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 )

Solution

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

reg_line <- lm(y_values ~ x_values)
reg_line
## 
## Call:
## lm(formula = y_values ~ x_values)
## 
## Coefficients:
## (Intercept)     x_values  
##     -14.800        4.257

Therefore, the regression line of the linear regression model is: \(y=−14.80 + 4.26x\)

Plot of the Regression line

plot(x_values, y_values, xlab="x", ylab="y")
abline(reg_line)
lines(c(5,9), -14.8+4.257*c(5,9), col="blue")

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

Solution

Partial derivatives:

\(f_x(x,y)=24−6y^2\)
\(f_y(x,y)=−12xy−24y^2\)

If \(24−6y2=0\)
\(y^2=4\ and\ y=±2\)

If \(y=2\) and \(−12xy−24y^2=0\),
then \(−24x=24×4\) hence \(x=−4\)
If \(y=2\) and \(−12xy−24y^2=0\),
then \(24x=24×4\) and \(x=4\)

f(x,y):

\(f_{(4,−2)}=24×4−6×4×(−2)2−8×(−2)3=64\)

\(f_{(4,−2)}=24×(−4)−6×(−4)×22−8×23=−64\)

Hence we have two critical points: \((4,−2,64)\) and \((−4,2,−64)\)

To determine if points are minimum, maximum or saddle, we use Second Derivative test:

Taking second partial derivatives:

\(f_{xx}=0\)
\(f_{yy}=−12x−48y\)
\(f_{xy}=−12y\)

And \(D(x,y)=f_{xx}f_{yy}−f_{xy}^2=−(−12y)^2=−144y^2\)

\(D(x,y)<0\) for all (x,y). As a result, according to the Second Derivative Test, the above two critical point are saddle points.

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.

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

Solution

\(R_{(x,y)} = Revenue\ from\ house\ brand + Revenue\ from\ name\ brand\)
\(R_{(x,y)} = (81- 21x + 17y)x + (40 + 11x - 23y)y\)
\(R_{(x,y)} = - 21x^2- 23y^2 + 28yx + 81x + 40y\)

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

Solution

compute_revenue <- function(x,y)
  {
   rev_func = - 21*x^2 - 23*y^2  + 28*y*x + 81*x + 40*y  
   return(c(rev_func))
  }
print(paste0("Revenue = $", compute_revenue(2.30,4.10)))
## [1] "Revenue = $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?

Solution

Cost function: \(C_{(x, y)} = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700\) \

Number of units: \[ x+y=96\\ x=96-y \]

We can now convert \(C_{(x, y)}\) into a univariate function:
\[ C_{(x, y)}=\frac{1}{6}x^{2}+\frac{1}{6}y^{2}+7x+25y+700\\ =\frac{1}{6}(96-y)^{2}+\frac{1}{6}y^{2}+7(96-y)+25y+700\\ =\frac{1}{6}(1536-192y-y^2)+\frac{1}{6}y^{2}+7(96-y)+25y+700\\ C(y)=\frac{1}{3}y^{2}-14y+2908 \]

Optimize (Derivative equals zero) \[ C'(y)=\frac{2}{3}y-14=0\\ \frac{2}{3}y=14\\ 2y=42\\ y=21 \]

Now we have: \(x=96-y=96-21=75\)
Hence, 21 units produced in the Denver plant and 75 units in the Los Angeles plant.

Question 5

Evaluate the double integral on the given region.

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

Solution

\[ \iint_{2}^{4}{{e}^{8x+3y}dxdy} \\ \iint_{ 2 }^{ 4 }{ { e }^{ 8x }{e}^{3y}dxdy } \ \iint_{ 2 }^{ 4 } ({e}^{3y}) \int _{ 2 }^{ 4 }(e^{8x})dx dy \\ \iint_{ 2 }^{ 4 } ({e}^{3y})(\frac{e^{32}-e^{16}}{8})dy \\ \frac{e^{32}-e^{16}}{8}\int _{ 2 }^{ 4 } ({e}^{3y})dy \\ [ \frac{e^{32}-e^{16}}{8}][\frac{e^{12}-e^{6}}{3}] \\ \frac{e^{44}-e^{38}-(e^{28}-e^{22})}{8*3} \\ \frac{e^{44}-e^{38}-e^{28}+e^{22}}{24} \\ \frac{1}{24}(e^{44}-e^{38}-e^{28}+e^{22}) \\ \frac{1}{24}(e^{44-38-28+22}) \\ \frac{1}{24}(e^{0}) \\ \frac{1}{24}(1) \\ \frac{1}{24} \]