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 )
data<-data.frame(x=c(5.6,6.3,7,7.7,8.4),y=c(8.8,12.4,14.8,18.2,20.8))
reg<-lm(y~x,data)
summary(reg)##
## Call:
## lm(formula = y ~ x, data = data)
##
## 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-05Therefore,the equation for the regression is y = -14.8 + 4.26x
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.
\[f(x,y)=24x-6xy^2-8y^3\]
first derivative of x and y,
\(fx=24-6y^2\)
\(fy=-12xy-24y^2\)
Critical point(min and max)
\(24-6y^2=0\)
\(y^2=4\)
\(y=+/-2\)
\(-12xy-24^2=0\)
\(xy=8\)
\(x=+/-4\)
Therefore, the critical point is (-4,2)(4,-2)
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).
\(R(x)=(81 - 21x + 17y)x\)
\(R(y)=(40 + 11x - 23y)y\)
then with combine R(x) and R(y) , the revenue function is \(R(x,y)=28xy−23y2−21x2+40y+81x\)
Step 2. What is the revenue if she sells the “house” brand for 2.30 dollar and the “name” brand for 4.10 dollar?
substitue x=2.3, and y=4.1,
\(R(x,y)=2.3∗(81−21∗2.3+17∗4.1)+4.1∗(40+11∗2.3−23∗4.1)\)
The revenue is 116.62 dollars.
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?
\(x+y=96\)
\(x=96-y\)
\(c(x,y)=\frac{1}{6}x^2+\frac{1}{6}y^2+7x+25y+700\)
\(=\frac{1}{6}(96-y^2)+\frac{1}{6}y^2+7(96-y)+25y+700\)
\(=\frac{1}{6}(y^2−192y+9216)+\frac{1}{6}y^2+672−7y+25y+700\)
\(=\frac{1}{6}y^2-32y+1526+\frac{1}{6}y^2+18y+1372\)
\(=\frac{1}{3}y^2-14y+2908\)
\(c'1(y)=\frac{2}{3}y-14=0\)
\(y=21, x=96-y\)
\(x=75\)
Therefore, to minimize the weekly cost, 75 units in Los Angeles and 21 units in Denver should be produced.
Evaluate the double integral on the given region.
\[\int_{}^{}\int_{}{}(e^{8x+3y})dA;R;2\le x\le 4 and 2 \le y \le4\] \(\int_{}^{}\int_{}{}(e^{8x+3y})dA=(\int_{2}^{4}e^{8x}dx)(\int_{2}^{4}e^{3y}dy)\)
\(=(\frac{1}{8}e^{8x}|_{x=2}^4)(\frac{1}{3}e^{3y}|_{y=2}^4)\)
\(=(\frac{e^{32}-e^{16}}{8})(\frac{e^{12}-e^6}{3})\)
\(=\frac{1}{24}(e^{32}-e^{16})(e^{12}-e^6)\)