Library
library(Deriv)Excercise 1
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)
data_lm <- lm(y~x)
summary(data_lm)##
## Call:
## lm(formula = y ~ x)
##
## 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-05Equation of the Regression Line: \(y=4.2571x - 14.8\)
Excercise 2
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. \[f ( x, y ) = 24x - 6xy^{2} - 8y^{3}\]
fx <- function(x){24*x - 6*x*y^2 - 8*y^3}
fy <- function(y){24*x - 6*x*y^2 - 8*y^3}
f_x <- Deriv(fx)
f_x## function (x)
## 24 - 6 * y^2f_y <- Deriv(fy)
f_y## function (y)
## -(y * (12 * x + 24 * y))\[24-6y^2=0{\longrightarrow} 24 = 6y^2{\longrightarrow} y= 2,-2 \] \[-12xy-24y^2\] \[-12x(2)-24(2)^2 = 0 {\longrightarrow} x = -4\] \[-12x(-2)-24(-2)^2 = 0 {\longrightarrow} x = 4\]
Critical Points: (2,-4) & (-2,4) \[f_{xx}=0\] \[f_{xy}=-12y\] \[D = f_{xy}^2=0 {\longrightarrow} (-12*-4)^2 ; (-12*4)^2 {\longrightarrow} D >0\]
The Second Derivative Test does not meet any of the 4 criteria and is therefore inconclusive.
Excercise 3
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 ). \[House = 81-21x+17y\] \[Name = 40+11x-23y\] \[R(x,y)=House\cdot x + Name\cdot y\] \[R(x,y)=(81-21x+17y) x + (40+11x-23y)y\] \[R(x,y)=81x-21x^2+17xy + 40y+11yx-23y^2\]
Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10? \[R(2.30,4.10)=81(2.3)-21(2.3)^2+17(2.3)(4.1) + 40(4.1)+11(4.1)(2.3)-23(4.1)^2\]
81*2.3-21*2.3^2+17*2.3*4.1+40*4.1+11*4.1*2.3-23*4.1^2## [1] 116.62Excercise 4
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 \(C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700\), 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? \[g(x,y) = 0{\longrightarrow} x + y -96 = 0\] \[C_x= \lambda g_x {\longrightarrow} \frac{2}{6}x +7 = \lambda\] \[C_y= \lambda g_y {\longrightarrow} \frac{2}{6}y + 25 = \lambda \] \[\frac{2}{6}x +7 = \frac{2}{6}y + 25\] \[x = y + 54\] \[(y+54) + y = 96\] \[y=21\] \[x = (21) + 54 = 75\]