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 )
regdata <- 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)
)

print(regdata)
##     x    y
## 1 5.6  8.8
## 2 6.3 12.4
## 3 7.0 14.8
## 4 7.7 18.2
## 5 8.4 20.8
model <- lm(y ~ x, data = regdata)
model
## 
## Call:
## lm(formula = y ~ x, data = regdata)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257
summary(model)
## 
## Call:
## lm(formula = y ~ x, data = regdata)
## 
## 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
plot(regdata$x,regdata$y)
abline(model)

Regression equation:

y=-14.80+4.247x

  1. Find all local maxima, local minima, and saddle points for the function given below. Write your answer(s) in the formv( x, y, z ). Separate multiple points with a comma. \(f ( x, y ) = 24x - 6xy^2 - 8y^3\) \[f_{x}(x,y)=24-6y^2\] \[f_{y}(x,y)=-12xy-24^2\] Set =0 and solve: \[24-6y^2=0\] \[y=+/-2\]

\[-12xy-24^2=0\] (4,-2) (-2,2)

  1. 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 ).

“house” cost = x “name” cost=y

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

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

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

-21*(2.3^2)+(81*2.3)+(28*2.3*4.1)+(40*4.1)-(23*4.1^2)
## [1] 116.62
  1. 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 \[C(x, y) = 1/6 x^2 + 1/6 y^2 + 7x + 25y + 700\] 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? \[C(x, y) = 1/6 x^2 + 1/6 y^2 + 7x + 25y + 700\] \[x+y=96\] \[y=96-x\] \[c(x,y)=1/6x^2+1/6*(96-x)^2+7x+25(96-x)+700\]

\[C(x,y)=1/6x^2+1/6((96^2)-192x+x^2)+7x+25*96-25x+700\] \[C_{x}(x,y)=1/2x-32+2x+7-25=0\] \[2.5=50\] \[x=20\] \[y=76\]

  1. Evaluate the double integral on the given region. \[\iint(e^{8x+3y})dA\] \[R: 2 \le x \le 4\quad and \quad2 \le x \le 4\]

Write your answer in exact form without decimals. \[\iint(e^{8x+3y})dxdy\] \[\int1/8(e^{8x+3y})dy\] eval at 2,4 \[1/8\int[e^{32+3y}-e^{16+3y}]dy\] \[1/8*1/3[e^{32+12}-1/3e^{16+12}-e^{32+6}+1/3e^{16+6}]\]