library(stats)
library(ggplot2)

Problem 1

Find the equation of the regression line for the given points. Round any final values to the nearest hundredth, if necessary.

\((5.6, 8.8), (6.3, 12.4), (7, 14.8), (7.7, 18.2), (8.4, 20.8)\)

Solution

df <- 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))
mod <- lm(y ~ x, data=df)
summary(mod)
## 
## 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
ggplot(df, aes(x, y)) +
  geom_point(color = "green", alpha=0.5, size = 3) +
  geom_abline(aes(slope = coef(mod)[2], intercept = coef(mod)[1], color="red"))

Equation: $y = -14.8 + 4.26x $

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.

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

Solution

Partial derivatives:

\(df/dx = 24-6y^2\) \(df/dy = -12xy-24y^2\)

\(d^2f/dx = 0\) \(d^2f/dy = -12x-48y\)

\(df/dx = 24-6y^2 = 0 -> 4-y^2=0\) \(df/dy = -12xy-24y^2 = 0 -> -xy-2y^2=0\)

y=+/-2. when y=2 then x=-4 and when y=-2 then x=4. So our points are (4,-2) and (-4,2).

for (4,-2) \(f(x,y) = 24*4-6*4*(-2)^2-8(-2)^3 = 64\) which is > 0

for (-4,2) \(f(x,y) = 24*-4-(6*-4*(2)^2)-8(2)^3 = -64\) which is < 0

Therefore, only (-4,2) is the saddle point.

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 \(R ( x, y )\). Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?

\[ Rev = price * units sold \\ R_1(x) = x * (81 - 21x + 17y) = (81x -21x^2 + 17xy) \\ R_2(y) = y * (40 + 11x -23y) = (40y + 11xy = 23y^2) \]

Revenue \[ R(x,y) = 81x - 21x^2 + 17xy + 40y + 11xy -23y^2 \\ R(x,y) = -21x^2 + 28xy - 23y^2 + 81x +40y \]

r.xy <- function(x, y) -21*x^2 + 28*x*y - 23*y^2 + 81*x + 40*y

r.xy(2.3, 4.1)
## [1] 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

Consider \(x+y=96\), then \(x=96-y\).

\[ \begin{split} C(x,y) = C(96-y,y) &= \frac{1}{6} x^2 + \frac{1}{6} y^2 + 7x + 25y + 700 \\ &=\frac{1}{6} (96-y)^2 + \frac{1}{6} y^2 + 7\times (96-y) + 25y + 700 \\ &=\frac{1}{6}(y^2 - 192 y + 9216) + \frac{1}{6}y^2+672-7y+25y+700\\ &= \frac{1}{6}y^2 - 32y+1536+\frac{1}{6}y^2+18y+1372\\ &= \frac{1}{3}y^2 - 14y + 2908\\ &=C_1(y) \end{split} \] \(C_1'(y) = \frac{2}{3}y-14\)

For minimum value, \(C_1'(y)=\frac{2}{3}y-14=0\), then, \(y=21\), then, \(x=96-y=75\).

So, in order to miniumize the cost, 75 units should be produced in Los Angeles and 21 units should be produced in Denver.

Problem 5

Evaluate the double integral on the given region.

\[ \int\int_R (e^{8x+3y}) dA, R:2\le x\le4\ and\ 2 \le y \le 4 \] Write your answer in exact form without decimals. #Solution

\[ \begin{split} \int_2^4\int_2^4 (e^{8x+3y})\ dy\ dx &= \int_2^4 (\frac{1}{3}e^{8x+3y})|_2^4\ dx\\ &= \int_2^4 ((\frac{1}{3}e^{8x+12})-(\frac{1}{3}e^{8x+6}))\ dx\\ &= \int_2^4 \frac{1}{3}e^{8x+6}(e^6-1)\ dx\\ &= \frac{1}{24}e^{8x+6}(e^6-1) |_2^4\\ &= \frac{1}{24}e^{32+6}(e^6-1)-\frac{1}{24}e^{16+6}(e^6-1)\\ &= \frac{1}{24}(e^6-1)(e^{38}-e^{22})\\ &= \frac{1}{24}(e^{44} - e^{38} - e^{28} + e^{22}) \end{split} \]