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 )


My best fit was


\[x^{1.7} - 2x\]

R linear fit is


\[0.234x + 3.48\]


Below Im comparing my equation to R’s Estimate Cefficients and lm() and nls().

a<-matrix(c(5.6,8.8,6.3,12.4,7,14.8,7.7,18.2,8.4,20.8), byrow = TRUE, ncol = 2)
a_df<-as.data.frame(a)
colnames(a_df) <- c("x","y")

s1<-summary(lm(a_df$y~a_df$x))
a2_df<-cbind(a_df, 0, 0)
colnames(a2_df) <- c("x","y", "y_estimate", "y_reg" )


# y_lm is R's
#  y_reg is mine
for (i in 1:nrow(a2_df))
{
  a2_df[i,3]<-round(s1$coefficients[2,1]*a2_df[i,1] + s1$coefficients[1,1],2)
  a2_df[i,4]<-1 + round(a2_df[i,1]^1.7 - a2_df[i,1] * 2,2 )
}


par(mfrow=c(1,2))


plot(a_df$x, a2_df$y, main = "Original. Estimate. lm()",
     col = "blue", 
     xlab ="x",
     ylab="y" ,type='b')

lines(a_df$x, a2_df$y_estimate, col="red", lty=1, lwd=1 )
abline(lm(a_df$y~a_df$x), col="green", lty=3, lwd=4 ) # non-linear best fit


legend("topleft", legend = c("Original", "Estimate", "lm()"), col = c("blue", "green", "red"),   lty = 1)


r_nls <- nls(a_df$y~a_df$x^power, data = a_df, start = list(power = 1), trace = F)
y_nls<-predict(r_nls,newdata = a_df$x )

plot(a_df$x, a2_df$y, main = "Original. Mine. nls()",
     col = "blue", 
     xlab ="x",
     ylab="y" ,type='b')


lines(a_df$x, a2_df$y_reg, col="red")
lines(a_df$x, y_nls, col="green")

legend("topleft",                                       # Add legend to plot
       legend = c("Original", "Mine", "nls()"),
       col = c("blue", "red", "green"),
       lty = 1)


Compare lm() to mine…


library(kableExtra)

kable(a2_df , caption="Linear Regressions",row.names = FALSE, booktabs=TRUE, table.attr = "style='width:80%;'") %>%
  kable_styling(font_size = 12)
Linear Regressions
x y y_estimate y_reg
5.6 8.8 9.04 8.50
6.3 12.4 12.02 11.25
7.0 14.8 15.00 14.33
7.7 18.2 17.98 17.74
8.4 20.8 20.96 21.46



  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.

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


Create the function. Calculate the partial derivatives (set x and y to constants)


library(Deriv)

f1<-function(x,y) {
  24*x -6 * x *y^2 + -8*y^3
}


f1_prime<-Deriv(f1) # not clear if you can parse this into seperate functions so i did it manually below

fx<-function(y) 24 - 6 * y^2
fy<-function(x,y) -12 * x * y - 24 * y^2


The derivatives are :


\[f_x\prime(x,y) \ = \ 24-6y^2\]

\[f_y\prime(x,y) \ = \ -24y^2 -12xy\]


Its clear that the roots pf \(f_x\prime()\) would be y=2 and y=-2


uniroot(fx,c(0,10))$root

polyroot(c(24,0,-6))
## [1] 1.99998
## [1]  2+0i -2+0i


Its also clear that (0,0) would be a root of \(f_y\prime()\) It seems like the roots \(f_y\prime()\) are infinite. Im guessing we would ignore infinite roots.


library(rootSolve)
multiroot(fy, c(0, 0), parms = c(0,0))
## $root
## [1] 0 0
## 
## $f.root
## [1] 0 0
## 
## $iter
## [1] 1
## 
## $estim.precis
## [1] 0


Points are (0,2) and (0,-2), Now multiply the 2 second derivatives and plug in those points.


\[f_x\prime\prime(x,y) \ = \ 0\]

\[f_y\prime\prime(x,y) \ = \ -48y -12x\]


Since one of our second derivatives equals zero ( which is a given since the maximum degree of x was 1), our test is inconclusive.



Animation with Theta 10 to Theta 180 



library(animation)


x = seq(-10, 10)
y = seq(-10, 10)
z <- outer(x, y, f1)


saveGIF({
ani.options(interval = 0.2, nmax = 50)

  
for (i in 1:180) {
 theta<-10*i
 persp(x,y, z, theta = theta, phi = 0, expand = 0.5, col = "lightblue", main="24x-6xy^2-8y^3",axes=T, ticktype="detailed")
 ani.pause() 
 Sys.sleep(1)
} }, movie.name ="persp.gif", 
 ani.width = 600, ani.height = 600)




  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 )



\[H \ = \ 81-21x+17y\]


\[N \ = \ 40+-11x-23y\]


# returns units sold
fx_units<-function(x,y) 81-21*x+17*y
fy_units<-function(x,y) 40 + 11*x -23*y

# ((81-21*x+17*y)*x)  + ((40 + 11*x -23*y)*y)

fxfy_rev<-function(x,y) 81*x-21*x^2  +  28*x*y +  40 * y    -   23*y^2

Deriv(fxfy_rev)
## function (x, y) 
## c(x = 28 * y + 81 - 42 * x, y = 28 * x + 40 - 46 * y)

\[Total Revenue = \frac{\partial(rev)}{\partial(x)} \ = \ 28y + 81 - 42x \ = 0\]


\[\frac{\partial(rev)}{\partial(x)} \ = \ 28x + 40 - 46y \ = 0\]
Isolate y


\[y =\frac{14x+20}{23}\]


Plug in expression to y and solve for x.

\[\ 28\frac{14x+20}{23} + 81 - 42x \ = 0\]


x=4.22 y=3.44


Now invoke our function to get total revenue.


fxfy_rev(4.22,3.44)
## [1] 239.7412


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

fxfy_rev(2.30,4.10)
## [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


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


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?


Note we can already isolate x and y.

\[x=96-y\]

The derivative roots are less than zero.


fxy<-function(x,y) x^2/6 + y^2/6 + 7*x + 25*y + 700

fxy_prime<-Deriv(fxy)
fxy_prime
## function (x, y) 
## c(x = 0.333333333333333 * x + 7, y = 0.333333333333333 * y + 
##     25)

\[f_x\prime(x,y) \ = \ \frac{x}{3} - 7 = 0\]


\[f_y\prime(x,y) \ = \ \frac{y}{3} - 25 = 0\]


Lowest cost : Los Angeles (x) 21. Denver (y) 75.


fxy(21,75)
## [1] 3733




  1. Evaluate the double integral on the given region.

\[\int_2^4 \int_2^4 ( e^{8x+3y}) \ dy \ dx\]


Evaluating dy dx…


  1. Integrate w.r.t y. Hold x constant


\[\int_2^4 ( e^{8x+3y}) \ dy \ = \ \frac{e^{8x+3y}}{3} \ = \ \frac{1}{3} \int_2^4 ( e^{8x+3y}) \]


  1. Plug bounds into y and subtract upper bounds from the lower bounds


\[ = \frac{1}{3} \ e^{8x+12} \ - \ e^{8x+6} \]


  1. Integrate w.r.t x


\[ \frac{1}{3} \int_2^4 e^{8x+12} \ - \ e^{8x+6} \ = \frac{1}{24} \int_2^4 e^{8x+12} - e^{8x+6}\]


  1. Plug bounds into x and subtract upper bounds from the lower bounds


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

e<-exp(1)

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