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)
summary(lm(y ~ x))##
## 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-05According to the model, the regression line is
\[y = -14.8 + 4.26x\]
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.
\[f(x,y) = 24x-6xy^2-8y^3\]
starting by computing the derivative \(f_x(x,y)\) and \(f_y(x,y)\):
\[f_x(x,y) = 24 -6y^2\] \[f_y(x,y) = -12xy -24y^2\]
now, compute \(f_x(x,y) = 0\) and \(f_y(x,y) = 0\):
\[\begin{align*} f_x(x,y) = 24 -6y^2 &= 0 \\ -6y^2 &= -24 \\ y^2 &= 4 \end{align*}\]
therefore, either \(y = 2\) or \(y = -2\) will satisfy the equation.
\[\begin{align*} f_y(x,y) = -12xy -24y^2 & = 0 \\ -12y(x+2y) &=0 \end{align*}\]
therefore, either \(y = 0\) or \(y = -\frac{1}{2}x\) will satisfy the equation.
when \(y = 2\), \(x = -4\)
when \(y = -2\), \(x = 4\)
z <- function(x,y){
24*x-6*x*y^2-8*y^3
}
x <- seq(-100, 100, by = 0.1)
y <- seq(-100, 100, by = 0.1)
# plot 3d
library(rgl)
plot3d(x = x, y = y, z = z(x,y))
rglwidget() # interaction\(f(-4, 2) = -64\)
a = -4
b = 2
24*a-6*a*b^2-8*b^3## [1] -64\(f(4, -2) = 64\)
a = 4
b = -2
24*a-6*a*b^2-8*b^3## [1] 64\(f(0, 0) = 0\)
a = 0
b = 0
24*a-6*a*b^2-8*b^3## [1] 0As a result, the local maxima is (4,-2, 64), minima is (-4, 2, -64), saddle point is (0, 0, 0)
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 “name” brand.
step1. find the revenue function \(R(x,y)\)
the revenue function will be the total units of “house” brand and “name” brand.
\[R(x,y) = x(81-21x+17y) + y(40+11x-23y)\]
step2. what is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?
\[R(2.30, 4.10) = 116.62\]
2.3*(81-21*2.3+17*4.1) + 4.1*(40+11*2.3-23*4.1)## [1] 116.624
A company has a plant in Los Angeles and a plant in Denver. The firm is commited 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?
since \(x+y = 96\), the function becomes:
\[\begin{align*} C(x)&=\frac{1}{6}x^2+\frac{1}{6}(96-x)^2+7x+25(96-x)+700 \\ &=\frac{1}{3}x^2-50x+4636 \end{align*}\]
from the function, there is a minima.
\[\begin{align*} \frac{dC}{dx} = \frac{2}{3}x-50 &= 0 \\ \frac{2}{3}x &= 50 \\ x &=75 \end{align*}\]
Therefore, 75 units should be produced in Los Angeles and 21 units should be produced in Denver each week to minimize the total weekly cost.
5
evaluate the double integeral on the given region.
\[\int \int_R (e^{8x+3y})dA; R: 2\le x \le 4, 2\le y \le 4\]
write your answer in exact form without decimals.
with the region, the integeral becomes:
\[\begin{align*} &\int_2^4 \int_2^4 (e^{8x+3y})dxdy \\ =&\int_2^4 \int_2^4 e^{8x} e^{3y}dxdy \\ =&\int_2^4 e^{8x}dx \int_2^4 e^{3y}dy \\ =&\frac{1}{8}e^{8x}|_2^4 \times \frac{1}{3}3e^{3y}|_2^4 \\ =&\frac{1}{8}(e^{32}-e^{16})\times \frac{1}{3}(e^{12}-e^6) \\ =&\frac{e^{44}-e^{38}-e^{28}+e^{22}}{24} \end{align*}\]