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 )

To find the equation of the regression line, we need to split the values for independent (x) and dependent (y). After, we will use linear regression model to define intercept and slope.
Using formula: \[k=\frac{\bar{x} \bar{y} - \overline{xy}}{\bar{x}^2-\overline{x^2}}, \\ b = \bar{y}-k\bar{x}\] \[\bar{x}=\frac{5.6+6.3+7+7.7+8.4}{5}=7, \\ \bar{x}^2=7^2=49, \\ \bar{x^2}=\frac{5.6^2+6.3^2+7^2+7.7^2+8.4^2}{5}=49.98, \\ \bar{y}=\frac{8.8+12.4+14.8+18.2+20.8}{5}= 15, \\ \overline{xy} = \frac{5.6\cdot8.8+6.3\cdot12.4+7\cdot14.8+7.7\cdot18.2+8.4\cdot20.8}{5} = 109.172\]

\[k=\frac{7 \cdot 15 - 109.172}{49-49.98}=4.257, \\ b = 15-4.257\cdot 7=-14.8\] The equation is \[y=4.26x -14.8\] Using R:

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

lm_xy = lm(y~x)
summary(lm_xy)
## 
## 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 is \[y=4.26x -14.8\]

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

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

Critical points: \[f_x(x,y)=24-6y^2, \\ f_y(x,y)=-12xy-24y^2,\]

\[ \begin{cases} 24-6y^2=0\\ -12xy-24y^2=0 \end{cases} => \begin{cases} y=\pm2\\ x=\mp4 \end{cases} \]

\[f(4,-2)=24\cdot4 - 6\cdot4\cdot(-2)^2 - 8\cdot(-2)^3=64, \\ f(-4,2)=24\cdot(-4) - 6\cdot(-4)\cdot2^2 - 8\cdot2^3=-64\] The critical points are (-4,2,-64), (4,-2,-64)

To define maxima, minima, we need second derivative and the second derivative test: \[D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f_xy^2(x_0,y_0), \\ f_{xx}(x,y)=0, \\ f_{yy}(x,y)=-12x-48y, \\ f_{xy}=-12y\] For points (-4,2,-64) and (4,-2,-64): \[D=0\cdot0-(-12\cdot2)^2=-144, \\ D=0\cdot0-(-12\cdot(-2))^2=-144, \\\] D<0, then these two points are saddle points.

Check derivatives using R:

f=expression(24*x - 6*x*y^2 - 8*y^3)
fx <- D(f,'x')
fy <- D(f,'y')
fxx <- D(fx,'x')
fyy <- D(fy,'x')
fxy <- D(fx,'y')
fx
## 24 - 6 * y^2
Simplify(fy)
## -(y * (12 * x + 24 * y))
fxx
## [1] 0
fyy
## -(6 * (2 * y))
Simplify(fxy)
## -(12 * y)

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 ).

We can find the revenue function by multiplying price by brands sold. \[R(x,y)=x\cdot(81 - 21x + 17y)+y\cdot(40 + 11x - 23y)=81x-21x^2+17xy+40y+11xy-23y^2= \\ =-21x^2+81x+28xy+40y-23y^2\]

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

\[R(2.3,4.1)=-21\cdot2.3^2+81\cdot2.3+28\cdot2.3\cdot4.1+40\cdot4.1-23\cdot4.1^2=116.62\] The revenue is 116.62 Checking the answer using R:

x <- 2.3
y <- 4.1
f <- -21*x^2+81*x+28*x*y+40*y-23*y^2
f
## [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) = 1/6 x^2 + 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?

The total is \[x+y=96,y=96-x, \\ C(x) = \frac{1}{6}x^2 + \frac{1}{6}(96-x)^2 + 7x + 25(96-x) + 700=\frac{1}{3}x^2 -50x + 4,636\] To find minimum, we will need to find the derivative: \[C'(x)=\frac{2}{3}x-50=0, \\ x=75, y=96-75=21\] To minimize the total weekly cost, there should be produced x=75 units in Los Angeles, y=21 units in Denver. Check using R:

C <- expression(1/3*x^2-50*x+4636)
Cx <- fx <- D(C,'x')
 Simplify(Cx)
## 0.666666666666667 * x - 50

5. Evaluate the double integral on the given region. Write your answer in exact form without decimals.

\[\iint_R e^{8x+3y}\;dA; R:2\leq x\leq 4, 2 \leq y \leq 4\] As R is , we will first integrate by x, then by y: \[\int_2^{4} \int_2^{4} e^{8x+3y}\;dxdy=\frac{1}{8}\int_2^{4}(e^{32+3y}-e^{16+3y})\;dy=\frac{1}{24}(e^{32+12}-e^{16+12}-e^{32+6}-e^{16+6})=\frac{1}{24}(e^{44}-e^{28}-e^{38}+e^{22})\] The answer is \[\frac{1}{24}(e^{44}-e^{28}-e^{38}+e^{22})\] Approximately,

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