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)
# build regression model
regr_mod <- lm(y ~ x)
regr_mod
## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257

The linear regression model is y = 4.26 * x - 14.8

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

## Warning: package 'emdbook' was built under R version 3.6.3

Start by taking the derivative with respect to x and with respect to y:

ddx=24−6y2 ddy=−12xy−24y2 or −12y(x+2y)

Find values where ddx=0 or 24−6y2=0. 6y2=24, so y2=4, so y=−2 or y=2.

Find values where ddy=0 or −12y(2y+x)=0. When y=−2, then ddy=−12y(x+2y)=24(x−4)=0, so x=4. When y=2, then ddy=−12y(x+2y)=−24(x+4)=0, so x=−4. So the two critical points are (4,−2) and (−4,2).

fxx=ddx(24−6y2)=0 fyy=ddy(−12xy−24y2)=−12x−48y fxy=ddy(ddx)=ddy(24−6y2)=−12y.

Use second derivative test for (4,−2): D=fxx(x,y)fyy(x,y)−f2xy(x,y)=(0∗48)−(−242)=576. Since fxx is zero, and D<0, then (4,−2) is a saddle point, but not a relative extrema.

Use second derivative test for (−4,2): D=fxx(x,y)fyy(x,y)−f2xy(x,y)=(0∗−48)−(−242)=−576. Since fxx is zero, and D<0, then (−4,2) is also a saddle point, but not a relative extrema.

Plugging in x and y to find z and displaying the saddle points in (x,y,z) form, we have:

(4,−2,64) and (−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 ).

h= number of “house” brand units sold n= number of “name” brand units sold

The revenue function is:

R=h(81−21x+17y)+n(40+11x−23y)

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(81−21x+17y)+4.1(40+11x−23y)=186.3−48.3x+39.1y+164+45.1x−94.3y=350.3−3.2x−55.2y

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) = 1/6x^2 + 1/6y^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?

If x is the umber of units produced in L.A., and y is the number of units produced in Denver, and there will be 96 total units produced, then x+y=96, which means x=96−y or y=96−x.

Since C(x,y)=16x2+16y2+7x+25y+700 and we know y=96−x, then C(x,y)=C(x,96−x). Substititing96−x in for y, we get:

=16x2+16(96−x)2+7x+25(96−x)+700 =16x2+16(9216−192x+x2)+7x+2400−25x+700 =16x2+1536−32x+16x2−18x+3100 =13x2−50x+4636

We want to find the minima, so we find ddx(13x2−50x+4636) and set it equal to zero:

C′(x)=23x−50=0 or 2x=150 or x=75. If x=75, then y=96−x=96−75=21. So to minimize weekly cost, 75 units should be produced in L.A. and 21 units should be produced in Denver.

Problem 5:

Evaluate the double integral on the given region.

∬(e8x+3y)dA;R:2≤x≤4 and 2≤y≤4

Write your answer in exact form without decimals.

First we calculate ∫42e(8x+3y)dx:

If we substitute u=8x+3y then dx=18du, so 18∫eudu. Using the exponential rule and substituting back, we get 18e(8x+3y) as the integral, so the definite integral is:

18e(32+3y)−18e(16+3y)

We now take the definite integral of the result with respect to y:

∫42(18e(32+3y)−18e(16+3y))dy, or ∫4218e(32+3y)dy−∫4218e(16+3y))dy, or ∫42124e(32+3y)−∫42124e(16+3y).

Evaluating the definite integrals:

=124(e44−e38)−124(e28−e22)

e <- exp(1)
# compute
answer <- ((e^44 - e^38)/24) - ((e^28 - e^22)/24)
# print without scientific notation
format(answer, scientific = FALSE)
## [1] "534155947497083840"