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
```

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

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:

-(6*x*2*2 + 8*3*(2)^2) = 0

- (24x + 96) = 0

-24x - 96 = 0

-24x = 96

x = -4

Solving fy = 0 when y = -2:

-(6*x*2*-2 + 8*3*(-2)^2) = 0

- (-24x + 96) = 0

24x - 96 = 0

24x = 96

x = 4

```
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

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`

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`

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

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`

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.

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

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`