Problem 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)

res<-lm(y~x)
res
## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257
summary(res)
## 
## Call:
## lm(formula = y ~ x)
## 
## Residuals:
##     1     2     3     4     5 
## -0.24  0.38 -0.20  0.22 -0.16 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -14.8000     1.0365  -14.28 0.000744 ***
## x             4.2571     0.1466   29.04 8.97e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.3246 on 3 degrees of freedom
## Multiple R-squared:  0.9965, Adjusted R-squared:  0.9953 
## F-statistic: 843.1 on 1 and 3 DF,  p-value: 8.971e-05
plot(x,y)
abline(res,col="blue")

The Output from the linear regression model shows that the regression line can be written as below

\[-14.8+4.257x\]

Problem 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.

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

To Solve this , we need to write derivatives and then substitute the values to get the saddle points.

\[f(x,y)=24x-6xy^{2}-8y^{3}\] \[fx=24-6y^{2}\] \[fy=-12xy-24y^{2}\] \[fxx=0\] \[fyy=-12x-48y\] \[fxy=-12y\] \[fyx=-12y\]

Lets start with fx=0

\[24-6y^{2}=0\] \[y=2,y=-2\]

Substituting the y-values in fy equation we get x = 4, x=-4

Substituting both x-values and y-values in the original equation we get the values 64, -64

Hence the Saddle Points are

(4,-2,64) (-4,2,-64)

Problem 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 ).

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

The Value of x is House Brand which is 2.3 $ and the value for name brand is 4.10$

x = 2.3 , y =4.1

Substiuting the values in revenue function would provide us

rev <- function(x,y) {81*x-21*x^2+28*x*y+40*y-23*y^2}
rev
## function(x,y) {81*x-21*x^2+28*x*y+40*y-23*y^2}
x=2.3
y=4.1
rev(x,y)
## [1] 116.62

The Output is 116.62 and the total Revenue is ~ 117 dollars.

Problem 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 \[C(x,y)=\dfrac {1}{6}x^{2}+\dfrac {1}{6}y^{2} +7x+25y+700\] 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?

To Solve this problem we have x+y=96 , where x being the plant in LA and y being the plant in Denver

y=x-96

Substiuting the value of y in equation we will get

\[\dfrac {1}{6}x^{2}+\dfrac {1}{6}(96-x)^{2} +7x+25(96-x)+700\] After calculating derivatives we will get

\[f`(x)=\dfrac {2}{3}x-50\]

This shows that the value of x is 75 and substiuting the x value in y=96-x=21 , the y-value is 21.

This means that the production units to minimize the total cost in LA is 75 and in Denver is 21.

Problem 5

\[\underset{R}{\iint} (e^{8x+3y})dA; R:2\leq x\leq 4 , 2\leq y\leq 4\]

\[=\int_{2}^{4} (e^{8x})dx\int_{2}^{4} (e^{3y})dy\]

Solving the above integrals separately we get the below equation

\[=\dfrac {1}{24}(e^{32}-e^{16})(e^{12}-e^{6})\]