Loading the necessary libraries:

library(Deriv) # for derivative
library(MASS) #to convert to fractions
library(matlib) # for integral

Exercie # 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)

First, let’s plot the data points:

# The data points to be plotted
x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)

# Creating the scatter plot
plot(x, y, 
     main = "Scatter Plot",   # Title of the plot
     xlab = "X-axis",         # Label for the x-axis
     ylab = "Y-axis ",        # Label for the y-axis
     col = "blue",            # Color of the points
     pch = 16)                # Shape of the points (16 for solid circles)

It seems that the points are following a linear, positive association, let’s fit the points with a linear regression model, then display the equation of the line of the best fit.

# The data points to be plotted
x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)

# Creating the scatter plot
plot(x, y, 
     main = "Scatter Plot",   # Title of the plot
     xlab = "X-axis",         # Label for the x-axis
     ylab = "Y-axis ",        # Label for the y-axis
     col = "blue",            # Color of the points
     pch = 16)                # Shape of the points (16 for solid circles)

# Fit linear regression
fit <- lm(y ~ x)

# Add regression line
abline(fit, col = "red")

# Display the summary of the linear regression model
summary(fit)
## 
## Call:
## lm(formula = y ~ x)
## 
## 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

The equation of the line of the best fit is \(y=4.2571x-14.8\) where the slope is the \(4.2571\) and the y-intercept of is \(-14.8\).

Rounding to the nearest hundredths: \(y=4.26x-14.8\)

Exercise # 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.

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

First, let’s find the partial derivatives; we can use the following R-code to do so:

# Define the function 
f_expr <- expression(24*x - 6*x*y^2 - 8*y^3)

# Compute the partial derivative of f with respect to x
df_dx <- Deriv(f_expr, "x")

# Print the expression for the partial derivative with respect to x
print(df_dx)
## expression(24 - 6 * y^2)

\(\Rightarrow: f_x(x,y)=24-6y^2\)

# Define the function 
f_expr <- expression(24*x - 6*x*y^2 - 8*y^3)

# Compute the partial derivative of f with respect to y
df_dy <- Deriv(f_expr, "y")

# Print the expression for the partial derivative with respect to y
print(df_dy)
## expression(-(y * (12 * x + 24 * y)))

\(\Rightarrow: f_y(x,y)=-12xy-24y^2\)

Now, we set the partial derivatives equal to \(0\) to find the critical points:

\[ \begin{align} 24-6y^2 &= 0\\ -6y^2 &= -24\\ y^2 &= 4\\ y&= \pm 2 \end{align} \]

For \(y=2\) :

\[ \begin{align} -12xy-24y^2 & = 0 \\ -12x(2)-24(2)^2 &= 0 \\ -24x -96 &= 0 \\ -24x &= 96 \\ x &= -4 \end{align} \]

For \(y=-2\) :

\[ \begin{align} -12xy-24y^2 & = 0 \\ -12x(-2)-24(-2)^2 &= 0 \\ 24x -96 &= 0 \\ 24x &= 96 \\ x &= 4 \end{align} \]

So the critical points are \((-4,2)\) and \((4,-2)\)

Second, let’s find the second partial derivatives and evaluate them at the critical points

2nd derivative with respect to x then x:

\[ \begin{align} f_{xx}(x,y) &= \frac{\partial}{\partial x} (24-6y^2)\\ &= 0 \end{align} \]

# Define the first derivative with respect to x 
first_dx <- expression(24-6*y^2)

# Compute the second partial derivative of f with respect to x, then x
df_dxx <- Deriv(first_dx, "x")

# Print the expression for the second partial derivative with respect to x, then x
print(df_dxx)
## expression(0)

2nd derivative with respect to y then y:

\[ \begin{align} f_{yy}(x,y) &= \frac{\partial}{\partial y} (-12xy-24y^2)\\ &= -12x-48y \end{align} \]

# Define the first derivative with respect to y 
first_dy <- expression(-12*x*y - 24*y^2)

# Compute the second partial derivative of f with respect to y, then y
df_dyy <- Deriv(first_dy, "y")

# Print the expression for the second partial derivative with respect to y, then y
print(df_dyy)
## expression(-(12 * x + 48 * y))

2nd derivative with respect to x then y:

\[ \begin{align} f_{xy}(x,y) &= \frac{\partial}{\partial y} (24-6*y^2)\\ &= -12y \end{align} \]

