1

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

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

Function is \(y = -14.8 + 4.257(x)\)

x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
points <- data.frame(x,y)

reg <- lm(y ~ x)
reg
## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257
ggplot(points, aes(x =x, y=y)) +
  geom_point(color = "blue") +
  geom_abline(slope = 4.257, intercept = -14.8, colour = "red")

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.

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

Partial Derivatives

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

\(24 - 6y^2 = 0\)

\(6y^2 = 24\)

\(y^2 = 4\)

\(y = (2, -2)\)

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

\(-12xy - 24y^2 = 0\)

When \(y=2, x=-4\)

When \(y = -2, x = 4\)

Critical Points are (-4, 2), (4, -2)

Second Derivatives

\(f_{xx} = 0\)

\(f_{yy} = -12 - 48y\)

\(f_{xy} = -12y\)

\(D(x, y) = f_{xx}f_{yy} - f_{xy}^2\)

\(D(-4,2) = (0)(84) - (24)^2 = -576\)

\(D(4, 2) = (0)(-108) - (-24)^2 = -576\)

Both critical points are saddle points

x <- c(4, -4)
y <- c(-2, 2)
df <- data.frame(x, y)

df$z <- 24*df$x - 6*df$y*(df$x^2) - 8*df$y^3

print(paste0("(", df$x, ",", df$y, ",", df$z, ")"))
## [1] "(4,-2,352)"  "(-4,2,-352)"

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.

Find the revenue function R ( x, y ).

Revenue = \(x(81 - 21x + 17y) + y(40 + 11x - 23y)\)

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

Partial Derivatives

\(f_x(x,y) = 81 - 42x +28y\)

\(f_y(x,y) = 40 + 28x - 46y\)

Find Critical Points

\(28x - 46y = -40\)

\(-42x + 28y = -81\)

m <- matrix(c( -42, 28, 28, -46), nrow = 2, byrow = T)
v <- c(-81, -40)


roots <- solve(m, v)

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

print(paste0("x = ", roots[1]))
## [1] "x = 4.2212543554007"
print(paste0("y = ", roots[2]))
## [1] "y = 3.4390243902439"

Function = \(R(x, y) = 50.86x + 7.3y\)

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

\(R = 50.86(2.3) + 7.3(4.1)\)

\(R = 145\)

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?

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

\(f_x(x,y) = \frac{1}{3}x + 7\)

\(f_y(x,y) = \frac{1}{3}x + 25\)

\(f_{xx} = \frac{1}{3}\)

\(f_{yy} = \frac{1}{3}\)

\(f_{xy} = 0\)

\(D = 175\)

Since 175 is greater than 1/3, these are the relative maximum of C(x,y)

\(x = 21\)

\(y = 75\)

The minimum cost seems to be the opposite. I’m not sure why. I’m missing something.

\(x = 75\)

\(y = 21\)

5

Evaluate the double integral on the given region.

\[\int \int (e^{8x+3y})dA ; \ R: 2\geq x \geq4 \ and\ 2 \geq y \geq 4\] \(\int^4_2 \int^4_2 (e^{8x+3y})dxdy\)

\(\int^4_2 \frac{1}{8}e^(3y+8x) |^4_2 dx = \int^4_2 \frac{1}{8}e^(3y+32)-e^(3y+16)\)

\(\frac{1}{3} \frac{1}{8}e^(3y+32)-e^(3y+16)|^4_2 dy = (\frac{e^{44}}{24}-\frac{e^{28}}{24})-(\frac{e^{38}}{24}-\frac{e^{22}}{24})\)

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