Final Assignment

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 )

df = data.frame(rbind(c( 5.6, 8.8 ), c( 6.3, 12.4 ), c( 7, 14.8 ), c( 7.7, 18.2 ), c( 8.4, 20.8 )))
model <- lm(df$X2 ~ df$X1, df)
plot(df$X2 ~ df$X1, data=df)
abline(model)

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.

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

\(f_{x}=24 - 6y^2\)
\(f_{y}= - 12xy -24y^2\)
\(f_{xx} = 0\)
\(f_{yy}= - 12x -48y\)
\(f_{xy}= - 12y\)

When \(f_{x} = 0 ,\ f_{y}=0\) Critical Points:
\(f_{x}=24 - 6y^2 = 0\)
\(y^2 = 4\)
\(y = 2 \ or -2\)
\(f_{y}= - 12xy -24y^2 = 0\)
\(- xy -2y^2 = 0\)

when \(y = 2, x = -2y , x = -4\)
when \(y = -2, x = -2y , x = 4\) So we have critical point (4, -2) and (-4, 2).

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 ).

Revenue for House brand:

Revenue for name brand:

Total Revenue:

\(R(x,y)= 81x - 21x^2 + 17xy + 40y + 11xy - 23y^2\)

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

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

Find R(2.3, 4.1):

x <- 2.3
y <- 4.1

z <- -21*(x^2) - 23*(y^2) + 28*x*y + 81*x +40*y
z
## [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?

Total unit per week = 96 units Number of LA = x Number of Denver = y

$ x + y = 96$

$ y = 96-x$

When $ y = 96-x$ in equation:

\(C(x,y) = \frac{1}{6} x^2 + \frac{1}{6} y^2 + 7x + 25y + 700\)
\(= x^2 + (96 - x)^2 + 7x + 25(96-x) + 700\)

\(= x^2 + 1536 - 32x + + 7x + 2400 - 25x + 700\)

\(= x^2 - 50x + 4636\)

Minimum critical point :

\(x = 75\)

when \(x = 75\) in $ x + y = 96$:

\(y = 21\)

Minimize the cost should be 75 units in LA, 21 unit in Denver.

Question 5

Evaluate the double integral on the given region.

\(\int \int_R (e^{8x + 3y}) dA; R: 2 \leq x \leq 4 \ { and } \ 2 \leq y \leq 4\)

Write your answer in exact form without decimals.

\({y = 2}^{y = 4} {x = 2}^{x = 4} (e^{8x + 3y}) dx dy\)
\(= \int_{2}^{4} \int_{2}^{4} (e^{8x}e^{3y}) \ dx \ dy\)
\(= \int_{2}^{4} \int_{2}^{4} (e^{8x}e^{3y}) \ dx \ dy\)
\(=\int_{2}^{4} \frac{1}{8}e^{3y+32}-\frac{1}{8}e^{3y+16} \ dy\)
\(=\int_{2}^{4} \frac{1}{8}\left(e^{16}-1\right)e^{3y+16} \ dy\)
\(=\frac{e^{44}-e^{28}}{24}-\frac{e^{38}-e^{22}}{24}\)
\(=\frac{1}{24}\left(e^{22}-e^{28}-e^{38}+e^{44}\right)\)