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)

r <- lm(y~x)
r
## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257

Based on the linear regression model, the regression line is \(y = -14.8 + 4.257x\).

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

Partial derivatives:

\(f_x(x, y) = 24 - 6y^2\)

\(f_y(x, y) = -12xy - 24y^2\)

\(f_x\) and \(f_y\) are never underfined.

If \(24-6y^2=0\), then \(y^2 = 4\) and \(y = \pm2\).

If \(y=2\) and \(-12xy - 24y^2=0\), then \(-24x = 24\times 4\) and \(x=-4\).

If \(y=-2\) and \(-12xy - 24y^2=0\), then \(24x = 24\times 4\) and \(x=4\).

Calculate \(f(x, y)\).

\(f(4,-2) = 24\times 4 - 6\times4\times (-2)^2 - 8 \times (-2)^3 = 64\)

\(f(-4,2) = 24\times (-4) - 6\times(-4)\times 2^2 - 8 \times 2^3 = -64\)

Critical Points**: \((4,-2,64)\) and \((-4, 2, -64)\).

Second Derivative - mins, max or saddle

Second partial derivatives:

\(f_{xx}=0\)

\(f_{yy}=-12x-48y\)

\(f_{xy}=-12y\)

Then \(D(x,y) = f_{xx} f_{yy}-f^2_{xy} = -(-12y)^2 = -144y^2\).

\(D(x,y)<0\) for all \((x, y)\), so per Second Derivative Test, any critical point is a saddle point.

  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.

\[ \begin{split} R(x,y) &= (81 - 21x + 17y)x + (40 + 11x - 23y)y \\ &= 81x-21x^2+17xy+40y+11xy-23y^2\\ &=81x+40y+28xy-21x^2-23y^2 \end{split} \]

\(R(2.3, 4.1)=81\times 2.3+40\times 4.1+28\times 2.3\times 4.1-21\times (2.3)^2-23\times (4.1)^2 = 116.62\)

  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?

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

\[ \begin{split} C(x,y) = C(96-y,y) &= \frac{1}{6} x^2 + \frac{1}{6} y^2 + 7x + 25y + 700 \\ &=\frac{1}{6} (96-y)^2 + \frac{1}{6} y^2 + 7\times (96-y) + 25y + 700 \\ &=\frac{1}{6}(y^2 - 192 y + 9216) + \frac{1}{6}y^2+672-7y+25y+700\\ &= \frac{1}{6}y^2 - 32y+1536+\frac{1}{6}y^2+18y+1372\\ &= \frac{1}{3}y^2 - 14y + 2908\\ &=C_1(y) \end{split} \]

\(C_1'(y) = \frac{2}{3}y-14\)

Minimal value –> \(C_1'(y)=\frac{2}{3}y-14=0\), then \(y=21\). Then \(x=96-y=75\).

L.A.: 75 units

Denver: 21 units

  1. Evaluate the double integral on the given region.

\[ \int\int_R (e^{8x+3y}) dA, R:2\le x\le4\ and\ 2 \le y \le 4 \]

\[ \begin{split} \int_2^4\int_2^4 (e^{8x+3y})\ dy\ dx &= \int_2^4 (\frac{1}{3}e^{8x+3y})|_2^4\ dx\\ &= \int_2^4 ((\frac{1}{3}e^{8x+12})-(\frac{1}{3}e^{8x+6}))\ dx\\ &= \int_2^4 \frac{1}{3}e^{8x+6}(e^6-1)\ dx\\ &= \frac{1}{24}e^{8x+6}(e^6-1) |_2^4\\ &= \frac{1}{24}e^{32+6}(e^6-1)-\frac{1}{24}e^{16+6}(e^6-1)\\ &= \frac{1}{24}(e^6-1)(e^{38}-e^{22})\\ &= \frac{1}{24}(e^{44} - e^{38} - e^{28} + e^{22}) \end{split} \]