1. Find the equation of the regression line for the given points. Round any final values to the nearest hundredth, if necessary.

x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)

model <- lm(y ~ x)
summary(model)
## 
## 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

Equation for the regression line from the above results is \[ y = -14.8 + 4.25x\]

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.

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

\[f_{x} = 24 - 6y^2\] \[f_{y} = -12xy - 24y^2\]

Identifying critical points for \(f_{x}\), \(f_y\) = 0 \[y = \sqrt {\frac{24}{6}} = \pm 2\] \[x = -2y = \pm 4\]

Therefore critical points are (-4,2) and (4,-2)

Finding the second derivative, \(f_{xx} = 0\), \(f_{yy} = -12x - 48y\) and \(f_{xy} = f_{x}(f_{y}) = -12y\)

\[D = f_{xx}(x_{0}, y_{0}).f_{yy}(x_{0},y_{0}) - f_{xy}^{2}(x_{0},y_{0}) < 0\]

\[D = 0.(-12x - 48y) - 144y^2\]

Applying for critical points (-4,2) and (4,-2), we get D<0 for both points. Hence both critical points are saddle points for f.

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 ).

Revenue for house brand = R(x) = x(81 - 21x + 17y) Revenue for name brand = R(y) = y(40 + 11x - 23y)

Total revenue function \[R(x,y) = R(x) + R(y) = x(81 - 21x + 17y) + y(40 + 11x - 23y)\] \[R(x,y) = 81x - 21x^2 + 17xy + 40y + 11xy - 23y^2 = 81x - 21x^2 + 28xy + 40y - 23y^2\]

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

R(2.3, 4.1) =

x <- 2.3
y <- 4.1
81*x - 21*x^2 + 28*x*y + 40*y - 23*y^2
## [1] 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?

\[x + y = 96 \] \[y = 96 -x \] \[C(x, y) = \frac{1}{6} x^2 + \frac{1}{6}y^2 + 7x + 25y + 700\] \[ = \frac{1}{6} x^2 + \frac{1}{6}(96-x)^2 + 7x + 25(96-x) + 700\] \[ = \frac{1}{3}x^2 + 4636 - 50x \]

\[C^{'}(x) = \frac{2}{3}x - 50\] x = 75 => y = 96 - x = 21

Hence, 75 units in LA and 21 units in Denver will deliver total units in minimum weekly cost

5. Evaluate the double integral on the given region

\[\int\int_{R}(e^{(8x+3y)}.dA;R:2 \le x \le4 \space and \space 2 \le y \le 4\] Write your answer in exact form without decimals.

\[A = \int _{2}^{4}\int _{2}^{4}e^{(8x+3y)}.dA \] \[ = \int _{2}^{4}\int _{2}^{4}e^{8x}.e^{3y}.dA \] \[\int _{2}^{4}e^{8x}.dx * \int _{2}^{4}e^{3y}.dy\] \[= \frac {1}{8}.e^{8x}|_{2}^{4} * \frac {1}{3}.e^{3x}|_{2}^{4}\] \[= \frac {1}{8}.\frac {1}{3}.e^{8x}|_{2}^{4} * e^{3x}|_{2}^{4}\] \[= \frac {1}{24}.(e^{32} - e^{16}) * (e^{12} - e^6)\]