Question 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)
(model <- stats::lm(y ~ x))
## 
## Call:
## stats::lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257
ggplot(data.frame(x,y), aes(x, y)) + geom_point() + geom_smooth(method = "lm")

cc <- model$coefficients
(eqn <- paste("Y =", paste(round(cc[1],2),"+", paste0(round(cc[-1],2), names(cc[-1])))))
## [1] "Y = -14.8 + 4.26x"

Question 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

\[\text{We will need the x, y, and mixed partial derivatives as a 1st step}\] \[\textrm{Constant Multiple Rule: }\frac{\partial}{\partial x}(c \cdot f)=c \cdot\frac{\partial}{\partial x}(f)\] \[\textrm{Power Rule: }\frac{\partial}{\partial x}(x^n)=n \cdot x^{-1+n}\]

\[\frac{\partial}{\partial x}(24x-6xy^2-8y^3)\] \[(\frac{\partial}{\partial x}24x-\frac{\partial}{\partial x}6xy^2-\frac{\partial}{\partial x}8y^3)\] \[24(1)-6y^2(1)-0\] \[\therefore \frac{\partial}{\partial x}(24x-6xy^2-8y^3)=24-6y^2\]

\[\frac{\partial}{\partial y}(24x-6xy^2-8y^3)\] \[(\frac{\partial}{\partial y}24x-\frac{\partial}{\partial y}6xy^2-\frac{\partial}{\partial y}8y^3)\] \[0-6x(2y^{-1+2})-8(3y^{-1+3})\] \[\therefore \frac{\partial}{\partial y}(24x-6xy^2-8y^3)=-12xy-24y^2\]

\[\text{For the mix partial we can start off from the partial derivative of x}\] \[\frac{\partial}{\partial y}(24-6y^2)\] \[\frac{\partial}{\partial y}(24)-(6\frac{\partial}{\partial y}(y^2))\] \[-6(2y^{-1+2})=-12y\] \[\therefore \frac{\partial^2}{\partial x\partial y}(24x - 6xy^2 - 8y^3)=-12y\]

\[\text{Now we can find critical points by 1st solving for 0 using the partial derivatives of x and y}\] \[24-6y^2=0\] \[y^2=4\] \[y= \pm 2\]

\[\text{Simplification of partial derivative of y set to 0}\] \[-12xy-24y^2=0\] \[xy+2y^2=0\]

\[\text{Now we can plug our y values to solve for x}\] \[\text{Let }y=2\ \text{for }xy+2y^2=0\] \[2x+2(2^2)=0\] \[2x=-8\] \[x=-4\]

\[\text{Let }y=-2\ \text{for }xy+2y^2=0\] \[-2x+2(-2^2)=0 \] \[-2x=-8\] \[x=4\]

\[\therefore x= \mp4\]

\[\therefore \text{critical points}= (-4,2) \ \ (4,-2)\]

\[\text{To test if the above points are maxima or minima we need the 2nd order partials of x and y}\] \[\frac{\partial}{\partial x}(24-6y^2)=0\] \[\frac{\partial}{\partial y}(-12xy-24y^2)=-12x-48y\]

\[\text{Now plug critical points to all 2nd order partials (x, y, and mixed partial)}\] \[f_{xx}=0\] \[f_{xx}(-4,2)=0\] \[f_{xx}(4,-2)=0\]

\[f_{yy} = -12x-48y\] \[f_{yy}(-4,2)=-12(-4)-48(2)=-48\] \[f_{yy}(4,-2)=-12(4)-48(-2)=144\]

\[f_{xy}=-12y\] \[f_{xy}(-4,2)=-12(2)=-24\] \[f_{xy}(4,-2)=-12(-2)=24\]

\[\text{Finally we will use the 2nd derivative test to find out whether or not the critical points local min, max, or neither}\] \[D(x,y) = f_{xx}(x,y)f_{yy}(x,y) - f_{xy}(x,y)^2\] \[D>0 \text{ can evaluate to max or min, }D < 0 \text{ is a saddle point, and } D=0 \text{ is inconclusive} \] \[\text{*Saddle Points are between min and max}\]

\[D\:(-4,2) = 0 \cdot -48-(-24)^2=-576\] \[D\:(4,-2) = 0 \cdot 144-(24)^2=-576\]

\[\therefore critical\ points\ (-4,2) \ \ (4,-2)=saddle\ points\]

Question 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 and the “name” brand for $4.10?

