Problem 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 y
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)

df <- data.frame(x,y)

model <- lm(y~x, df)

summary(model)
## 
## Call:
## lm(formula = y ~ x, data = df)
## 
## 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

\[y=4.2571x-14.8\]

Problem 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.

\(f(x,y)=24x-6xy^2-8y^3\)

Partial Derivatives

\(f_{x}=24-6y^2\)

\(f_{y}=-12xy-24y^2\)

Critical Points

Solve the system of equations

\(f_{x}=24-6y^2=0\) — (1)

\(f_{y}=-12xy-24y^2=0\) —- (2)

(1): \(24-6y^2=0\)

\(24=6y^2\)

\(y^2=4\)

\(y=\pm 2\)

(2): \(-12xy-24y^2=0\)

\(12xy=24y^2\)

\(x=2y\)

If \(y=-2\), then \(x=2y=2(-2)=-4\)

If \(y=2\), then \(x=2y=2(2)=4\)

Local Minima, Local Maxima, or Saddle Points

knitr::include_graphics("~/Desktop/SecondDerivativeTest.png")

\(f_{x}=24-6y^2\)

\(f_{xx}=0\)

\(f_{xy}=-12y\)

\(f_{y}=-12xy-24y^2\)

\(f_{yy}=-12x-48y\)

Critical Point

\(D=f_{xx}(x_{0},y_{0)}f_{yy}(x_{0},y_{0})-f^2_{xy}(x_{0},y_{0})\)

\(D=[0\cdot(-12x-48y)]-(-12y)^2=-144y^2\)
(4, -2) \(D=-[(-12)(-2)]^2=-576\) Since D<0, then \(f\) has a saddle point at P |
(-4, 2) \(D=-[(-12)(2)]^2=-576\) Since D<0, then \(f\) has a saddle point at P |

Problem 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 ).

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

\(R ( x, y )=-21x^2+81x+28xy+40y-23y^2\)

Step 2

What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?

\(R ( 2.3, 4.1 )=-21(2.3)^2+81(2.3)+28(2.3)(4.1)+40(4.1) -23(4.1)^2\)

revenue <- function(x,y){-21*x^2+81*x+28*x*y+40*y-23*y^2}

revenue(2.3, 4.1)
## [1] 116.62

The revenue is $116.62.

Problem 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 1

Rewrite the Cost Function in terms of one variable

\(x+y=96\) — (1)

\(C(x, y) = \frac{1}{6}x^2 + \frac{1}{6} y^2 + 7x + 25y + 700\) — (2)

Rewrite \(x+y=96\) in terms of one term

(1): \(x+y=96\)

\(x=96-y\) — (3)

Substitute x

  1. –> (2):

\(C(x, y) = \frac{1}{6}(96-y)^2 + \frac{1}{6} y^2 + 7(96-y) + 25y + 700\)

\(C(y) = \frac{1}{3}y^2 -14y + 2908\)

Partial Derivative

\(C_{y}=\frac{2}{3}y -14\)

\(C_{y}=\frac{2}{3}y -14=0\)

\(\frac{2}{3}y =14\)

\(y=21\)

Solve for x

(3): \(x=96-y\)

\(x=96-21=75\)

Cost

To minimize the cost, the firm must produce 21 units in Los Angeles and 75 units in Denver.

cost <- function(x,y){(1/6)*x^2 + (1/6)*y^2 + 7*x + 25*y + 700}

cost(21,75)
## [1] 3733

The cost is $3733.

Solution 2

\(C(x, y) = \frac{1}{6}x^2 + \frac{1}{6} y^2 + 7x + 25y + 700\)

Partial Derivatives

\(C(x, y) = \frac{1}{6}x^2 + \frac{1}{6} y^2 + 7x + 25y + 700\)

\(C_{x}=\frac{1}{3}x + 7\)

\(C_{y}=\frac{1}{3} y + 25\)

Critical Points

Solve the system of equations

\(C_{x}=\frac{1}{3}x + 7=0\) — (1)

\(C_{y}=\frac{1}{3} y + 25=0\) — (2)

(1): \(\frac{1}{3}x + 7=0\)

\(\frac{1}{3}x = - 7\)

\(x=-21\)

(2): \(\frac{1}{3} y + 25=0\)

\(\frac{1}{3} y = -25\)

\(y=-75\)

Critical Point: (-21,-75)

Local Minima, Local Maxima, or Saddle Points

knitr::include_graphics("~/Desktop/SecondDerivativeTest.png")

\(C_{x}=\frac{1}{3}x + 7\)

\(C_{xx}=\frac{1}{3}\)

\(C_{xy}=0\)

\(C_{y}=\frac{1}{3} y + 25\)

\(C_{yy}=\frac{1}{3}\)

Critical Point

\(D=f_{xx}(x_{0},y_{0)}f_{yy}(x_{0},y_{0})-f^2_{xy}(x_{0},y_{0})\)

\(D=[\frac{1}{3}\cdot\frac{1}{3}]-0^2=\frac{1}{9}\)

 \(f_{xx}(x_{0},y_{0})\)
(-21, -75) \(D=\frac{1}{9}\) \(C_{xx}(x_{0},y_{0)=\frac{1}{3}\) Since D>0 and \(f_{xx}(x_{0},y_{0})>0\), then f has a relative minimum at P. |

Cost

To minimize the cost, the firm must produce 21 units in Los Angeles and 75 units in Denver.

cost <- function(x,y){(1/6)*x^2 + (1/6)*y^2 + 7*x + 25*y + 700}

cost(21,75)
## [1] 3733

Problem 5

Evaluate the double integral on the given region. \(\int \int_{R} \left ( e^{8x+3y}\right ) dA\); R:\(2\leq x \leq 4\) and \(2\leq y \leq 4\)

Write your answer in exact form without decimals.

\(\int_{2}^{4}\int_{2}^{4}e^{8x+3y} dydx\)

\(\int_{2}^{4}\left (\int_{2}^{4}e^{8x+3y} dy \right) dx\)

\(\int_{2}^{4}\left( \left [ \frac{1}{3} e^{8x+3y} \right ]^{4}_{2} \right) dx\)

\(\int_{2}^{4}\left( \frac{1}{3} \left(e^{8x+12}-e^{8x+6} \right) \right) dx\)

\(\int_{2}^{4}\left[ \frac{1}{3} \left(e^{8x+12}(e^6-1) \right) \right] dx\)

\(\left [ \frac{1}{24} (e^6-1) e^{8x+12} \right ]^{4}_{2}\)

\(\frac{1}{24}(e^6-1) \left [e^{32+12}- e^{16+12}\right]\)

\(\frac{1}{24}(e^6-1) \left [e^{32+12}- e^{16+12}\right]\)

\(\frac{1}{24}(e^6-1) \left [e^{44}- e^{28} \right]\)