rm(list=ls())
  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)

lr<- lm(y~x)
lr
## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257

y=14.8+4.257x

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

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

Partial derivatives:

fx(x,y)=24-6y^2

fy(x,y)=-12xy-24y^2

fx and fy are never underfined.

If 24-6y^2=0, then y^2=4 and y=±2.

If y=2 and -12xy-24y^2=0, then -24x=24×4 and x=-4.

If y=-2 and -12xy-24y^2=0, then 24x=24×4 and x=4.

Calculate f(x,y).

f(4,-2)=24×4-6×4×(-2)2-8×(-2)3=64

f(-4,2)=24×(-4)-6×(-4)×22-8×23=-64

Two critical points: (4,-2,64) and (-4,2,-64).

Use Second Derivative test to determine if points are minimum, maximum or saddle.

Second partial derivatives:

fxx=0

fyy=-12x-48y

fxy=-12y

Then D(x,y)=fxxfyy-f2xy=-(-12y)2=-144y^2.

D(x,y)<0 for all (x,y), so per Second Derivative Test, any critical point is a saddle point.

  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.

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?

R(x,y)=(81-21x+17y)x+(40+11x-23y)y=81x-21x2+17xy+40y+11xy-23y2=81x+40y+28xy-21x2-23y2

R(2.3,4.1)=81×2.3+40×4.1+28×2.3×4.1-21×(2.3)2-23×(4.1)2=116.62

  1. 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 6 x 2 + 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?

Consider x+y=96, then x=96-y.

C(x,y)=C(96-y,y)=1/6x2+1/6y2+7x+25y+700

=1/6(96-y)2+1/6y2+7×(96-y)+25y+700

=1/6(y2-192y+9216)+1/6y2+672-7y+25y+700

=1/6y2-32y+1536+1/6y2+18y+1372=1/3y^2-14y+2908=C1(y)

C’1(y)=2/3y-14

To find the minimal value consider C’1(y)=2/3y-14=0, then y=21. Then x=96-y=75.

There should be 75 units produced in Los Angeles and 21 units produced in Denver.

  1. Evaluate the double integral on the given region. ff(e^8x+3y)dA ; R: 2 £ x £ 4 and 2 £ y £ 4

Write your answer in exact form without decimals.

f4 2f4 2(e^8x+3y) dy dx=???4 2(1/3e^8x+3y)|4 2 dx

=f4 2((1/3e8x+12)-(1/3e8x+6)) dx

=f4 2 1/3e8x+6(e6- 1) dx

=1/24e^8x+6 (e^6- 1)|4 2

=1/24e32+6(e6- 1)-1/24e^16+6 (e^6- 1)

=1/24(^e6- 1)(e38-e22)

=1/24(e44-e38-e28+e22)