WebTest Name:(Test)

  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))

Answer: y=3.489004+0.2340668*x

x<-c(5.6, 6.3,7,7.7,8.4)
y<-c(8.8, 12.4,14.8,18.2,20.8)
summary(lm(x~y))
## 
## Call:
## lm(formula = x ~ y)
## 
## Residuals:
##        1        2        3        4        5 
##  0.05121 -0.09143  0.04681 -0.04901  0.04241 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 3.489004   0.125618   27.77 0.000102 ***
## y           0.234066   0.008061   29.04 8.97e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.0761 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
plot(lm(x~y))

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. \(f\left( x,y \right) \quad =24x-6x{ y }^{ 2 }-8{ y }^{ 3 }\)

Answer: The maximum and the minimum mast occur at either a critical point or on the boundary of the region.

\(\frac { df }{ dx } \quad =24-6{ y }^{ 2 }=0,\quad y=2,-2\) \(\frac { df }{ dy } \quad =-12x{ y }^{ }-\quad 24{ y }^{ 2 }=\quad -12*y(x+2y)\quad =\quad 0,\quad y=0,\quad x=-2y\)

So critical points are (4,2) and (-4,-2).

Maximum mast: \(f\left( 4,-2 \right) =24*4-6*4*{ (-2) }^{ 2 }-8*({ -2) }^{ 3 }=64\)

Minimum mast: \(f\left( -4,2 \right) =24*(-4)-6*(-4)*{ (2) }^{ 2 }-8*({ 2) }^{ 3 }=-8\)

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 ).

Answer: R ( x, y ) = (81- 21x + 17y)x + (81- 21x + 17y)y

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 Answer: R(2.3,4.1) = 655.36

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

Answer: units produced in Los Angeles : \(\frac { dC }{ dx } \quad =\frac { 1 }{ 3 } { x }^{ }\quad +\quad 7\quad =\quad 0\quad ,\quad x\quad =\quad -21\)

units produced in Denver: \(\frac { dC }{ dy } \quad =\frac { 1 }{ 3 } y^{ }\quad +\quad 25\quad =\quad 0\quad ,\quad y\quad =\quad -75\)

  1. Evaluate the double integral on the given region. \(\iint _{ R }^{ }{ { (e }^{ 8x+3y } } )dA;\quad R:\quad 2\quad \le \quad x\quad \le 4\quad and\quad 2\quad \le \quad y\quad \le 4\)

Write your answer in exact form without decimals.

Answer: \(\iint _{ R }^{ }{ { (e }^{ 8x+3y } } )dA\quad =\quad \int _{ 2 }^{ 4 }{ \int _{ 2 }^{ 4 }{ { (e }^{ 8x+3y })\quad dxdy } } \\ \qquad \qquad \qquad \qquad =\quad \int _{ 2 }^{ 4 }{ \frac { { e }^{ 8*4+3y } }{ 8 } } -\quad \frac { { e }^{ 8*2+3y } }{ 8 } dy\\ \qquad \qquad \qquad \qquad =\quad \frac { { e }^{ 8*4+3*4 } }{ 8*3 } -\frac { { e }^{ 8*2+3*4 } }{ 8*3 } -\frac { { e }^{ 8*2+3*4 } }{ 8*3 } +\frac { { e }^{ 8*2+3*2 } }{ 8*3 } \\ \qquad \qquad \qquad \qquad =\quad \frac { 1 }{ 24 } \quad ({ e }^{ 44 }-{ e }^{ 38 }-{ e }^{ 28 }-{ e }^{ 28 })\)