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)

#build linear regression
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

According to the summary statistics, the variable ‘X’ is statistically significant since it’s p-value is less that the level of significance of 5%.

The model is described by the following equation:

\(Y = -14.8000+4.2571*X\)

The intercept of -14.8000 specifies that Y equals to -14.8000 when X equals to 0. The slope coefficient of 4.2571 specifies that each unit increase in X increases Y by 4.2571.

plot(Y ~ X,main="Scatterplot of the relationships between Y and X", xlab="X",ylab="Y",col="blue")
abline(model,col="red")

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

  1. Find the partial derivatives

\(f_x(x,y)= 24 -6y^2\) \(f_y(x,y)= -12xy-24y^2\)

  1. Set each derivative equal to 0 and solve for x and y

\(24 -6y^2 =0, y1 = 2 and y_2 = -2\) \(-12xy-24y^2=0\)

Let’s substitute y with y_1=2

\(-12*2*x-24*2^2=0\) \(-24x-96 =0\) \(x_1= -4\)

Let’s substitute y with y_2=-2

\(-12*(-2)*x-24*(-2)^2=0\) \(24x-96 =0\) \(x_2= 4\)

  1. Calculate f(x,y)

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

There are two critical points: (-4,2,-64) and (4,-2,64)

  1. Find partial second derivatives

\(f_{xx}(x,y)=0\)

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

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

  1. Calculate D

\(D(x,y) = f_{xx} f_{yy}-f^2_{xy} = 0*(-12x-48y)-(-12y)^2 = -144y^2\)

D(x,y)<0 for all (x,y)

So, Second Derivative Test showed that both critical points (-4,2,-64) and (4,-2,64) are 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 ).

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

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

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

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

Total number of units produced is \(x+y=96\), \(x=96-y\)

Let’s substitute x with \(96-y\)

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

In order to find minimal value we should find derivative

\(C'(96-y,y)=\frac{2}{3}y - 14\)

Set derivative equal to 0 and solve for y

\(\frac{2}{3}y - 14=0\)

\(\frac{2}{3}y = 14\)

\(y=21\)

Calculate x

\(x=96-y=96-21=75\)

We can conclude that 75 units should be produced in LA while 21 units should be produced in Denver.

Evaluate the double integral on the given region.

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

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