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. \[( 5.6, 8.8 ), ( 6.3, 12.4 ), ( 7, 14.8 ), ( 7.7, 18.2 ), ( 8.4, 20.8 )\]
x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
regression_line <- lm(y~x)
(variables<-summary(regression_line))
## 
## 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

\[f(x) = 4.26x -14.8\]

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

\[\begin{cases}(4,-2): f(x,y) = 24*4-6*4*(-2)^2-8(-2)^3 = 64 > 0 \space [saddle \space point] \\ (-4,2): f(x,y) = 24*-4-(6*-4*(2)^2)-8(2)^3 = -64 < 0 \end{cases}\]

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

Step 1. \(R(x,y) = x(81 - 21x + 17y) + y(40 + 11x - 23y) \\ = 81x−21x^2+17xy+40y+11xy−23y^2 \\ = -21x^2 - 23y^2 + 28xy + 81x + 40y\)

x <- 2.3; y <- 4.1
(total <- -21*x^2 - 23*y^2 + 28*x*y + 81*x + 40*y)
## [1] 116.62

Step 2. \(R(2.3, 4.1) =\) 116.62

  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?

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

\(C(96-y, y) = \frac{1}{6}(96-y)^2 + \frac{1}{6}y^2 + 7(96-y) + 25y + 700 \\ = \frac{1}{6}(9216−192y+y^2)+ \frac{1}{6}y^2 + 672 -7y +25y+700 \\ = \frac{1}{6}y^2+1536−32y+\frac{1}{6}y^2+18y+1372 \\ = \frac{1}{3}y^2−14y+2908\)

\(C^{'}(y) = \frac{2}{3}y-14\) => \(\frac{2}{3}y-14 = 0\) => \(y = 21\)

\(x=96-y=96-21=75\)

To minimize the total weekly cost: \(\begin{cases} 75 \space units \space in \space LA \\ 21 \space units \space in \space Denver \end{cases}\)
  1. Evaluate the double integral on the given region. \[\int_{R}\int(e^{8x + 3y})dA; R: 2 \leq x \leq 4 \space and \space 2 \leq y \leq 4\] Write your answer in exact form without decimals.

Let \(u = 8x + 3y\) => \(du = 8dx\) => \(dx = \frac{1}{8}du\)

\[ \int_2^4\int_2^4 e^{(8x + 3y)} dA\\ = \int_2^4\int_2^4 e^{(8x+3y)} dxdy \\ = \frac{1}{8}\int_2^4\int_2^4 e^udu \\ = \frac{1}{8}\int_2^4 [e^{(8x+3y)}]_2^4 dy\\ = \frac{1}{8}\int_2^4 [e^{(32+3y)}-e^{(16+3y)}] dy\\ = \frac{1}{8}\int_2^4 e^{(32+3y)}dy -\frac{1}{8}\int_2^4e^{(16+3y)}dy \\ = \frac{1}{24} [e^{(32+3y)}-e^{(16+3y)}]_2^4 \\ = \frac{1}{24} \Bigg[e^{44}-e^{38}-6^{28}+e^{22}\Bigg] \]

1/24*(exp(44) - exp(38) - exp(28) + exp(22))
## [1] 5.341559e+17