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 )
I will use R to determine the regression equation.
x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
linreg <- lm(y ~ x)
linreg##
## Call:
## lm(formula = y ~ x)
##
## Coefficients:
## (Intercept) x
## -14.800 4.257The regression line for the given points has the equation \(y=4.26x-14.8\)
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.
\(f(x,y)=24x-6xy^2-8y^3\)
If \(f(x,y)=24x-6xy^2-8y^3\), then \(f_x(x,y)=24-6y^2\) and \(f_y(x,y)=-12xy-24y^2\)
The critical points of \(f\) occur when \(24-6y^2=0\) and \(-12xy-24y^2=0\). From \(24-6y^2=0\) we get \(y=2\) or \(y=-2\), we can then substitute each of those into \(-12xy-24y^2=0\) to determine the associated \(x\)-value.
When \(y=2\):
\(-12x(2)-24(2)^2=0\\-24x-96=0\\-24x=96\\x=-4\)
Substituting back into \(f(x,y)\) results in:
\(f(-4,2)=24(-4)-6(-4)(2)^2-8(2)^3=-96+96-64=-64\), so there is a critical point at \((-4,2,-64)\).
When \(y=-2\):
\(-12x(-2)-24(-2)^2=0\\24x-96=0\\24x=96\\x=4\)
Substituting back into \(f(x,y)\) results in \(f(4,-2)=24(4)-6(4)(-2)^2-8(-2)^3=96-96+64=64\), so there is a critical point at \((4,-2,64)\).
To determine if the points are minimums, maximums, or saddle points, we find to find the discriminant of \(f(x,y)\), as follows:
\(D(x,y)=f_{xx}f_{yy}-(f_{xy})^2\\D(x,y)=(0)(-12x-48y)-(-12y)^2\\D(x,y)=0-144y^2\\D(x,y)=-144y^2\)
Therefore, when \(y=2\), \(D(x,y)=-144(2)^2=-576\), so \(D(x,y)<0\), so \((-4,2,-64)\) is a saddle point. Additionally, when \(y=-2\), \(D(x,y)=-144(-2)^2=-576\), so \(D(x,y)<0\), so \((4,-2,64)\) is a saddle point.
Final Answer:
Local Maxima: { }
Local Minima: { }
Saddle Points: {\((-4,2,-64),(4,-2,64)\)}
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.
Find the revenue function \(R(x,y)\).
\(R(x,y)=R_h+R_n\), where \(R_h\) is the revenue from the “house” brand and \(R_n\) is the revenue from the “name” brand.
Then \(R_h=x(81-21x+17y)=81x-21x^2+17xy\), and \(R_n=y(40+11x-23y)=40y+11xy-23y^2\).
Therefore:
\(R(x,y)=R_h+R_n\\R(x,y)=81x-21x^2+17xy+40y+11xy-23y^2\\R(x,y)=-21x^2-23y^2+81x+28xy+40y\)
What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?
We can find the total revenue by substituting 2.30 for \(x\) and 4.10 for \(y\) in the revenue function found in Solution 3A.
\(R(2.3,4.1)=-21(2.3)^2-23(4.1)^2+81(2.3)+28(2.3)(4.1)+40(4.1)=116.62\)
The revenue will be $116.62.
We can confirm this result using R:
U_h <- function(x,y) { 81 - 21*x + 17*y } #Units sold of "house" brand
U_n <- function(x,y) { 40 + 11*x - 23*y } #Units sold of the "name" brand
#The total revenue is the price of the "house" brand times "house" units sold, plus the price of the "name" brand times "name" units sold.
revenue <- function(x,y) { (x * U_h(x,y)) + (y * U_n(x,y)) }
revenue(2.3,4.1)## [1] 116.62A 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?
Since the company is committed to produce a total of 96 units each week, we know that \(x+y=96\), and therefore \(y=96-x\). We can therefore rewrite \(C(x,y)\) as a function only of \(x\), as follows:
\(C(x)=\frac{1}{6}x^2+\frac{1}{6}(96-x)^2+7x+25(96-x)+700\\C(x)=\frac{1}{6}x^2+\frac{1}{6}(9216-192x+x^2)+7x+2400-25x+700\\C(x)=\frac{1}{6}x^2+1536-32x+\frac{1}{6}x^2+7x+2400-25x+700\\C(x)=\frac{1}{3}x^2-50x+4636\)
We can now find the value of \(x\) that minimizes weekly cost by setting \(C'(x)=0\), as follows:
\(C'(x)=\frac{2}{3}x-50\\0=\frac{2}{3}x-50\\50=\frac{2}{3}x\\75=x\)
Therefore, the weekly cost is minimized when 75 units are manufactured in Los Angeles, meaning there are 21 manufactured in Denver.
We can confirm this result using R:
library(tidyverse)## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr 1.1.2 ✔ readr 2.1.4
## ✔ forcats 1.0.0 ✔ stringr 1.5.0
## ✔ ggplot2 3.4.4 ✔ tibble 3.2.1
## ✔ lubridate 1.9.2 ✔ tidyr 1.3.0
## ✔ purrr 1.0.1
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errorslibrary(knitr)
los_angeles_units <- c(0:96)
denver_units <- 96-los_angeles_units
production_combinations <- data.frame(los_angeles_units, denver_units)
minimized_costs <- production_combinations %>%
mutate(total_cost = 1/6*los_angeles_units^2 + 1/6*denver_units^2 + 7*los_angeles_units + 25*denver_units + 700) %>%
filter(total_cost == min(total_cost))
kable(minimized_costs, col.names = c("Los Angeles Units", "Denver Units", "Cost"), align = "ccc")| Los Angeles Units | Denver Units | Cost |
|---|---|---|
| 75 | 21 | 2761 |
Evaluate the double integral on the given region.
\(\int\limits_R\int(e^{8x+3y})dA;R:2\leq x\leq4\space\text{and}\space 2\leq y\leq4\)
Write your answer in exact form without decimals.
\(\int\limits_R\int(e^{8x+3y})dA;R:2\leq x\leq4\space\text{and}\space 2\leq y\leq4\\=\int\limits_2^4(\int\limits_2^4e^{8x+3y}dx)dy\\=\int\limits_2^4(_2^4[\frac{1}{8}e^{8x+3y}]dy\\=\int\limits_2^4(\frac{1}{8}e^{32+3y}-\frac{1}{8}e^{16+3y})dy\\=_2^4[\frac{1}{24}e^{3y+32}-\frac{1}{24}e^{3y+16}]\\=(\frac{1}{24}e^{44}-\frac{1}{24}e^{28})-(\frac{1}{24}e^{38}-\frac{1}{24}e^{22})\\=\frac{1}{24}(e^{44}+e^{22}-e^{28}-e^{38})\)