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 )
datadf <- data.frame(xvalues=c(5.6, 6.3, 7, 7.7, 8.4), yvalues=c(8.8, 12.4, 14.8, 18.2, 20.8))
print (datadf)
##   xvalues yvalues
## 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
library(ggplot2)
ggplot(data=datadf, aes(x=xvalues, y=yvalues)) + geom_point()

m1 = lm(yvalues ~ xvalues, data=datadf)
summary(m1)
## 
## Call:
## lm(formula = yvalues ~ xvalues, data = datadf)
## 
## 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 ***
## xvalues       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
ggplot(datadf, aes(xvalues, yvalues)) + geom_point(colour="blue", size=2) + 
    geom_abline(aes(slope=round(m1$coefficients[2], 2), intercept=round(m1$coefficients[1], 2))) +
    xlab("xvalues") + ylab("yvalues") + labs(title = "xvalues vs. yvalues")

  1. 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

Partial Derivative

f_x=24−6y^2 f_y=−12xy−24y^2

Critical Points Set f_x=0

24−6y^2=0 24=6y^2 4=y^2 ±2=y

Substitute y-values into f_y=0

−12xy−24y^2=0 xy+2y^2=0 Substitute y=2

x(2)+2(2)^2=0

2x+8=0 2x=−8 x=−4

Substitute y=-2

x(−2)+2(−2)^2=0

−2x+8=0 −2x=−8 x=4 Substitute (x,y) values for z

f(x,y)=24x−6xy2−8y3

Substitute (-4,2)

z=24(−4)−6(−4)(2)2−8(2)3

z=−96+96−64 z=−64 Substitute (4,-2)

z=24(4)−6(4)(−2)2−8(−2)3 z=96−96+64 z=64 Critical points are (-4,2,-64) and (4,-2,64)

fxx=0 fyy=−12x−48y fxy=−12y

Plug into formula

D=f_xx(x,y)∗f_yy(x,y)−f_xy^2(x,y)

D=0(x,y)∗(−12x−48y(x,y))−(−12y)^2(x,y) D=0−144y^2(x,y) D=−144y^2(x,y) f we plug in y=2 or y=-2, we get D<0 concluding critical points (-4,2,-64) and (4,-2,64) are both saddle points. 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 ).

House brand: R(x)=x∗(81−21x+17y)

Name brand: R(y)=y∗(40+11x−23y)

Total = $R(x,y)=x∗(81−21x+17y)+y∗(40+11x−23y) -> - 21x^2 - 23y^2 + 28xy + 81x + 40y $

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

x = 2.3
y = 4.1
total <- -21*x^2 - 23*y^2 + 28*x*y + 81*x + 40*y

print(total)
## [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) = 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?

Solution Consider x+y=96, then x=96−y.

C(x,y)=C(96−y,y)=

= 1/6x2+1/6y2+7x+25y+700

= 1/6(96−y)2+16y2+7×(96−y)+25y+700

= 1/6(y2−192y+9216)+1/6y2+672−7y+25y+700

= 1/6(y^2−32y+1536+1/6y2+18y+1372

= 13y^2−14y+2908

= C1(y)

C′1(y) = 2/3y−14

Finding minimal value by considering C′1(y)=2/3y−14=0, then y=21. Then x=96−y=75.

There will be production of 75 units in Los Angeles and 21 units in Denver.

  1. Evaluate the double integral on the given region.

    _R (e^{8x+3y}) dA ; R: 2 £ x £ 4 and 2 £ y £ 4

Write your answer in exact form without decimals.

Answer: _24_24 (e^{8x+3y}) dy dx &= _2^4 (e^{8x+3y})|_2^4 dx

&= _2^4 ((e{8x+12})-(e{8x+6})) dx

&= _2^4 e{8x+6}(e6-1) dx

&= e{8x+6}(e6-1) |_2^4

&= e{32+6}(e6-1)-e{16+6}(e6-1)

&= (e6-1)(e{38}-e^{22})

&= (e^{44} - e^{38} - e^{28} + e^{22})