Question 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 ).

#place in df
x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
df <- data.frame(x, y)
df
#lm/coefficent/slope
model <- lm(y ~ x, data = df)
model
## 
## Call:
## lm(formula = y ~ x, data = df)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257
#place lm/coefficent/slope into equation
coefficients <- coef(model)
intercept <- round(coefficients[1], 2)  
slope <- round(coefficients[2], 2)      

equation <- paste("y =", slope, "x +", intercept)
print(equation)
## [1] "y = 4.26 x + -14.8"
plot(df$x, df$y, main = "Plot of LM", xlab = "X", ylab = "Y", pch = 19, col = "blue")
abline(model, col = "lightblue", lwd = 2)

Question 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

\[f ( x, y ) = 24x - 6xy^2 - 8y ^3\] First we find the partial derivatives which are \[f(x) = 24−6y^2\]

\[f(y) = −12xy−24y^2\] Next set each derivative to zero to find the critical points: \[24−6y^2=0\] \[y^2=4\]

\[y = ± 2\]

\[−12xy−242=0\]

\[xy = 8\]

\[x = ± 4\] The critical points would be (-4, 2)(4, -2)

Question 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 ).
h = # of house brand units sold
n = # of name brand units sold
The revenue function is: \[R = h(81 - 21x + 17y) + n(40 + 11x - 23y)\]

Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?

\[R = h(81 - (21*2.30) + (17*4.10)) + n(40 + 11*2.30 - (23*4.10))\] \[2.30* (81 - 48.30 + 69.70) + 4.10 * (40+25.30-94.30)\] \[2.30* (102.4) + 4.10 * (-28) \] \[235.52 -114.90\] The revenue is:\[$120.62\]

Question 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?

We know: \[x+y=96\] \[x=96-y\] \[y=96-x\] Therefore \[C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}(96-x)^2 + 7x + 25(96-x) + 700\] \[C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}(96 - x^2) + 7x + 2400 - 25x + 700\] \[C(x, y) = \frac{1}{6}x^2 + 1536 - 32x - 18x + 3100\] \[C(x, y) = \frac{1}{3}x^2 - 50x + 4636\] \[\frac{dc}{dx} = \frac{2}{3}x - 50 = 0\] \[x = \frac{2}{3} * 50 = 75 \] So there are 75 units produced in LA and 21 (y=91-75=21) produced in Denver

5) Evaluate the double integral on the given region. \[ \iint_R { { e }^{ 8x+3y }dA } \\ 2\le x \le 4\\ 2 \le y \le 4 \]

\[ \iint _{ 2 }^{ 4 }{ { e }^{ 8x+3y }dxdy } \\ \iint _{ 2 }^{ 4 }{ { e }^{ 8x }{e}^{3y}dxdy } \\ \int _{ 2 }^{ 4 } ({e}^{3y}) \int _{ 2 }^{ 4 }(e^{8x})dx dy\\ \int _{ 2 }^{ 4 } ({e}^{3y})(\frac{e^{32}-e^{16}}{8})dy\\ \frac{e^{32}-e^{16}}{8}\int _{ 2 }^{ 4 } ({e}^{3y})dy\\ [ \frac{e^{32}-e^{16}}{8}][\frac{e^{12}-e^{6}}{3}]\\ e^{16}e^{6}(\frac{e^{16}-1}{8})(\frac{e^{2}-1}{3})\\ \frac{e^{16}e^{6}}{24}(e^{16}-1)(e^{2}-1) \]