Assignment 15

  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)

regression <- lm(y~x)

summary(regression)
## 
## 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
regression$coefficients
## (Intercept)           x 
##  -14.800000    4.257143

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.

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

Step 1 [First and Second Derivatives]

equation = function(x,y){
  z = 24*x - 6*x*y^2 - 8*y^3
  return(c(x,y,z))
}

z <- expression(24*x - 6*x*y**2 - 8*y**3)

# first
dx <- D(z, "x")
dx
## 24 - 6 * y^2
dy <- D(z, "y")
dy
## -(6 * x * (2 * y) + 8 * (3 * y^2))
#second
dx2 <- D(dx, "x")
dx2
## [1] 0
dy2 <- D(dx, "y")
dy2
## -(6 * (2 * y))

Step 2

Determine critical points by solving for fx(x,y)=0 and fy(x,y)=0

\(0 = 24 - 6y^2\)
\(y = \sqrt\frac{24}{6}\)
\(y = \pm2\)

x where y = 2:

\(0 = -12xy - 24y^2\)
\(0 = -12(2)x - 24(2)^2\)
\(0 = -24x - 96\) \(x = -4\)

x where y = -2 \(0 = -12xy - 24y^2\)
\(0 = -12(-2)x - 24(-2)^2\)
\(0 = 24x - 96\) \(x = 4\)

\(x= \pm4\)

Critical points are (-4,2) and (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.

Step 1. Find the revenue function R (x, y).

House -> \(R(x) = (81 - 21x + 17y) * \$x\) Name -> \(R(y) = (40 + 11x - 23y) * \$y\)

Combined -> \(R(x, y) = R(x) + R(y)\)

\(R(x,y) = x(81-21x+17y) + y(40+11x-23y) -> 81x - 21x^2 + 28xy + 40y - 23y^2\)

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

2.3(81 - 21(2.3) + 17(4.1) + 4.1(40 + 11(2.3) - 23(4.1))

revenue = 2.3 * (81-21*2.3+17*4.1) + 4.1 * (40 + 11*2.3 - 23*4.1)
paste0("She will earn $",revenue," in revenue.")
## [1] "She will earn $116.62 in revenue."
  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:

\(C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700\)

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?

\(x+y = 96\)

\(y = 96 - x\) \(x = 96 - y\)

Rewrite equation for x

\(C(x) = \frac{1}{6}x^2 + \frac{1}{6}(96 - x)^2 + 7x + 25(96 - x) + 700\)

\(C(x) = \frac{1}{3}x^2 + 50x + 4636\)

\(\frac{dc}{dx} = \frac{2}{3}x - 50\)

x = 75 y = 21

75 units should be produced in Los Angeles and 21 units should be produced in Denver to minimize the total weekly cost.

  1. Evaluate the double integral on the given region.

\(\iint_g e^{8x + 3y}dA ; R: 2\leq x \leq 4\) and \(2 \leq y \le 4\)

\(\int_2^4\int_2^4 e^{8x + 3y}dxdy\)

\(\int_2^4 e^{8x}dx\) * \(\int_2^4 e^{3y}dy\)

not sure how to continue from here