.4 20.8
q1_df <- data.frame(q1)
model1 <- lm(q1_df$X2 ~., q1_df)
summary(model1)
##
## Call:
## lm(formula = q1_df$X2 ~ ., data = q1_df)
##
## 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 ***
## X1 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-05
p <- ggplot(model1, aes(q1_df$X2, q1_df$X1))
p <- p + geom_point() +
stat_smooth(method="lm")
p <- ggplotly(p)
p
- 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.
\[f ( x, y ) = 24x - 6xy^2 - 8y^3\]
\[\frac{d f}{d x} = 24 - 6y^2\] \[\frac{df}{d y} = -12xy - 24y^2\]
\[\frac{d f}{d x} = 24-6y^2 = 0 -> 4-y^2=0\]
\[\frac{d f}{d y}= -12xy-24y^2 = 0 -> -xy-2y^2=0\]
\[for (4,-2) f(x,y) = 24*4-6*4*(-2)^2-8(-2)^3 = 64\]
\[for (4,-2)f(x,y) = 24*-4-(6*-4*(2)^2)-8(2)^3 = -64\] (-4,2) the 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.
\[R (x, y) = x(81 - 21x + 17y) + y(40 + 11x - 23y)\] \[R (x, y) = -21x^2 + 81x + 28xy + 40y - 23y^2\]
x <- 2.3
y <- 4.1
-21 * x^2 + 81 * x + 28 * x * y + 40 * y - 23 * y^2
## [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)= 1/6x2+1/6y2+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?
x + y = 96 so we substitute y = 96 − x
C(x, y) gives us x2 − 50 ∗ x + 4636
x = 75 which means that y = 21
- Evaluate the double integral on the given region. Write your answer in exact form without decimals
\[e^(8x+3y) = e^(8*x)+e^(3*y)\] \[(e^12-e^6)(e^32-e^16)/24\]