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

The linear regression line is y=4.26x-14.8

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

\(\frac{df}{dx}=24-6y^2\) => \(0=24-6y^2\) => y=-2, 2

\(\frac{df}{dy}=12xy-24y^2\) => \(0=12xy-24y^2=y(12x-24y)\) => \(12x=24y\) => \(x=2y\), so x=4, -4

The local maxima and minima are: (4, 2), (-4, -2)

  1. 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.

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

  2. \(R(2.30,4.10)=2.30*(81-21*2.30 + 17*(4.1))+4.1*(40 + 11*2.3-23*4.1)=117\)

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

\(\frac{df}{dy}=\frac{-96+y}{3}+\frac{y}{3}-7+25=-32+\frac{2y}{3}+18=\frac{2y}{3}-14\) => \(\frac{2}{3}y=14\) => y=21, so x=75.

21 units need to be prodiced in Denver adn 75 need to be produced in LA.

  1. Evaluate the double integral on the given region. Write your answer in exact form without decimals.

\(\int\int_Re^{8x+3y}dA; R:2\leq x \leq 4\) and \(2\leq y\leq4\)

Start with \(\int_2^4e^{8x+3y}dy\) and let \(u=8x+3y\) and \(\frac{1}{3}du=dy\), so we have \(\frac{1}{3}\int_2^4e^{u}du=\frac{1}{3}(e^{u})]_2^4=\frac{1}{3}(e^{8x+3y})]_2^4=\frac{1}{3}e^{8x+12}-\frac{1}{3}e^{8x+6}\) then we get \(\int_2^4\frac{1}{3}e^{8x+12}-\frac{1}{3}e^{8x+6}dx\)= \(\int_2^4\frac{1}{3}e^{8x+12}dx-\int_2^4\frac{1}{3}e^{8x+6}dx\)=\(\frac{1}{24}(e^{8x+12}-e^{8x+6})]_2^4=\frac{1}{24}(e^{44}-e^{38}-e^{28}+e^{22})\)