Solution

\[R(x,y) = (81 - 21x + 17y)x + (40 + 11x - 23y)y\]

\[R(2.30, 4.10) = (81 - 21(2.30) + 17(4.10))2.30 + (40 + 11(2.30) -23(4.10)) 4.10\]

\[R(2.30, 4.10) = 116.62\]

Question 4

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

\[x=\text{units produced in LA}\] \[y=\text{units produced in Denver}\] \[\text{total units per week firm is commited to produce}=96\] \[\therefore\ x+y=96\] \[\text{Total weekly cost is given below as:}\] \[C(x,y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700\] \[x+y=96 \Rightarrow y=96-x\] \[\text{Now we have a univariate function after substitution:}\] \[\frac{1}{6}x^2 + \frac{1}{6}(96-x)^2 + 7x + 25(96-x) + 700\] \[\frac{1}{6}x^2 + \frac{1}{6}(x^2-192x+9216) + 7x + 2400-25x + 700\] \[\frac{1}{6}x^2 + \frac{1}{6}x^2-32x+1536 + 7x + 2400-25x + 700\] \[C(x)=\frac{x^2}{3} -50x+4636\] \[\text{Next step is to differentiate with respect to x}\] \[C'(x)=\frac{d}{dx}(\frac{x^2}{3} -50x+4636)\] \[C'(x)=\frac{1}{3}\cdot\frac{d}{dx}(x^2)-50\frac{d}{dx}(x)+\frac{d}{dx}(4636)\] \[C'(x)=\frac{2x}{3}-50\]

\[\text{Next we find the critical points by evaluatiing the derivative at 0}\] \[\frac{2x}{3}-50=0\] \[2x=150\] \[x=75\] \[\text{Finally we plug this back into our original equation to get the optimal y value}\] \[75+y=96\] \[y=21\] \[\therefore x=75=\text{units should be produced in LA}\] \[\therefore y=21=\text{units should be produced in Denver}\]

Question 5

Evaluate the double integral on the given region. Write your answer in exact form without decimals.

\[\iint \limits_R (e^{8x + 3y})\:dA\ ;\ R: 2 \leq x \leq 4 \ \ \text{and } \ 2 \leq y \leq 4\]

Solution

\[\int\limits_{2}^{4} \int\limits_{2}^{4} (e^{8x + 3y}) \ dx \ dy \] \[\int\limits_{2}^{4} \Bigg[ \int\limits_{2}^{4} (e^{8x + 3y}) \ dx \Bigg]\ dy \]

\[\text{Substitution Rule let }u=8x, \text{then }du=8\: dx, dx=\frac{1}{8}du\]

\[\int\limits_{2}^{4} \Bigg[\frac{1}{8} \int\limits_{2}^{4} (e^{8x + 3y}) \ dx \Bigg]\ dy \]

\[\int\limits_{2}^{4} \Bigg[\frac{1}{8} (e^{8x + 3y})|_2^4 \Bigg]\ dy \]

\[\int\limits_{2}^{4} \Bigg[\frac{1}{8} (e^{32 + 3y})-(e^{16 + 3y})\Bigg]\ dy \]

\[\frac{1}{8}\Big(e^{32}-e^{16} \Big) \int\limits_{2}^{4} e^{3y}\ dy \]

\[\text{Substitution Rule let }u=3y, \text{then }du=3\: dy, dy=\frac{1}{3}du\]

\[\frac{1}{8}\Big(e^{32}-e^{16} \Big) \frac{1}{3}\ e^{3y}|_2^4 \]

\[\frac{1}{8}\Big(e^{32}-e^{16} \Big) \frac{1}{3}\Big(\ e^{12}-e^6 \Big) \] \[\frac{1}{24} \Big(e^{44}-e^{38}-e^{28}+e^{22} \Big)\]

(format(1/24*(exp(44)-exp(38)-exp(28)+exp(22)),scientific=F))
## [1] "534155947497085184"