Find the equation of the regression line for the given points. Round any final values to the nearest hundredth, if necessary.
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))
lm.1 <- lm(y ~ x, data)
summary(lm.1)##
## 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\[\widehat{y}=-14.8x+4.26\]
Find all local maxima, local minima, and saddle points for the function given below. \(f(x,y)=24x-6xy^2-8y^3\)
To begin we need to find the partial derivatives of the equation:
\(f_x(x,y)=24-6y^2\) and \(f_y(x,y)=-24y^2-12xy\)
Next, we find the critical points by solving the system of equations. This provides the two critical points we need to examine. \((4,-2)\) and \((-4,2)\)
In order to determine what type of points these are, we need to find the second partial derivatives.
\(f_{xx}(x,y)=0\) and \(f_{yy}(x,y)=-12(4y+x)\) and \(f_{xy}(x,y)=-12y\)
We will substitute each critical point into the following equation to determine what type of point it is.
For \((4,-2)\), \(D=f_{xx}(4,-2)f_{yy}(4,-2)-f^2_{xy}(4,-2) < 0\). When \(D<0\) the point is a saddle point.
For \((-4,2)\), \(D=f_{xx}(-4,2)f_{yy}(-4,2)-f^2_{xy}(-4,2) < 0\). When \(D<0\) the point is a saddle point.
In conclusion, \((4,-2,0)\) and \((-4,2,0)\) are saddle points. There are no local maxima or minima.
A grocery store sells two brands of a produce, 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” \(40+11x-23y\) units of the “name” brand.
Find the revenue function \(R(x,y)\)
Revenue is found by multiplying units sold and the cost summed across all items. Thus, \(R(x,y)=(81-21x+18y)x + (40+11x-23y)y\)
What is the revenue if she sells the “house” brand for \(\$2.30\) and the “name” brand for \(\$4.10\)?
\(R(2.30, 4.10)=(81-21(2.30)+18(4.10))(2.30) + (40+11(2.30)-23(4.10))(4.10)=126.05\)
The revenue is $126.05.
(81-21*2.3+18*4.1)*2.3 + (40+11*2.3-23*4.1)*4.1## [1] 126.05A 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 prodcued in each plant to minimize the total weekly cost?
Using the cost function in the question and the additional information that \(x+y=96\), we can substitute into the equation and get
\(C(x, 96-x)=\frac{1}{6}x^2+\frac{1}{6}(96-x)^2+7x+25(96-x)+700\)
Simplifying from here gives:
\(\frac{1}{3}(x^2-150x+13908)\)
The minimum of this equation can be found by taking the derivative and solving for \(0\).
\(\frac{d}{dx}(\frac{1}{3}(x^2-150x+13908))=\frac{2(x-75)}{3}\)
Solving this equation gives
\(\frac{2(x-75)}{3}=0\), \(x=75\). Using the original unit equation above \(y=21\).
Thus, the most effective distribution of work is 75 units in Los Angeles and 21 units in Denver for a total cost of 2761.
Evaluate the double integral on the given region. \(\int\int(e^{8x+3y})dA; R: 2\leq x\leq 4, 2\leq y \leq 4\)
The problem can be rewritten as follows \(\int_2^4\int_2^4 (e^{8x+3y})dydx\). Extracting the inner problem and solving:
\(\int_2^4 e^{8x+3y}dy\)
\(\frac{e^{8x+3y}}{3}\bigg|_2^4\)
\(\frac{e^{8x+12}-e^{8x+6}}{3}\)
Substituting this back into the original equation
\(\int_2^4(\frac{e^{8x+12}-e^{8x+6}}{3})dx\)
\(\frac{e^{8x+12}-e^{8x+6}}{24}\bigg|_2^4\)
\(\frac{e^{44}-e^{38}-e^{28}+e^{22}}{24}\)