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 )

Solution:

Firstly, we create two vectors X and Y to organize the given data points.

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

Now we apply for simple regression model

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

Observing we the Coefficients of intercept and slope from above we conclude

\(y = 4.26x - 14.80\)

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.

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

Solution:

We start with finding the first derivative and then find the critical points. We then calculate the second derivative to classify the critical points.

Finding the first derivative

\(f_x(x,y)=24−6y^2\)

\(f_y(x,y)=−12xy−24y^2\)

Finding the critical points

If \(24−6y^2=0\) then we get \(y = ±2\)

When \(y = 2\) and \(−12xy−24y^2=0\) then

\(-12*x*2 -24*4 = 0\\ 24x = -96\\ x = -4\)

When \(y = -2\) and \(−12xy−24y^2=0\) then

\(-12*x*-2 -24*4 =0\\ 24x = 96 x=4\)

\(f(x,y)\)

\(f(4,-2) =24*4 -6*4*4 - 8*-8\\ =96 - 96 + 64\\ =64\)

\(f(-4,2) = 24*-4 - 6*-4*4 - 8*8\\ = -96 + 96 -64 =-64\)

We have 2 critical points which are : \((4, -2, 64)\) and \((-4, 2, -64)\)

Classification of the critical points

We calculate the second derivative

\(f_{xx}=0\)

\(f_{yy}=−12x−48y\)

\(f_{xy}=−12y\)

We calculate the discriminant D

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

So here D is less than 0 so the critical points are classified as 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 \(R ( x, y )\)

Solution:

Revenue function = product of the sale price times the number of units sold summed over both products.

So here we get

\(R(x,y) =(81-21x+17y)x+ (40+11x-23y)y\\ =81x−21x^2+17xy+40y+11xy−23y^2\\ =−21x^2−23y^2+28xy+81x+40y\)

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

Solution:

R = function(x,y){
   - 21*x^2 - 23*y^2  + 28*y*x + 81*x + 40*y
}
R(2.3,4.10)
## [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?

Solution:

We have the cost function given

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

Committed to produce a total of 96 units of a product each week, so

\(x+y=96\\x=96-y\)

We will eliminate \(x\) in the cost function \(C(x,y)\) to obtain a univariate function \(f(x)=C(x(y),y\).

\(C(x,y)=\frac{1}{6}(96−y)^2 + \frac{1}{6}y^2+7(96−y)+25y+700\\ =\frac{1}{6}(1536−192y−y^2)+ \frac{1}{6}y^2 + 672 - 7y+25y+700\\ =13y^2−14y+2908\)

Calculating the first derivative

\(C′(y)=\frac{2}{3}y−14=0\\ =\frac{2}{3}y=14\\ y=21\)

We know that,

\(x+y=96\)

replacing \(y\) value we get \(x\)

\(x=96-21=75\)

Thus, to minimize the total weekly cost, we should produce \(x=75\) units in Los Angeles and \(y=21\) units in Denver.

5. Evaluate the double integral on the given region.

\(∬_R(e^{8x+3y})dA; R:2≤x≤4 and 2≤y≤4\)

Write your answer in exact form without decimals.

Solution:

\(∬_R(e^{8x+3y})dA = (∫_2^4e^{8x}dx)(∫_2^4e^{3y}dy)\\ = (\frac{1}{8}e^{8x}|_{x=2}^4)(\frac{1}{3}e^{3y}|_{y=2}^4)\\ =(\frac{e^32 - e^16}{8})(\frac{e^12 - e^6}{3})\\ =\frac{1}{24}(e^32 - e^16)(e^12 - e^6)\)