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:

Apply the Linear Model:

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(x, y)
df
##     x    y
## 1 5.6  8.8
## 2 6.3 12.4
## 3 7.0 14.8
## 4 7.7 18.2
## 5 8.4 20.8
reg_model <- lm(y~x, data=df)

reg_model
## 
## Call:
## lm(formula = y ~ x, data = df)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257

Visualize the Plot:

plot(x,y, xlab="", ylab="")
abline(reg_model)
lines(c(5,9), -14.8+4.257*c(5,9), col="green")

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:

To all local maxima, local minima, and saddle points for the function, we have to first calculate the Determinant:

Finding Partial Derivatives:

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

\[ fx = \frac{\partial f}{\partial x}=24 - 6y^2 \]

\[ fy = \frac{\partial f}{\partial y}=12xy - 24y^2 \]

Finding Critical Points:

\[ fx = \frac{\partial f}{\partial x}=24 - 6y^2 = 0\]

\[ 6y^2 = 24\]

\[ y^2 = 4\]

\[ y = \pm 2\]

Substituting the values to solve for x

\[ \frac{\partial f}{\partial y}=12xy - 24y^2=0 \]

\[ xy + 2y^2=0 \]

\[ y(x + 2y)=0 \]

When \[ y=2; 2(x+2*2)=0 \]

\[ 2x+8=0 \]

\[ x=-4 \]

When \[ y=-2; -2(x+2*(-2))=0 \]

\[ -2x+8=0 \]

\[ x=4\]

There are two critical points at (−4,2) and (4,−2)

At (−4,2)

\[ z=24(-4)-6(-4)2^2-8(2)^3=-64\]

and at (4,−2)

\[ z=24(4)-6(4)(-2)^2-8(-2)^3=-64\]

two critical points are (−4,2,−64) and (4,−2,64), but to find out if they are minima, maxima or saddle points, we need to find the determinant, D.

Finding the Determinant

\[ fxx =0 \]

\[ fyy =-12x-48y \]

\[ fxy =-12y \]

Now \[ fxxfyy-f^2xy=-(-12y)^2 =-144y^2 \]

D(x,y)<0 for all (x,y), On the basis of the Second Derivative Test, any critical point 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.

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?

Solution:

\[R(x,y)=(81-21x+17y)x+(40+11x+23y)y\]

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

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

\[R(2.3,4.1)=116.62\]

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:

Consider x+y=96, then x=96−y

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

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

\[C(x,y)= C(96-y,y) =\frac{1}{3}y^2-14y+2908\]

\[=C_1(y)\]

\[=C'_1(y)=\frac{2}{3}y-14\]

To find minimal value by \[=C'_1(y)=\frac{2}{3}y-14=0\]

\[y=21\]

\[x=96−y=75\]

The company needs to produce 75 units in LA and 21 units in Denver each week in order to meet their production commitment and minimize costs.

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.

Solution:

\[\iint_R(e^{8x+3y}dydx=\int_{2}^{4}(\frac{1}{3}e^{8x+3y})|_{2}^{4}dx\]

\[\iint_R(e^{8x+3y}dydx=\int_{2}^{4}(\frac{1}{3}e^{8x+3y})|_{2}^{4}dx\]

\[=\int_{2}^{4}((\frac{1}{3}e^{8x+12})-(\frac{1}{3}e^{8x+6}))dx\]

\[=\int_{2}^{4}\frac{1}{3}e^{8x+6}(e^6-1)dx\]

\[=\frac{1}{24}e^{8x+6}(e^6-1)|_{2}^{4}\]

\[=\frac{1}{24}e^{32+6}(e^6-1)-\frac{1}{24}e^{16+6}(e^6-1)\]

\[=\frac{1}{24}(e^{6}-1)(e^{38}-e^{22})\]

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