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)
data <- as.data.frame(cbind(x,y))
lm <- lm(y~x, data = data)
summary(lm)##
## 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-05
Regression line equation y = 4.257*x - 14.8
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
\[z = 24x - 6xy^2 - 8y^3\] \[\frac {dz}{dx} = 24 - 6y^2\] \[\frac {dz}{dy} = - 12xy - 24y^2\]
\[\frac {dz}{dydx} = - 12y\]
\[ 24 - 6y^2 = 0, \space 6y^2 = 24, \space y^2 = 4, \space y = (+2,-2)\]
\[- 12xy - 24y^2 = 0, \space -xy -2y^2 = 0, \space x = -2y, \space x = (+4,-4)\] We have critical point (4, -2) and (-4, 2).
D = Second derivative test
- if D(x,y) > 0 and f(x,y) > 0 then f(x,y) is local minimum
- if D(x,y) > 0 and f(x,y) < 0 then f(x,y) is local maximum
- if D(x,y) < 0 then f(x,y) is saddle point
\[ \frac {d^2z}{d^2x} = 24 - 6y^2 = 0\]
\[ \frac {d^2z}{d^2y} = - 12xy - 24y^2\] \[ \frac {d^2z}{d^2y} = - 12x - 48y\]
\[D(x,y) = fxx(x,y)fyy(x,y)−fxy(x,y)^2 = 0 * (- 12x - 48y) - (-12y)^2 = -144y^2 (which \space is \space < 0)\] \[for (4, -2) = -144 *(-2^2) = -144 * 4 = - 576 < 0\]
\[for (-4, 2) = -144 *(2^2) = -144 * 4 = - 576 < 0\] So, saddle points: (4, -2) and (-4, 2).
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 ).
Step 2. What is the revenue if she sells the “house” brand for 2.30 dollar and the “name” brand for $4.10?
Solution
Step1:
Revenue(x,y) = house brand sells* dollar + name brand sell * dollar
\[R(x,y) = x*(81 - 21x + 17y) + y*(40 + 11x - 23y)\] Step 2:
x <- 2.30
y <- 4.10
revenue <- function(x,y){
return(x*(81 - 21*x + 17*y) + y*(40 + 11*x - 23*y))
}
revenue(x,y)## [1] 116.624. 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
given, x = y = 96 so, y = 96 - x \[C(x, y) = \frac {1}{6} x^2 + \frac{1}{6}y^2 + 7x + 25y + 700\] \[C(x, 96-x) = \frac {1}{6} x^2 + \frac{1}{6}(96-x)^2 + 7*x + 25*(96-x) + 700\] Minima critical point : C′(x,96-x)=0
## 1/6 * (2 * x) - 1/6 * (2 * (96 - x)) + 7 - 25f <- function(x) (1/6 * (2 * x) - 1/6 * (2 * (96 - x)) + 7 - 25)
uniroot(f, lower=0.1, upper=100000000)$root## [1] 75x = 75, y = 96 -75 = 21
5. Evaluate the double integral on the given region. \[\int_R \int e^{(8x + 3y)}dA; R: 2<=x<=4 \space and \space 2<=y<=4\] Write your answer in exact form without decimals.
Solution
\[\int_R \int e^{(8x + 3y)}dA; R: 2<=x<=4 \space and \space 2<=y<=4\] \[\int_{2}^{4} \int_{2}^{4} e^{(8x + 3y)} dxdy = \int_{2}^{4} e^{(3y)} \int_{2}^{4} e^{(8x)}dxdy\]
\[ = \int_{2}^{4} e^{(3y)} \frac {1}{8}(e^{32} -e^{16})dy = \frac {1}{8} (e^{32} -e^{16}) \int_{2}^{4} e^{(3y)} dy\]
\[ = \frac {1}{8} (e^{32} -e^{16}) \frac {1}{3} e^{3y}|_{4}^{2} = \frac {1}{8} (e^{32} -e^{16}) \frac {1}{3} (e^{12}-e^{6})\]
\[ = \frac {1}{24} (e^{32} - e^{16})(e^{12} - e^{6})\]
## [1] "534155947497085120"