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

x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
regression <- lm(y~x)
summary(regression)
## 
## 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 of the regression line:

y = -14.80 + 4.26 * x

the linear model is as follow: y = -14.8 + 4.25 x

# Plot the chart.
plot(y,x, main = "The regression line for the given points")
abline(lm(x~y),cex = 1.3,pch = 10, xlab = "x",ylab = "y")

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

partial derivatives:

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

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

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

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

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

When \(f_x=0, f_y=0\), The critical points:

\(f_x(x,y)=24-6y^2=0\)

\(y=\pm2\)

\(f_y(x,y)=-12xy-24y^2=0\)

\(-xy-2y^2=0\)

when \(y = 2, x = -4\)

when \(y = -2, x = 4\)

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

Using the Second Derivative Test:

\(D = D(x,y) = f_{xx}(x,y) f_{yy}(x,y) - f_{xy}(x,y)^2\)

saddle point: D > 0 is max or D < 0 is min

\(f_{xx}(x,y)=0, f_{yy}(x,y)=-12x-48y, f_{xy}(x,y)=-12y\)

\(D=(0)*(-12x-48y)-(-12y)^2=-144y^2\)

at \((4, -2)\):

\(D=-576<0\)

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

at \((-4, 2)\):

\(D=-576<0\)

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

So (4, -2) and (-4, 2) are 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 + 11 x - 23y units of the name brand.

Step 1. Find the revenue function \(R(x,y)\).

Solution

\(Revenue = (Units Sold) x (Sales Price)\) \(R(x,y)=x-(81-21x+17y)+y-(40+11x-23y)\) \(R(x,y)=81x-21x2+17xy+40y+11xy-23y^2\) \(R(x,y)=-21x^2-23y^2+81x+40y+28xy\)

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

Solution

fun<- function(x,y){-21*(x^2) - 23*(y^2) + 81*x + 40*y + 28*x*y}

revenue <- fun(2.30, 4.10)
print(paste0("Revenue: $", round(revenue, 2)))
## [1] "Revenue: $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

Total number of units produced is 96, x from Los Angeles and y from Denver.

\(x+y=96\)

\(y=96-x\)

So

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

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

\(C(x)=\frac{1}{6}x^2+\frac{9216-192x+x^2}{6}+7x+2400-25x+700\)

\(C(x)=\frac{1}{6}x^2+1536-32x+\frac{x^2}{6}+7x+2400-25x+700\)

\(C(x)=\frac{1}{3}x^2-50x+4636\)

Find the critical points in function \(C(x)\):

\(C_x(x)=\frac{2}{3}x-50\)

\(\frac{2}{3}x-50=0\)

\(x=75\)

So we know:

\(y=21\) the critical point: (75,21)

So the conlcusion is 75 units should be produced in Los Angeles, and 21 units should be produced in Denver to minimize the costs.

5. Evaluate the double integral on the given region.

\(\iint_R(e^{8x+3y})\,\mathrm{d}A;R:2≤x≤4\) and \(2≤y≤4\)

Write your answer in exact form without decimals.

Solution

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

\(=\int_2^4\int_2^4(e^{8x}e^{3y}) dx\, dy\)

\(=\int_2^4\int_4^2(e^{8x}e^{3y}) dx\, dy\)

\(=\int_2^4\frac{1}{8}e^{3y+32}-\frac{1}{8}e^{3y+16}\, dy\)

\(=\int_2^4\frac{1}{8}(e^{16}-1)e^{3y+16}\, dy\)

\(=\frac{e^{44}-e^{28}}{24}-\frac{e^{38}-e^{22}}{24}\)

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