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)
## Create df
points <- data.frame(x,y)
model <- lm(y~x,data=points)
summary(model)##
## Call:
## lm(formula = y ~ x, data = points)
##
## 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-05The equation for this set of data is: \(y=4.25x - 14.8\)
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 ).
The revenue function can be condensed this way:
\(R(x,y) = x(81-21x + 17y)+y(40+11x-23y)\)
Step 2. What is the revenue if she sells the “house” brand for 2.30 dollars and the “name” brand for 4.10 dollars?
This step can be done by simply substituting in the house brand and name brand prices into the x’s and y’s of the revenue function.
\(R(x,y) = (2.30)(81-21(2.30) + 17y(4.10)+(4.10)(40+11(2.30)-23(4.10))\)
x<-2.3
y<-4.1
eq<-x*(81-21*x + 17*y)+y*(40+11*x-23*y)
print(eq)## [1] 116.62A 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?
First start by subbing in the limit for the cost function, \(x+y=90\)
Then take this constraint and substitute in for y \(y=96-x\)
\(C(x,y) = \frac{1}{6}x^2+\frac{1}{6}(96-x)^2+7x+25(96-x)+700\)
Which simplifies to:
\(C(x,y) = \frac{1}{6}x^2+\frac{1}{6}(9216-192x+x^2)+7x+25(96-x)+700\)
and then to
\(C(x,y) = \frac{1}{3}x^2-50x+4636\)
Then to the first derivative:
\(f'(x)=\frac{2}{3}x-50\)
and
\(x=75\)
after subbing in x
\(y=21\)
Thus: to minimize the total cost, 75 units should be produced in Los Angeles and 21 should be produced in Denver
Evaluate the double integral on the given region.
\(\int\int(e^{8x+3y})\) where dA;R: 2 GEQ x GEQ 4 and 2 GEQ y GEQ 4
or written as \(\int_2^4 \int_2^4(e^{8x+3y})\)
Start by integrating the inner integral \(\int_2^4 (e^{8x+3y})\)
\[ \int_2^4 ((\frac{1}{3}e^{8x+3y})|_2^4)dx\\ \int_2^4 ((\frac{1}{3}e^{8x+12})-(\frac{1}{3}e^{8x-6}))dx\\ ((\frac{1}{24}e^{8x+6})-(e^{6}-1))|_2^4\\ =\frac{1}{24}(e^{38}-e^{22})(e^6-1)\\ =\frac{1}{24}(e^{44}-e^{38}-e^{28}+e^{22}) \]