# Define the first derivative with respect to x 
first_dxy <- expression(24-6*y^2)

# Compute the second partial derivative of f with respect to x, then y
df_dxy <- Deriv(first_dxy, "y")

# Print the expression for the second partial derivative with respect to x, then y
print(df_dxy)
## expression(-(12 * y))

Now, let’s compute \(D\) the second derivative test:

\(D=f_{xx}(x_0,y_0)f_{yy}(x_0, y_0) - f^2_{xy} (x_0,y_0)\)

\((x_0,y_0)=(-4,2) \quad \Rightarrow\) the first term of \(D\) is \(0\) since \(f_{xx}(x,y)=0\) so let’s evaluate the critical points on \(f_{xy}\):

\(f_{xy}(-4,2) = -12*2 = -24 \quad \Rightarrow D=(-24)^2 >0\)

\((-4,2)\) is a local minima

\(f_{xy}(4,-2) = -12*(-2) = 24 \quad \Rightarrow D=(24)^2 >0\)

\((4,-2)\) is a local minima

# Define the function
f <- function(x, y) {
  return(24*x - 6*x*y^2 - 8*y^3)
}

# Generate data for plotting
x <- seq(-10, 10, length.out = 100)
y <- seq(-10, 10, length.out = 100)
z <- outer(x, y, f)

# Create 3D plot
persp(x, y, z, 
      main = "Surface plot of f(x, y)",
      xlab = "x", ylab = "y", zlab = "f(x, y)",
      theta = 30, phi = 30)

Exercise # 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 \(R ( x, y )\).

\(R(x,y) = P(x,y) \times D(x,y)\) Where \(P(x,y)\) is the price of goods, in this case each brand of products. And \(D(x,y)\) is the the demand or the number of units will be sold.

  • For the “house” brand: \(P_h(x,y) = x\) and \(D_h(x,y) = 81-21x+17y\)

\[ \begin{align} R_h(x,y) &= P_h(x,y) \times D_h(x,y)\\ &= x(81-21x+17y)\\ &= 81x - 21x^2 + 17yx \end{align} \]

  • For the “name” brand: \(P_n(x,y) = y\) and \(D_n(x,y) = 40 + 11x -23y\)

\[ \begin{align} R_n(x,y) &= P_n(x,y) \times D_n(x,y)\\ &= y(40 + 11x -23y)\\ &= 40y + 11xy - 23y^2 \end{align} \]

$$ The revenue function \(R ( x, y )\) will be the total \(R_h(x,y) + R_n(x,y)\)

\[ \begin{align} R_h(x,y) + R_n(x,y) &= 81x - 21x^2 + 17yx + 40y + 11xy - 23y^2\\ R(x,y)&= 81x-21x^2+28xy+40y-23y^2 \end{align} \]

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

We substitute \(x=2.30\) and \(y=4.10\) into \(R(x,y)\) function:

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

# Define the values of x and y
x <- 2.30
y <- 4.10

# Calculate the Revenue with the given prices
Revenue <- R_function(x,y)
cat("Revenue at x =", x, "and y =", y, "is ", Revenue, "dollars", "\n")
## Revenue at x = 2.3 and y = 4.1 is  116.62 dollars

Exercise # 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?

First, let’s find the partial derivatives \(\frac{\partial}{\partial x} C(x,y)\) and \(\frac{\partial}{\partial y} C(x,y)\)

# Define the function 
Cost_expr <- expression(1/6 * x^2 + 1/6 * y^2 + 7*x + 25*y +700)

# Compute the partial derivative of f with respect to x
dC_dx <- Deriv(Cost_expr, "x")

# Print the expression for the partial derivative with respect to x
print(dC_dx)
## expression(0.333333333333333 * x + 7)

\(\Rightarrow \quad C_x(x,y) = \frac{1}{3} x + 7\)

# Compute the partial derivative of f with respect to y
dC_dy <- Deriv(Cost_expr, "y")

# Print the expression for the partial derivative with respect to y
print(dC_dy)
## expression(0.333333333333333 * y + 25)

\(\Rightarrow \quad C_x(x,y) = \frac{1}{3} y + 25\)

Second, identify and classify the critical points:

\[ \begin{align} \frac{1}{3} x + 7 &= 0 \\ x &= -21\\ \frac{1}{3} y + 25 &=0 \\ y &= -75 \end{align} \] The critical point is \((-21, -75)\)

