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 <- data.frame (x,y)

# Round to the nearest hundredth.
round(coef(lm(y ~ x, regression)), 2)
## (Intercept)           x 
##      -14.80        4.26

Answer: The equation of the regression line is \(y = -14.8 + 4.26x\)

 

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.

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

Answer:

Partial derivatives:

Set \(f_x\) to 0 and solve for \(y\):

Set \(f_y\) to 0 and substitute \(y = 2\):

Substitute \(y = -2\):

Substitute \((x,y)\) for \(z\):

Substitute \((-4,2)\) and \((4,-2)\):

The critical points are \((-4, 2, -64)\) and \((4, -2, 64)\)

Now use the Second Derivative test to determine if the points are minimum, maximum, or saddle.

Answer: As per the Second Derivative Test, \(D(x,y) < 0\) for both \((-4,2,-64)\) and \((4,-2,64)\), so therefore both critical points are saddle points.

 

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:

The revenue function is unit cost multiplied by units sold for the house/name brand and can be defined as follows:

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

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

Answer:

We can find the revenue by substituting \(x\) in the above function with the price of the house brand ($2.30), and \(y\) with the price of the name brand ($4.10).

x <- 2.3
y <- 4.1

revenue <- (x * (81 - (21 * x) + (17 * y))) + (y * (40 + (11 * x) - (23 * y)))
revenue
## [1] 116.62

The revenue is $116.62.

 

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?

If \(x + y = 96\), then \(y = 96 - x\).

\[C(x,y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700\] \[C(x,y) = \frac{1}{6}x^2 + \frac{1}{6}(96-x)^2 + 7x + 25(96-x) + 700\] \[C(x,y) = \frac{1}{6}x^2 + \frac{1}{6}(9216 - 192x + x^2) + 7x + 2400 - 25x + 700\] \[C(x,y) = \frac{1}{6}x^2 + 1536 - 32x + \frac{x^2}{6} + 7x + 2400 - 25x + 700\] \[C(x,y) = \frac{1}{3}x^2 - 50x + 4636\] \[C' = \frac{2}{3}x - 50\] \[0 = \frac{2}{3}x - 50\] \[50 = \frac{2}{3}x\] \[75 = x\]

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

 

5. Evaluate the double integral on the given region.

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

Answer:

\[\int_2^4\int_2^4(e^{8x + 3y})dx dy\] \[int_2^4(1/8)(e^{16} - 1)e^{3y + 16} dy\] \[e^{16} - 1) / 24 e^{3y + 16} + C\] \[(e^{44} - e^{28}) / 24 - (e^{38} - e^{22}) / 24\] \[(1 / 24)(e^{22} - e^{28} - e^{38} + e^{44})\]