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 )

df <- data.frame(X = c(5.6, 6.3, 7, 7.7, 8.4), Y = c(8.8, 12.4, 14.8, 18.2, 20.8))
md1 <- lm(Y ~ X, data = df)
s1 <- summary(md1)
s1
## 
## Call:
## lm(formula = Y ~ X, data = df)
## 
## 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-05

y = 4.26x -14.80

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.

\[f\left( x,y \right) =24x-6x{ y }^{ 2 }-8{ y }^{ 3 }\]

\[\frac { df }{ dx } =24-6{ y }^{ 2 }\]

\[\frac { df }{ dy } =-12xy-24{ y }^{ 2 }\]

\[if\_ y\Rightarrow 0=24-6{ y }^{ 2 }\Rightarrow { y }^{ 2 }=4\Rightarrow y\pm 2\]

\[if\_ y\Rightarrow \pm 2...\& ...-12xy-24{ y }^{ 2 }=0\Rightarrow \pm { 24 }x=24x4\Rightarrow x=\pm 4\]

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).

\[R(x,y) = (81 - 21x + 17y)x + (40 + 11x - 23y)y\]

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

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

x=2.30
y=4.10

s2 <- -21*x^2 +81*x +28*x*y +40*y -23*y^2
s2
## [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 \(C\left( x,y \right) =\frac { 1 }{ 6 } { x }^{ 2 }+\frac { 1 }{ 6 } { y }^{ 2 }+7x+25y+700\) 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?

\[ \begin{align} C(x,y) & = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700 & \text{ } \\ f(x) & = \frac{1}{6}x^2 + \frac{1}{6}(96-x)^2 + 7x + 25(96-x) + 700 & \text{substitute y=96-x} \\ f'(x)& = \frac{1}{6}2x + \frac{1}{6}2(96-x)(-1) + 7 + 25(-1) & \\ f'(x)& = \frac{1}{3}x - \frac{1}{3}(96-x) - 18 & \\ f'(x)& = \frac{2}{3}x - 32 - 18 & \\ f'(x)& = \frac{2}{3}x - 50 & \\ f''(x) &= \frac{2}{3} > 0 \end{align} \]

\(f'(x) = 0\) -> \(x = 75\)

Values of x:

\[0 \leq x \leq 96\]

\[y= 96-x = 21\]

\[y = 75; y = 21\]

5. Evaluate the double integral on the given region.

\[\int \int_{R} e^{8x + 3y} dA \text{ where } R:=\bigg \lbrace 2 \leq x \leq 4; 2 \leq y \leq 4\bigg \rbrace\]

Write your answer in exact form without decimals.

\[ \begin{align} \int \int_{R} e^{8x + 3y} dA & = & \bigg(\int_{2}^{4} e^{8x}dx \bigg)\bigg( \int_{2}^{4}e^{3y}dy \bigg) \\ & = & \bigg( \frac{1}{8}e^{8x} \bigg \lvert_{x=2}^{4}\bigg) \bigg( \frac{1}{3} e^{3y} \bigg \lvert_{y=2}^{4} \bigg) \\ & = & \bigg( \frac{ e^{32} - e^{16}}{8} \bigg) \bigg( \frac{ e^{12} - e^{6} }{ 3} \bigg) \\ & = & \frac{1}{24}(e^{32}-e^{16})(e^{12}- e^{6}) \end{align} \]