** DATA_605_Assignment 15 - Calculus 3 - MultiVariable **

if (!require(stats)) install.packages("stats")
library(stats)

if (!require(pracma)) install.packages("pracma")
## Loading required package: pracma
## Warning: package 'pracma' was built under R version 3.3.3
library(pracma)

if (!require(Deriv)) install.packages("Deriv")
## Loading required package: Deriv
## Warning: package 'Deriv' was built under R version 3.3.3
library(Deriv)

Assignment 15:

** Question-1 **

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 )

# check using summary function

# obtain the regression model
a <- c(5.6, 6.3, 7, 7.7, 8.4)
b <- c(8.8, 12.4, 14.8, 18.2, 20.8)
regression1 <- lm(b~a)
summary(regression1)
## 
## Call:
## lm(formula = b ~ a)
## 
## 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 ***
## a             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
# print the data and regression line
print(regression1)
## 
## Call:
## lm(formula = b ~ a)
## 
## Coefficients:
## (Intercept)            a  
##     -14.800        4.257
plot(a,b, main = "X vs. Y", xlab = "X", ylab = "Y")
abline(regression1, col = 'blue', lty = 2)

# answer-1a: equation of the regression line (from summary info)
#b = -14.80 + 4.26 * a

** Question-2 **

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.

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

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

partial derivative with respect to x:

\[\frac{\partial z}{\partial x} = 24 - 6y^2\] partial derivative with respect to y:

\[\frac{\partial z}{\partial y} = -12xy - 24y^2\] critical points, when \[\nabla z = \vec{0}\]

with respect to x: \[\frac{\partial z}{\partial x} = 24 - 6y^2 = 0\] \[4 - y^2 = 0\] \[ y^2 = 4\] \[ y = -2 \text{ and } 2\]

with respect to y: \[\frac{\partial z}{\partial y} = -12xy - 24y^2 = 0\] \[xy + 2y^2 = 0\] sub in \[y = -2\]

\[x(-2) + 2(-2)^2 = 0\] \[-2x + 8 = 0\] \[x = 4\]

sub in \[y = 2\]: \[x(2) + 2(2)^2 = 0\] \[2x + 8 = 0\] \[x = -4\]

The critical points to this equation are: \[(-4, 2)\]

\[(4, -2)\]

Check critical points For minima, maxima or saddles. Take the partial second derivative for each critical point to find the concavity. For concavity = positive, the critical point is a minima For concavity = negative, the critical point is a maxima.

with respect to x: \[\frac{\partial^2 z}{\partial x^2} = 0\]

with respect to y: \[\frac{\partial^2 z}{\partial y^2} = -12x - 48y^2\] Now let’s find \[f_{xy}\]

\[f_{xy} = -12y\] Check if these points are maxima, minima, or saddle points. Use the equation below:

\[D = f_{xx}(x,y) f_{yy}(x,y) - f_{xy}(x,y)^2 \text{ where D > 0: max or min; D < 0: saddle point}\] \[D = 0 * (-12x - 48y^2) - (-12y)^2 = 12y^2\]

AT critical point \[(x,y) = (-4,2)\]

\[D = 12(2)^2 > 0 \text{ and } \frac{\partial^2 z}{\partial y^2}(-4,2) = -12(-4) - 48(2)^2 = -144\]

So \((-4,2)\) has a negative concavity and \(D > 0\), this point is a maxima.

At critical point \((x,y) = (4,-2)\):

\[D = 12(-2)^2 > 0 \text{ and } \frac{\partial^2 z}{\partial y^2}(4,-2) = -12(4) - 48(-2)^2 = -240 \]

\[(4,-2)\]

has a negative concavity and \[H >0\]

this point is also another maxima.

** Question-3 **

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 ).

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

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

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

Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?

x <- 2.3
y <- 4.1

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

func_xy(x, y)
## [1] 116.62
# Answer-3
# [1] 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) = 1/6 x2 + 1/6 y2 + 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?

\[x + y = 96\]

\[y = 96 - x\]

Sub y into the formula \(C(x,y)\).

\[C(x,y) = \frac{1}{6} x^2 + \frac{1}{6} y^2 + 7x + 25y + 700\] \[C(x,y) = \frac{1}{6} x^2 + \frac{1}{6} (96 - x)^2 + 7x + 25(96-x) + 700\] \[C(x) = \frac{1}{6} x^2 + \frac{9216 -192x + x^2}{6} + 7x + 2400 - 25x + 700\] \[C(x) = \frac{1}{6} x^2 + 1536 - 32x + \frac{x^2}{6} + 7x + 2400 - 25x + 700\] \[C(x) = \frac{1}{3} x^2 - 50x + 4636\] Find the critical points in function \[C(x)\]: \[C'(x) = \frac{2}{3} x - 50\] \[\frac{2}{3} x - 50 = 0\] \[x = 75\] Therefore, plug in for x:

\[75 + y = 96\] \[y = 21\]

# check in R:
C_x <- function(x) 1/6 * x^2 + 1/6 * (96 - x)^2 + 7 * x + 25 * (96 - x) + 700
C_x_pr <- Deriv(C_x)

s <- seq(-500, 500, by = 1)
plot(s, C_x_pr(s), type = "l", col = "red")
abline(0, 0)

(x <- uniroot(C_x_pr, c(-500, 500))$root)
## [1] 75
## [1] 75

** Question-5 **

Evaluate the double integral on the given region.

\[\int \int_R (e^{8x + ey}) dA; R: 2 \leq x \leq 4 \text{ and } 2 \leq y \leq 4\]

Write your answer in exact form without decimals.

Solution:

show the double integral as: \[\int_{y = 2}^{y = 4} \int_{x = 2}^{x = 4} (e^{8x + ey}) \text{ }dx \text{ }dy = \int_{y = 2}^{y = 4} \int_{x = 2}^{x = 4} (e^{8x}e^{ey}) \text{ }dx \text{ }dy\] with \[dA = dx \text{ } dy\]

complete the integration with U substitution. \[\text{Let u } = 8x\]

\[\frac{du}{dx} = 8 \text{ equals } du = 8 \text{ }dx\] sub, \[\int_{y = 2}^{y = 4} \int_{u = 16}^{u = 32} e^{ey}e^u * \frac{1}{8} du \text{ } dy = \int_{y = 2}^{y = 4} \frac{e^{ey}}{8} \int_{u = 16}^{u = 32} e^u du \text{ }dy\]

integrate with respect to u:

\[\int_{u = 16}^{u = 32} e^u du = e^u|_{16}^{32} = e^{32} - e^{16}\] integrate with respect to y with u substitution:

\[\int_{y=2}^{y=4} \frac{e^{32} - e^{16}}{8} e^{ey} \text{ }dy = \frac{e^{32} - e^{16}}{8} \int_{y=2}^{y=4} e^{ey} \text{ } dy\] and \[u = ey\]

Therefore, \[\frac{du}{dy} = e\]

and, \[du = e \text{ } dy\]

\[\frac{e^{32} - e^{16}}{8} \int_{u=2e}^{u=4e} \frac{1}{e}e^{u} \text{ } du = \frac{e^{32} - e^{16}}{8e} \int_{u=2e}^{u=4e} e^{u} \text{ } du\] \[= (\frac{e^{31} - e^{15}}{8}) e^u|_{2e}^{4e} = (\frac{e^{31} - e^{15}}{8}) (e^{4e}-e^{2e}) + c\] Solution:

\[(\frac{e^{31} - e^{15}}{8}) (e^{4e}-e^{2e}) + c\]

** END **