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, 7.0, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)

line <- lm(y~x)
summary(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

The equation for the regression line is:

\[y = 4.2571x - 14.8\] ## 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.

\[f(x, y) = 24x - 6xy^{2} - 8y^{3}\] First take first derivatives with respect to x and y. Then we set each to zero and solve for the other variable.

\[f_{x} = 24 - 6y^{2}\] \[0 = 24 - 6y^{2}\] \[ y^{2}= 4 \] \[ y = +/- (2) \]

\[f_{y} = -12xy - 24y^{2}\] \[0 = -12xy - 24y^{2}\] \[-xy = 2y^{2}\] \[x = -2y\] We can substitute both y = 2 and y = -2 to get our critical points:

(4, -2), (-4, 2)

Finally to determine saddle points we need second derivatives.

\[f_{xx} =0\] \[f_{yy} = -12x - 48y\] \[f_{xy} = -12y\] \[H = f_{xx}f_{yy} - f_{xy}^{2}\] \[H = -144y^{2}\] \[H = -144*(-2)^2 = -576\] \[H = -144*(2)^2 = -576\] Both critical points are saddle points.

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).

\[R(x,y) = x(81 - 21x + 17y) + y(40 + 11x -23y)\]

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

rev_function <- function(x,y){
  x_units = 81 - 21*x + 17*y
  y_units = 40 + 11*x - 23*y
  
  x_rev <- x*x_units
  y_rev <- y*y_units
  
  return(x_rev + y_rev)
}
rev_function(2.3, 4.1)
## [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:

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

We know that x+y=96.

Therefore, y=96-x. We can substitute this value into our total cost equation above.

\[C(x) = \frac{1}{6}x^{2} + \frac{1}{6}(96-x)^{2} + 7x + 25(96-x) + 700\] Then we find the derivative with respect to x and solve for zero.

\[C'(x) = \frac{2}{6}x - \frac{2}{6}(96-x) -18\] \[C'(x) = \frac{4}{6}x - 50\] \[\frac{4}{6}x = 50\] \[x = 75\] \[y = 96 - 75 = 21\]

Question 5

Evaluate the double integral on the given region:

\[\int\int(e^{8x + 3y})dA\] where R: 2<=x<=4 and 2<=y<=4

\[A = \int_{2}^{4}e^{8x}dx * \int_{2}^{4}e^{3y}dy\]

\[A = \frac{1}{8}e^{8x}[4,2] * \frac{1}{3}e^{3x}[4,2]\] \[A = \frac{1}{24}(e^{32}-e^{16})(e^{12}-e^{6})\]

A = (1/24)*(exp(32)-exp(16))*(exp(12)-exp(6))
A
## [1] 5.341559e+17