1. Find the equation of the regression line for the given points. Round any final values to the nearest hundredth, if necessary.

\[ ( 5.6, 8.8 ), ( 6.3, 12.4 ), ( 7, 14.8 ), ( 7.7, 18.2 ), ( 8.4, 20.8 ) \]

Sol:

x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
(regression <- lm(y~x))
## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257
# plot
plot(x,y)
abline(regression)
lines(c(5,9), -14.8+4.257*c(5,9), col="yellow")

Based on the linear regression model, the regression line is \(y=−14.8+4.257x\).

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.

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

Sol:

\(f(x,y)=24x−6xy2−8y3\)

\(fx=24−6y2\)

\(fy=−12xy−24y2\)

\(24−6y2=0\) & \(−12xy−24y2=0\)

\(y2= 4\) & \(x=2y\)

\(y= ±2\) & \(x= ∓4\)

crit_finder <- function(x,y){
   z = 24*x-6*x*y^2-8*y^3
   return(c(x,y,z))
 }
crit_finder(-4,2)
## [1]  -4   2 -64
crit_finder(4,-2)
## [1]  4 -2 64
min_max <- function(x,y){
  D = (24 - 6*y^2)*(-12*x*y - 24*y^2) - 144*y^2
  if(D>0){
    return("maximum or minimum")
  }
  if(D<0){
    return("Saddle point")
  }
  else{
    return("Inconclusive")
  }
}
min_max(-4,2)
## [1] "Saddle point"
min_max(4,-2)
## [1] "Saddle point"

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.

Sol:

If the manager sells the “house” brand for x dollars, she is able to sell 81−21x+17y.

If the manager sells the “name” brand for y dollars, she is able to sell 40+11x−23y

Revenue = price X # of units

Therefore we can write the following:

\(R1(x)=x∗(81−21x+17y)=(81x−21x2+17xy)\)

\(R2(y)=y∗(40+11x−23y)=(40y+11xy−23y2)\)

Step 1. Find the revenue function R ( x, y )

Sol:

Now the total Revenue function R(x,y) would be:

\(R(x,y)=81x−21x2+17xy+40y+11xy−23y2\)

\(R(x,y)=−21x2+28xy−23y2+81x+40y\)

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

Sol:

x = 2.3
y = 4.1
(total <- -21*x^2 - 23*y^2 + 28*x*y + 81*x + 40*y)
## [1] 116.62

The total revenue if the manager sells the “house” brand for $2.30 and “name” brand for $4.10 is $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) = 16 x 2 +16 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?

Sol:

The total weekly cost is given by the following function:

\(C(x,y)=16x2+16y2+7x+25y+700\)

where x is the number of units produced in Los Angeles and y is the number of units produced in Denver.

Since the company is committed to produce a total of 96 units of a product each week, x+y=96

x=96−y

\(C(96−y,y)=16(96−y)2+16y2+7(96−y)+25y+700\)

\(C(96−y,y)=16(y2−192y+9216)+16y2+18y+1372\)

\(C(96−y,y)=13y2−14y+2908\)

To find the minimum value, we need to differentiate the function C(x,y) and equate it to 0.

\(C′(96−y,y)=ddy(13y2−14y+2908)=23y−14\)

\(23y−14=0\)

\(y=21\)

Substituting the value of y in the equation, we get x=96−y

\(x=96−21\)

\(x=75\)

Therefore, the company needs to product 75 units in Los Angeles plant and 21 units in Denver plant to minimize the total weekly cost.

5. Evaluate the double integral on the given region. Write your answer in exact form without decimals.

\[ ∫∫R(e8x+3y)dA,R:2≤x≤4 and 2≤y≤4 \]

Sol:

\[∫lim(4,2)∫lim(4,2)(e8x+3y) dy dx=∫lim(4,2)(13e8x+3y)|42dx\]

\[=∫lim(4,2) ((13e8x+12)−(13e8x+6))dx\]

\[=∫lim(4,2)(13e8x+6(e6−1))dx\]

\[=1/24(e8x+6(e6−1))|42\]

\[=1/24(e32+6(e6−1)−1/24(e16+6(e6−1)\]

\[=1/24(e6−1)(e38−e22)\]

\[=1/24(e44−e38−e28+e22)\]

(result <- 1/24*(exp(44)-exp(38)-exp(28)+exp(22)))
## [1] 5.341559e+17