Elina Azrilyan

December 6th, 2019

Homework 15

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)

m1<- lm(y~x)
summary(m1)
## 
## 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
Answer: y = 4.26x - 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

Solution:

f(x) and f(y)

f=expression(24*x - 6*x*(y^2) - 8*(y^3))
#F(x)
fx<-D(f,'x')
fx
## 24 - 6 * (y^2)
#F(y)
fy<-D(f,'y')
fy
## -(6 * x * (2 * y) + 8 * (3 * y^2))

Solving fx = 0:

24 - 6y^2 = 0

6y^2 = 24

6y^2 = 4

y = +/- 2

Solving fy = 0 when y = 2:

-(6x22 + 83*(2)^2) = 0

  • (24x + 96) = 0

-24x - 96 = 0

-24x = 96

x = -4

Solving fy = 0 when y = -2:

-(6x2-2 + 83*(-2)^2) = 0

  • (-24x + 96) = 0

24x - 96 = 0

24x = 96

x = 4

Critical points are (4,-2) and (-4,2)
Second derivative test
f=expression(24 - 6 * (y^2))
#F(xx)
D(f,'x')
## [1] 0
#F(xy)
D(f,'y')
## -(6 * (2 * y))
f=expression(-(6 * x * (2 * y) + 8 * (3 * y^2)))
#F(xy)
D(f,'x')
## -(6 * (2 * y))
#F(yy)
D(f,'y')
## -(6 * x * 2 + 8 * (3 * (2 * y)))

Simplifying:

f(xx) = 0

f(xy) = -12y

f(yy) = -12x + 48y

D=fxxfyy−(fxy)^2

At(-4,2)

D = 0 - (12*(-2))^2 = -576

This is a sadle point because D<0.

f=expression(24*x - 6*x*(y^2) - 8*(y^3))
x<-4
y<--2
eval(f)
## [1] 64
At (4,-2):

D = 0 - (12*2)^2 = -576

This is a sadle point because D<0.

f=expression(24*x - 6*x*(y^2) - 8*(y^3))
x<--4
y<-2
eval(f)
## [1] -64
Answer: 2 sadle points below

(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).

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

Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?

x<-2.3
y<-4.1
R<-expression(-21*(x^2) - 23*(y^2) + 28*x*y + 81*x + 40*y)
eval(R)
## [1] 116.62
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?

y = 96-x

C(x,y) = 1/6x^2 + 1/6(96-x)^2 + 7x + 25(96-x) + 700 = 1/6x^2 + 1539 -32x + 1/6x^2 + 7x + 2400 - 25x + 700 = 1/3x^2 - 50x + 4639

Let’s find a derivative:

f=expression((1/3)*x^2 - 50*x + 4639)
#F(x)
fx<-D(f,'x')
fx
## (1/3) * (2 * x) - 50

Simplfied:

2/3*x - 50

Let’s find critical points:

2/3*x - 50 = 0

2/3x = 50

x = 75

y = 96 - 75 = 21.

Answer

To minimize production costs 75 units need to produced in Los Angeles and 21 in Denver.

Problem 5.

Evaluate the double integral on the given region.

f <- function(x) {exp(8*x+3*y)}
f2<-integrate(f, lower = 2, upper = 4)
f2
## 2.16848e+18 with absolute error < 3.3e+09