Second Derivative Test: \(D=f_{xx}(x_0,y_0)f_{yy}(x_0, y_0) - f^2_{xy} (x_0,y_0)\)

# Compute the second derivative with respect to x then y
dC_dxy <- Deriv(dC_dx, "y")

# Print the second derivative with respect x then y
print(dC_dxy)
## expression(0)

\(\Rightarrow \quad D=0\) so the test is inconclusive. Something is wrong! Let’s do another way where we can use the information given that both plants has to produce 96 units;

We know that “The firm is committed to produce a total of 96 units of a product each week” meaning that \(x+y=96\) \(\Rightarrow \quad y=96-x\)

Let’s substitute the expression for y into the \(C(x,y)\) function:

\[ \begin{align} C(x) &= \frac{1}{6} x^2 + \frac{1}{6} (96-x)^2 +7x +25(96-x)+700\\ &= \frac{1}{6} x^2 + \frac{1}{6} (9216-192x+x^2) +7x +2400-25x+700\\ &= \frac{1}{6} x^2 + 1536-32x+\frac{1}{6}x^2 +7x +2400-25x+700\\ &= \frac{1}{3} x^2 -50x + 4636 \end{align} \] Now. let’s find the first derivative of \(C(x)\) with respect with x:

# Define the function
C_exp <- expression(1/3 * x^2 -50*x + 4636)

# Find the first derivative of C(x) with respect to x
First_Deriv <- Deriv(C_exp, "x")

# Print the first derivative of C(x)
print(First_Deriv)
## expression(0.666666666666667 * x - 50)

\(C'(x)=\frac{2}{3}x-50\)

Let’s find the critical point:

\[ \begin{align} \frac{2}{3}x-50 &= 0\\ \frac{2}{3}x &= 50\\ 2x &= 150\\ x &=75 \end{align} \]

To classify the critical point with \(x=75\), we need to find the second derivative \(C''(x)= (\frac{2}{3}x-50)' = \frac{2}{3}\)

Since the \(C''(x) = \frac{2}{3}>0\), then \(x=75\) is a minimum value; meaning that \(x=75\) will minimize the cost. Next, we can find the corresponding value of \(y\) that minimizes the cost:

We have: \(y=96-x\) So: \(y=96-75\) \(\Rightarrow \quad y= 21\)

To minimize the total weekly cost, Los Angeles has to produce 75 units, while Denver should produce 21 units.

Exercise # 5:

Evaluate the double integral on the given region.

\(\int \int_R (e^{(8x+3y)} dA)\) ; \(R: 2 \leq x \leq 4\) and \(2 \leq y \leq 4\)

Write your answer in exact form without decimals.

Let’s evaluate the inner integral with respect to y first:

\(\int_{2}^{4} e^{(8x+3y)}dy\) using substitution:

\(u=8x+3y\) and \(du=3dy\)

\(\Rightarrow \quad \frac{1}{3} \int_{2}^{4} e^u du = \frac{1}{3} (e^{(8x+3y)})_2^4 = \frac{1}{3} [e^{8x+12} - e^{8x+6}]\)

Now, We’ll evaluate integral of \(\frac{1}{3} [e^{8x+12} - e^{8x+6}]\) with respect to x:

\[ \begin{align} \int_2^4 (\frac{1}{3} [e^{8x+12} - e^{8x+6}])dx &= \frac{1}{3} [\int_2^4e^{8x+12}dx - \int_2^4e^{8x+6}dx] \\ \end{align} \] By substitution:

let: \(u=8x+12\), \(du=8dx\) and \(v=8x+6\), \(dv=8dx\)

We substitute:

\[ \begin{align} \frac{1}{3} [\int_2^4e^{8x+12}dx - \int_2^4e^{8x+6}dx] &= \frac{1}{3} [\int_2^4 \frac{1}{8} e^u du - \int_2^4 \frac{1}{8}e^vdv] \\ &= \frac{1}{24} (e^{8x+12})_2^4 - \frac{1}{24} (e^{8x+6})_2^4\\ &= \frac{1}{24} [(e^{44}-e^{28})-(e^{38}-e^{22})] \end{align} \]

So the exact form without decimals of \(\int \int_R (e^{(8x+3y)} dA)\) ; \(R: 2 \leq x \leq 4\) and \(2 \leq y \leq 4\) is \(\frac{1}{24} (e^{44}-e^{28}-e^{38}+e^{22})\)