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

\(\color{red}{( 5.6, 8.8 ), ( 6.3, 12.4 ), ( 7, 14.8 ), ( 7.7, 18.2 ), ( 8.4, 20.8 )}\)

##     x    y
## 1 5.6  8.8
## 2 6.3 12.4
## 3 7.0 14.8
## 4 7.7 18.2
## 5 8.4 20.8
## 
## Call:
## lm(formula = y ~ x, data = mydf)
## 
## 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
## (Intercept)           x 
##  -14.800000    4.257143

From the stats above.. Equation is : y = (4.26 * x) - 14.8

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

First derivative:

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

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

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

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

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

Replacing x and y

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

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

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

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

\(f_{xx}=0\)

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

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

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

\(D(x,y)<0\) for all \((x, y)\), any critical point is a saddle point.

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 ).

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

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

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

Total revenue: $116.62.

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

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

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

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

\(C(x,y)=\frac{1}{3}y^2 - 14y + 2908\)

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

For minimal value of cost:- \(C_1'(y)=\frac{2}{3}y-14=0\), so \(y=21\) or \(x=96-y=75\).

75 units produced in Los Angeles and 21 units produced in Denver.

Evaluate the double integral on the given region. \(\color{red}{\int\int_R e(^{8x+3y})dA ; R: 2 \leq x \leq 4}\) and \(\color{red}{2 \leq y \leq 4}\).

Write your answer in exact form without decimals.

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

\(\int_{2}^{4}\int_{2}^{4}(e^{8x + 3y})DA\)

\(\int_{2}^{4}\int_{2}^{4}(e^{8x + 3y}) \frac {3dy}{3}.dx\)

\(=\frac{1}{3}\int_{2}^{4}\bigg[(e^{8x + 3y})\bigg]_{2}^{4}.dx\)
\(=\frac{1}{3}\int_{2}^{4} (e^{8x+12} -e^{8x +6} )dx\)

\(=\frac{1}{3}\int_{2}^{4} (e^{8x+12} -e^{8x +6} ). \frac{8}{8}dx\)

\(=\frac{1}{24} \bigg[ (e^{8x+12} -e^{8x +6} ) \bigg] _{2}^{4}\)

\(=\frac{1}{24} \bigg[ e^{44} -e^{38} -(e^{28} - e^{22}) \bigg]\)

\(=\frac{1}{24} \bigg[ e^{44} -e^{38} -e^{28} + e^{22}) \bigg]\)

\(=\frac{1}{24} . 1\)