library(ggplot2)
library(dplyr)

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)

reg_line <- lm(y~x)
reg_line
## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257

Per regression model, the line is y = -14.8 + 4.26x

Below is the line plot.

reg_line %>% 
  ggplot(aes(x, y)) +
  geom_point() +
  geom_smooth(method = lm, se = F)
## `geom_smooth()` using formula 'y ~ x'

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.

\[f(x, y) = 24x - 6xy^2 - 8y^3\]

Partial derivatives:

\[f_x(x, y) = 24 - 6y^2\]

\[f_y(x, y) = -12xy - 24y^2\]

For Critical Points:

\(f_x = 0\)

If \(24-6y^2=0\), then \(y^2 = 4\) and \(y = \pm2\)

Substitute y values in \(f_y=0\):

If \(y=2\) and \(-12xy - 24y^2=0\), then \(-24x = 24 * 4\) and \(x=-4\)

If \(y=-2\) and \(-12xy - 24y^2=0\), then \(24x = 24 * 4\) and \(x=4\)

Calculate \(f(x, y)\).

\[f(4,-2) = 24\times 4 - 6\times4\times (-2)^2 - 8 \times (-2)^3 = 64\]

\[f(-4,2) = 24\times (-4) - 6\times(-4)\times 2^2 - 8 \times 2^3 = -64\]

Two critical points: \((4,-2,64)\) and \((-4, 2, -64)\).

Now Lets use second derivative test to determine if points are minimum, maximum or saddle.

Second partial derivatives:

\[f_{xx}=0\]

\[f_{yy}=-12x-48y\]

\[f_{xy}=-12y\]

Not Putting the values in formula

\[D(x,y) = f_{xx} f_{yy}-f^2_{xy} = -(-12y)^2 = -144y^2\]

If we replace y=2 or y=-2, we get D<0 hence critical points (-4,2,-64) and (4,-2,64) are both saddle points.

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.

Revenue = Units sold * price In this case

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

\[R(2.3, 4.1)=81\times 2.3+40\times 4.1+28\times 2.3\times 4.1-21\times (2.3)^2-23\times (4.1)^2 = 116.62\]

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?

Consider x + y = 96 where x is the number of units produced in Los Angeles and y is the number of units produced in Denver.

Replacing x = 96 - y

\[C(x, y) = C(96 - y, y) = \frac{1}{6} (96 - y)^2 + \frac{1}{6}y^2 + 7(96 - y) + 25y + 700\]

\[=\frac{1}{6} (96 - y)^2 + \frac{1}{6}y^2 + 7(96 - y) + 25y + 700\]

\[=\frac{1}{6} (y^2-192y+9216) + \frac{1}{6}y^2 + (672-7y) + 25y + 700\]

\[=\frac{1}{3} y^2 -14y + 2908\]

\[=C(y)\]

To find the minumum value, \(C^{'}(y) = 0\)

\[\frac {2}{3} y - 14 = 0\]

\[y = 21\]

Now replacing y value in x = 96 - y = 96 -21

\[x = 75\]

So to minimize the total weekly cost, the number of units produced in Los Angeles should be 75 and in Denver be 21.

5. Evaluate the double integral on the given region

\[\iint_R (e^{8x+3y})dA;R: 2 \leq x \leq and 2 \geq x \geq 4\]

Write your answer in exact form without decimals.

\[ \begin{split} \int_2^4\int_2^4 (e^{8x+3y})\ dy\ dx &= \int_2^4 (\frac{1}{3}e^{8x+3y})|_2^4\ dx\\ &= \int_2^4 ((\frac{1}{3}e^{8x+3*4})-(\frac{1}{3}e^{8x+3*2}))\ dx\\ &= \int_2^4 ((\frac{1}{3}e^{8x+12})-(\frac{1}{3}e^{8x+6}))\ dx\\ &= \int_2^4 \frac{1}{3}e^{8x+6}(e^6-1)\ dx\\ &= \frac{1}{3*8}e^{8x+6}(e^6-1) |_2^4\\ &= \frac{1}{24}e^{8*4+6}(e^6-1)-\frac{1}{24}e^{8*2+6}(e^6-1)\\ &= \frac{1}{24}(e^6-1)(e^{38}-e^{22})\\ &= \frac{1}{24}(e^{44} - e^{38} - e^{28} + e^{22}) \end{split} \]