Question 1

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

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

Answer:

x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
model <- lm(y~x)
model
## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257

According to the above, the equation of the regression line is: \[y = -14.80 + 4.26x\]

If we plot this, we’ll get:

library(ggplot2)
ggplot(model, aes(x, y)) + 
  geom_point() + stat_smooth(method="lm") + 
  labs(title="Regression Line: y = 4.26x - 14.80")

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.

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

Answer:

Partial derivatives:

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

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

\(f_x\) and \(f_y\) are never underfined.

If \(24-6y^2=0\), then \(y^2 = 4\) and \(y = \pm2\).

If \(y=2\) and \(-12xy - 24y^2=0\), then \(-24x = 24\times 4\) and \(x=-4\).

If \(y=-2\) and \(-12xy - 24y^2=0\), then \(24x = 24\times 4\) and \(x=4\).

Calculate \(f(x, y)\).

\(f(4,-2) = 24\times 4 - 6\times4\times (-2)^2 - 8 \times (-2)^3 = 64\)

\(f(-4,2) = 24\times (-4) - 6\times(-4)\times 2^2 - 8 \times 2^3 = -64\)

Two critical points: \((4,-2,64)\) and \((-4, 2, -64)\).

Use Second Derivative test to determine if points are minimum, maximum or saddle.

Second partial derivatives:

\(f_{xx}=0\)

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

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

Then \(D(x,y) = f_{xx} f_{yy}-f^2_{xy} = -(-12y)^2 = -144y^2\).

\(D(x,y)<0\) for all \((x, y)\), so any critical point is a saddle point.

Therefore critical points (−4,2) and (4,−2) 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 \(R ( x, y )\). Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?

Answer:

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

\[ \begin{split} R(x,y) &= [HB Price×HB Units Sold]+[NB Price×NB Units Sold] \\ &= x(81 - 21x + 17y) + y(40 + 11x - 23y) \\ &= 81x-21x^2 +17xy +40y +11xy -23y^2\\ &=81x+40y+28xy-21x^2-23y^2 \end{split} \] Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?

\(R(2.3, 4.1)=81\times 2.3+40\times 4.1+28\times 2.3\times 4.1-21\times (2.3)^2-23\times (4.1)^2 = 116.62\)

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

revenue(2.30, 4.10)
## [1] 116.62

The revenue is: $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?

Answer:

We know \(x+y=96\), so \(x=96-y\).

\[ \begin{split} C(x,y) &= C(96-y,y) \\ &= \frac{1}{6} x^2 + \frac{1}{6} y^2 + 7x + 25y + 700 \\ &=\frac{1}{6} (96-y)^2 + \frac{1}{6} y^2 + 7\times (96-y) + 25y + 700 \\ &=\frac{1}{6}(y^2 - 192 y + 9216) + \frac{1}{6}y^2+672-7y+25y+700\\ &= \frac{1}{6}y^2 - 32y+1536+\frac{1}{6}y^2+18y+1372\\ &= \frac{1}{3}y^2 - 14y + 2908\\ &=C(y) \end{split} \]

To find the minimal value, we take the derivative and setting it to zero: \(C'(y) = \frac{2}{3}y-14\)

\(C'(y)=\frac{2}{3}y-14=0\), gives us \(y=21\).

\(x=96-y\) gives us \(x=75\).

Therefore we should have: 75 units produced in Los Angeles and 21 units produced in Denver.

Question 5

Evaluate the double integral on the given region.

\[ \int\int_R (e^{8x+3y}) dA, R:2\le x\le4\ and\ 2 \le y \le 4 \] Write your answer in exact form without decimals.

Answer:

\[ \begin{split} \int_2^4\int_2^4 (e^{8x+3y})\ dy\ dx &= \int_2^4 (\frac{1}{3}e^{8x+3y})|_2^4\ dx\\ &= \int_2^4 ((\frac{1}{3}e^{8x+12})-(\frac{1}{3}e^{8x+6}))\ dx\\ &= \int_2^4 \frac{1}{3}e^{8x+6}(e^6-1)\ dx\\ &= \frac{1}{24}e^{8x+6}(e^6-1) |_2^4\\ &= \frac{1}{24}e^{32+6}(e^6-1)-\frac{1}{24}e^{16+6}(e^6-1)\\ &= \frac{1}{24}(e^6-1)(e^{38}-e^{22})\\ &= \boxed{\frac{1}{24}(e^{44} - e^{38} - e^{28} + e^{22})} \end{split} \]