Question 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)

s1<- lm(y~x)
s1
## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257
plot(x,y)
abline(s1)
lines(c(5,9), -14.8+4.257*c(5,9))

Question 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.

\[f\left( x,y \right) =24x-6x{ y }^{ 2 }-8{ y }^{ 3 }\]
\[\frac { df }{ dx } =24-6{ y }^{ 2 }\] \[\frac { df }{ dy } =-12xy-24{ y }^{ 2 }\] \[if\_ y\Rightarrow 0=24-6{ y }^{ 2 }\Rightarrow { y }^{ 2 }=4\Rightarrow y\pm 2\] \[if\_ y\Rightarrow \pm 2...\& ...-12xy-24{ y }^{ 2 }=0\Rightarrow \pm { 24 }x=24x4\Rightarrow x=\pm 4\]

s2 = function(x,y){
  a = 24*x-6*x*y^2-8*y^3
  return(c(x,y,a))
  }
print(rbind(s2(-4,2),s2(4,-2)))
##      [,1] [,2] [,3]
## [1,]   -4    2  -64
## [2,]    4   -2   64

Question 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 ).

Pulling the formulas into one location:
House Brand = 81 - 21x + 17y
Name Brand = 40 + 11x - 23y
Giving us: \[R(x,y)=x(81-21x+17y)+y(40+11x-23y)\] \[R(x,y)=-21{ x }^{ 2 }+81x+28xy+40y-23{ y }^{ 2 }\]

Step 2.

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

x=2.30 #House Brand
y=4.10 #Named Brand

(Q3_2 <- -21*x^2 +81*x +28*x*y +40*y -23*y^2)
## [1] 116.62

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

Pulling the formula down to work with it: \[C\left( x,y \right) =\frac { 1 }{ 6 } { x }^{ 2 }+\frac { 1 }{ 6 } { y }^{ 2 }+7x+25y+700\] We know that, a total of 96 units of a product each week where x is from LA and y is from Denver: \[x+y=96\Rightarrow y=96-x\] \[C(x,y)\Rightarrow C(96-y,y)=\frac { 1 }{ 6 } { x }^{ 2 }+\frac { 1 }{ 6 } { y }^{ 2 }+7x+25y+700\] \[=\frac { 1 }{ 6 } { (96-y) }^{ 2 }+\frac { 1 }{ 6 } { y }^{ 2 }+7(96-y)+25y+700\] \[=\frac { 1 }{ 6 } { ({ y }^{ 2 }-192+9216) }+\frac { 1 }{ 6 } { y }^{ 2 }+672-7y+25y+700\] \[=\frac { 1 }{ 6 } { { y }^{ 2 }-32y+1536 }+\frac { 1 }{ 6 } { y }^{ 2 }+18y+1372\] \[=\frac { 1 }{ 3 } { y }^{ 2 }+14y+2908\] \[{ C }_{ 1 }^{ ` }=\frac { 2 }{ 3 } { y }-14y\] \[y=21\] \[21=96-x\Rightarrow x=75\] 75 product is from LA 21 product is from Denver

Question 5

Evaluate the double integral on the given region.

\[\iint { ({ e }^{ 8x+3y })dA;R:2\le x\le 4...and...2\le y\le 4 } \] Write your answer in exact form without decimals.
\[\iint _{ 2 }^{ 4 }{ { e }^{ 8x{ e }^{ 3y } } } dxdy\] \[\int { { e }^{ 8x }dx*\int { { e }^{ 3y }dy } } \] \[\begin{matrix} 4 \\ 2 \end{matrix}|\frac { 1 }{ 8 } { e }^{ 8x }*\begin{matrix} 4 \\ 2 \end{matrix}|{ \frac { 1 }{ 3 } e }^{ 3y }\] \[\begin{matrix} 4 \\ 2 \end{matrix}|\frac { 1 }{ 24 } { e }^{ 8x }*\begin{matrix} 4 \\ 2 \end{matrix}|{ e }^{ 3y }\] \[\frac { 1 }{ 24 } ({ e }^{ 32 }-{ e }^{ 16 })({ e }^{ 12 }-{ e }^{ 6 })\]

1/24*((exp(32)+exp(16))*(exp(12)+exp(6)))
## [1] 5.368107e+17