Week 15

Exercise 1

x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)

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

# Get coefficients of the regression line
intercept <- coef(model)[1]
slope <- coef(model)[2]

# Print the equation of the regression line
cat("Regression equation: y =", round(intercept, 2), "+", round(slope, 2), "x\n")

## Regression equation: y = -14.8 + 4.26 x

Exercise 2

1. Partial derivative with respect to \(x\): \[ \frac{\partial f}{\partial x} = 24 - 6y^2 \] Setting it equal to zero: \[ 24 - 6y^2 = 0 \] \[ y^2 = 4 \] \[ y = \pm 2 \]

2. Partial derivative with respect to \(y\): \[ \frac{\partial f}{\partial y} = -12xy - 24y^2 \] Setting it equal to zero: \[ -12xy - 24y^2 = 0 \] \[ -12x - 24y = 0 \] \[ x = -2y \]

Substitute the values of \(y\) into \(x\) from step 1: \[ x = -2(2) \] or \[ x = -2(-2) \] \[ x = -4 \] or \[ x = 4 \]

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

Exercice 3

# Define the prices
x <- 2.30  
y <- 4.10  

# Calculate the revenue function
R <- function(x, y) {
  return(x * (81 - 21*x + 17*y) + y * (40 + 11*x - 23*y))
}

# Calculate revenue
revenue <- R(x, y)
revenue

## [1] 116.62

Exercice 4

The constraint equation is \(x + y = 96\), so \(y = 96 - x\). Substitute \(y = 96 - x\) into the cost function \(C(x, y)\):

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

Simplify \(C(x)\):

\[ C(x) = \frac{1}{6}x^2 + \frac{1}{6}(9216 - 192x + x^2) + 7x + 2400 - 25x + 700 \] \[ C(x) = \frac{1}{3}x^2 - 50x + 4636 \]

Find the derivative of \(C(x)\) with respect to \(x\) and set it equal to zero:

\[ C'(x) = \frac{2}{3}x - 50 = 0 \] \[ \frac{2}{3}x = 50 \] \[ x = 75 \]

Use the constraint equation \(x + y = 96\) to find \(y\):

\[ y = 96 - x = 96 - 75 = 21 \]

Therefore, to minimize the total weekly cost, the company should produce 75 units in Los Angeles and 21 units in Denver.

Exercice 5

To evaluate the given double integral over the region \(R: 2 \leq x \leq 4\) and \(2 \leq y \leq 4\), we integrate the function \(e^{8x + 3y}\) with respect to \(x\) and \(y\) over the specified region.

The integral can be expressed as:

\[ \iint_R e^{8x + 3y} \, dA \]

Let’s first integrate with respect to \(x\) and then with respect to \(y\):

\[ \int_{2}^{4} \int_{2}^{4} e^{8x + 3y} \, dx \, dy \]

Let’s perform these integrations:

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

\[ = \int_{2}^{4} \left( \frac{1}{8} e^{8x + 3y} \Big|_{x=2}^{x=4} \right) \, dy \]

\[ = \int_{2}^{4} \left( \frac{1}{8} (e^{32 + 3y} - e^{16 + 3y}) \right) \, dy \]

\[ = \frac{1}{8} \int_{2}^{4} (e^{32 + 3y} - e^{16 + 3y}) \, dy \]

\[ = \frac{1}{8} \left( \int_{2}^{4} e^{32 + 3y} \, dy - \int_{2}^{4} e^{16 + 3y} \, dy \right) \]

Now, we evaluate each integral individually:

\[ = \frac{1}{8} \left( \frac{1}{3} e^{32 + 3y} \Big|_{y=2}^{y=4} - \frac{1}{3} e^{16 + 3y} \Big|_{y=2}^{y=4} \right) \]

\[ = \frac{1}{8} \left( \frac{1}{3} (e^{44} - e^{38}) - \frac{1}{3} (e^{28} - e^{22}) \right) \]

\[ = \frac{1}{24} (e^{44} - e^{38} - e^{28} + e^{22}) \]

So, the exact value of the given double integral is \(\frac{1}{24} (e^{44} - e^{38} - e^{28} + e^{22})\).

…