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 )

x1 <- c(5.6, 6.3, 7, 7.7, 8.4)
y1 <- c(8.8, 12.4, 14.8, 18.2, 20.8)

lm_m1 <- lm(x1 ~ y1)
print(lm_m1)
## 
## Call:
## lm(formula = x1 ~ y1)
## 
## Coefficients:
## (Intercept)           y1  
##      3.4890       0.2341
lm_m2 <- lm(y1 ~ x1)
print(lm_m2)
## 
## Call:
## lm(formula = y1 ~ x1)
## 
## Coefficients:
## (Intercept)           x1  
##     -14.800        4.257
dset = matrix(c(5.6,8.8,6.3,12.4,7,14.8,7.7,18.2,8.4,20.8),ncol=2,byrow=TRUE)
colnames(dset) = c("X","Y")
dset = data.frame(dset)
dset$XY<- dset$X *dset$Y
dset$XY<- dset$X *dset$Y
dset$Xsq<- dset$X^2
dset$Ysq<- dset$Y^2

kable(dset)
X Y XY Xsq Ysq
5.6 8.8 49.28 31.36 77.44
6.3 12.4 78.12 39.69 153.76
7.0 14.8 103.60 49.00 219.04
7.7 18.2 140.14 59.29 331.24
8.4 20.8 174.72 70.56 432.64 
#summary of columns
dsum <- (colSums(dset[,]))
dsum
##       X       Y      XY     Xsq     Ysq 
##   35.00   75.00  545.86  249.90 1214.12
x  <-dsum[1]
y  <-dsum[2]
xy <-dsum[3]
xsq<-dsum[4]
ysq<-dsum[5]
#(sum(y) * sum(x^2)) - ((sum(x) * sum(xy)) / 5(sum(x^2)- sum(x)^2)
a1=( (y * xsq ) - (x * xy) )/ (5* (xsq - x^2))

#(5(sum(xy) -(sum(x) * sum(y)))/(5(sum(x^2) - sum(x)^2)

b1 = ( (5*xy - (x*y) )/(5*(xsq - x^2)) )

a1
##          Y 
## 0.07437186
b1
##          XY 
## -0.02139268
#y = a + b*x

\[Y = 0.07 - 0.02 * X\]

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.

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

compute the values of \[ \frac {df}{dx} and \frac{df}{dy} \]

\[ \frac {df}{dx} = \frac {d}{dx} (24x -6xy^2 -8y^3)\]

\[ = 24 -6y^2 \]

\[ \frac {df}{dy} = \frac {d}{dy} (24x -6xy^2 -8y^3) \]

\[ = -12xy -24y^2\]

To find the critical points, equate \[ \frac {df}{dx} and \frac{df}{dy} \] to zeros:

\[ 24 - 6y^2 = 0 \]

\[ 24 = 6y^2 \]

\[ 4 = y^2 \]

\[ y = 2, -2 \]

\[ -12xy -24y^2 = 0 \]

\[ -12xy = 24y^2 \]

\[ -xy = 2y^2 \]

\[ -x = 2y \]

so the critical points are \[ (-4,2) and (4,-2) \]

According to the second derivative test, the value of the discriminant is

\[ D=f_{xx}f_{yy} - f``_{xy} \]

And the second derivate test asserts the following:

  1. If D(a,b) > 0 and f_{xx}(a,b) >0 then (a,b) is a local minimum of f.

  2. If D(a,b) > 0 and f_{xx}(a,b) <0 then (a,b) is a local maximum of f.

  3. If D(a,b) < 0 then (a,b) is a saddle point of f.

  4. If D(a,b) > 0 then this test is inconclusive.

So, calculate the partial derivatives as below:

\[ f_{xx} = \frac {d}{dx}(f_x) = \frac {d}{dx}(24 -6y^2) = 0 \]

\[ f_{yy} = \frac {d}{dy}(f_y) = \frac {d}{dy}(-12xy -24y^2) = -12x -48y \]

\[ f_{xy} = \frac {d}{dx}(f_y) = \frac {d}{dx}(-12xy -24y^2) = -12y \]

so, the value of the discriminant is

D=(0)(-12x-48Y) -(-12Y)^2

\[D =-144Y^2\] At the critical point \[ (-4,2) and (4,-2) \] The discriminant is

\[ D = -144*4 = -576 \] Since D < 0 the critical point is saddle point of f..

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 ).

Total revenue is given by

R = Revenue from brand X + Revenue from brand Y

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

\[=81x - 21x^2 + 17xy + 40y + 11xy -23y^2\]

\[ = -21x^2 -23y^2 + 28xy + 81x +40y \] now find out partial derivatives

\[\frac {dR}{dx} = -42x + 28y + 81 \]

\[\frac {dR}{dy} = -46y + 28x + 40 \]

So critical points are roots of following equations

-42x + 28y + 81 = 0

28x - 46y + 40 = 0

y= -2.4, x = 0.32

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

Rev2= -21*(2.30)^2 -23*(4.10)^2 + 28*(2.3)*(4.1) + 81*(2.3) + 40*(4.1) 

Rev2
## [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

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

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?

\[ x + y =96 \] \[ C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700 \] \[ C`(X) = \frac{1}{3}x + 7 \] \[ 0 = \frac{1}{3}x + 7 \]

\[ -7 = \frac{1}{3}x \] \[ -21 = x \] \[ C`(Y) = \frac{1}{3}y + 25 \]

\[ 0 = \frac{1}{3}y + 25 \] \[ - 25 = \frac{1}{3}y \]

\[ -75 = y \] (x,y) = (75,21) are units to be produced in each of the units.

5. Evaluate the double integral on the given region.

The double integral of the given region can be represented as below.

\[\int\int(e^{8x + 3y})DA: R:2<x<=4 and 2<y<=4 \]

\[\int_{2}^{4}\int_{2}^{4}(e^{8x + 3y})DA \]

\[\int_{2}^{4}\int_{2}^{4}(e^{8x + 3y}) \frac {3dy}{3}.dx \]

\[=\frac{1}{3}\int_{2}^{4}\bigg[(e^{8x + 3y})\bigg]_{2}^{4}. dx\]

\[=\frac{1}{3}\int_{2}^{4} (e^{8x+12} -e^{8x +6} )dx \]

\[=\frac{1}{3}\int_{2}^{4} (e^{8x+12} -e^{8x +6} ). \frac{8}{8}dx \]

\[=\frac{1}{24} \bigg[ (e^{8x+12} -e^{8x +6} ) \bigg] _{2}^{4}\]

\[=\frac{1}{24} \bigg[ e^{44} -e^{38} -(e^{28} - e^{22}) \bigg] \]

\[=\frac{1}{24} \bigg[ e^{44} -e^{38} -e^{28} + e^{22}) \bigg] \]

\[=\frac{1}{24} . 1 \]