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 )

library(tidyverse)
library(ggthemes)
X <- c(5.6, 6.3, 7, 7.7, 8.4)
y = c(8.8, 12.4, 14.8, 18.2, 20.8)
z <- rep(0, 5)
df = data_frame(X, y, z)

ggplot(data = df, aes(x=X, y=y)) +
  geom_point() + 
  theme_tufte() +
  ggtitle('Scatter Plot')

mod = lm(y ~ X, data = df)
sum1 <- summary(mod)
ans1 <- sum1$coefficients[1]

The answer is -14.8.

#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.

\[ f(x,y) = 24x-6xy^2 -8y^3 \\ \frac{\partial f}{\partial x}=24 - 6y^2 \\ \frac{\partial f}{\partial y}=-12xy-24y^2 \]

Now we must find where both partials are equal to zero. For x:

\[ 24-6y^2=0 \\ 4-y^2=0 \\ y^2=4 \\ y = (2, \quad -2) \]

For y:

\[ -12xy-24y^2=0 \\ xy-2y^2=0 \\ xy=2y^2 \\ x=2y \]

Now plug in the values from the partial WRT x:

\[ y=(2,-2) \\ x=2y \\ x=(4,-4) \]

Thus the cirtical points (inflection) points are:

\[ (-4,2) \quad \& \quad (4,-2) \]

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.

Find the revenu function and the revenue at x=2.3 and y = 4.1 First we need to multiply each equation by the variable that it represents:

$$ R(x,y) = (81- 21x + 17y)x + (40 + 11x - 23y)y \ = 81x- 21x^2 + 17yx + 40y + 11xy - 23y^2 \ = 81x - 21x^2 + 28yx + 40y - 23y^2 \

$$

x = 2.3
y = 4.1
81 * x - 21* x^2 + 28 *y*x + 40*y  - 23*y^2

## [1] 116.62

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?

This can be written as a univariate function because \(x=96-y\)

$$ C(x, y) = x^2 + y^2 + 7x + 25y + 700 \ C(96-y, y) = (96-y)^2 + y^2 + 7(96-y) + 25y + 700 \ = (9216 - 192y + y^2) + y^2 + 7(96-y) + 25y + 700 \ = y^2 -32y + 1536 + y^2 + 672-7y + 25y + 700 \ C(y)= y^2 -14y +2908 \ C’(y) = y - 14

$$

Costs are minimized when \(C'(y)=0\)

\[ \frac{2}{3}y - 14=0 \\ \frac{2}{3}y =14 \\ y =21 \\ \therefore \\ x = 96-21=75 \]

5

Evaluate the double integral on the given region. \[ \int \int (e^{8x+3y}dA);\quad R:2 \leq x \leq 4 \quad \&\quad \leq y \leq 4 \]

Write your answer in exact form without decimals.

\[ \int_2 ^4 \int_2 ^4 (e^{8x+3y}dx dy) \\ = \int_2 ^4 \frac{1}{3}e^{8x+3y}|_2^4 dx \\ = \int_2 ^4 (\frac{1}{3}e^{8x+12}) - (\frac{1}{3}e^{8x+6}) dx \\ = (\frac{1}{24}e^{8x+12}) - (\frac{1}{24}e^{8x+6}) |_2 ^4 \\ = ((\frac{1}{24}e^{44}) - (\frac{1}{24}e^{38})) - ((\frac{1}{24}e^{28}) - (\frac{1}{24}e^{22})) \\ = \frac{1}{24}e^{44} - \frac{1}{24}e^{38} - \frac{1}{24}e^{28} + \frac{1}{24}e^{22} \\ = \frac{1}{24}(e^{44} - e^{38} - e^{28} + e^{22}) \]