Question 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)

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

xy.lm <- lm(y ~ x, data = df)
xy.lm
## 
## Call:
## lm(formula = y ~ x, data = df)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257
plot(df$x, df$y)
abline(xy.lm, col="blue")

The equation of the line is:

\[y = -14.8 + 4.257 \times x\]

Question 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.

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

\[ f_x(x,y) = 24-6y^2\\ f_y(x,y) = -12xy - 24y^2 \]

\[\begin{aligned} 24 - 6y^2 = 0\\ 4(4-y^2) = 0\\ y^2 = 4\\ y = \pm 2\\\\\\ -12x(-2 )- 24(-2)^2 = 0\\ 24x - 96 = 0\\ x = 4\\\\ -12x(2) - 24(2)^2 = 0\\ -24x - 96 = 0\\ x = -4 \end{aligned}\]

Therefore the critical points are (4, -2) and (-4, 2)

We must then use the Second Derivave Test to obtain whether equation has local minima, maxima and saddle points.

\[f_{_{xx}}(x,y) = 0\\ f_{_{yy}}(x,y) = -12x - 48y\\ f_{_{xy}}(x,y) = -12y\\\\\]

\[D = f_{_{xx}}(x,y).f_{_{yy}}(x,y)-[f_{_{xy}}(x,y)]^2\\ = 0 (-12x - 48y) - (12y)^2\\ =-144y^2\\\\\\\]

\[D(4,-2) = -576 < 0 \therefore (4,-2) \ is \ a \ saddle \ point\\ D(-4,2) = -576 < 0 \therefore (-4,2) \ is \ a \ saddle \ point\]

Question 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. 3.

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

To find pending revenue we multiply the amount of units to sell times the dollar amount.

Therefore the revenue function is like so:

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

Simplfied we get:

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

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

\[-21(2.30)^2 -23(4.10)^2 + 28(2.30)(4.10) + 81(2.30) + 40(4.10)\\ -111.09 - 386.63 + 264.04 + 186.3 + 164\\ =116.62\]

Question 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?

\[f_x(x,y) = \frac{1}{3}x + 7\\ \frac{1}{3}x + 7 =0\\ x = -21\\\]

\[f_y(x,y) = \frac{1}{3}y + 25\\ \frac{1}{3}y + 25 = 0\\ y = -75\]

Using the absolute values of x and y, the amount of units for minimum cost weekly is 21 and 75 respectively.

Question 5

Evaluate the double integral on the given region.

\[ \iint_R (e^{8x + 3y})dA ; R:2 \le x \le 4 \ and \ 2 \le y \le 4\]

Write your answer in exact form without decimals.

\[\iint_R (e^{8x + 3y})dA=\int_2^4 \int_2^4 e^{8x + 3y} \\ = \int_2^4 \bigg[\frac{1}{8} e^{8x + 3y}\bigg]_2^4\\ = \int_2^4 \frac{1}{8} \bigg( e^{32 + 3y}- e^{16 + 3y}\bigg)\\ = \frac{1}{8} \int_2^4 e^{16}- 1 \bigg(e^{16 + 3y}\bigg)\\ = \frac{1}{8} \bigg(e^{16} - 1 \bigg) \int_2^4 e^{16 + 3y}\\ = \frac{1}{8} \bigg(e^{16} - 1 \bigg).\frac{1}{3} \bigg[ e^{16 + 3y}\bigg]_2^4\\ = \frac{1}{24} e^{16} - 1 \bigg(e^{16 + 12} - e^{16 + 6}\bigg)\\ = \frac{1}{24} e^{16} - 1 \bigg(e^{28} - e^{22}\bigg)\\ = \frac{1}{24} \bigg(e^{44} - e^{38} - e^{28} + e^{22}\bigg)\]