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 <- 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)
df
mod <- lm(y ~ x, data=df)
mod
## 
## Call:
## lm(formula = y ~ x, data = df)
## 
## Coefficients:
## (Intercept)            x  
##      -14.80         4.26

The equation of the regression line for the given points is: \(y = 4.26x - 14.8\)

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.

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

Find Partial Derivatives

\[ \begin{align} f(x,y) &= 24x - 6xy^2 -8y^3 \\ f_x = \frac{\partial f}{\partial x} &= 24 - 6y^2 \\ f_y = \frac{\partial f}{\partial y} &= -12xy - 24y^2 \end{align} \]

Find Critical Points

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

There are two possible \(y\) values at \(\pm 2\). Substituting these into \(\frac{\partial f}{\partial y}\) we can solve for \(x\).

\[ \begin{align} \frac{\partial f}{\partial y} = -12xy - 24y^2 &= 0 \\ xy + 2y^2 &= 0 \\ y(x + 2y) &= 0 \\ \\ \text{when } y = 2 \ ; 2(x + 2\times 2) &= 0 \\ 2x + 8 &= 0 \\ 2x &= -8 \\ x &= -4 \\ \text{when } y = -2 \ ; -2(x + 2\times (-2)) &= 0 \\ -2x + 8 &= 0 \\ -2x &= -8 \\ x &= 4 \end{align} \]

There are two critical points at \((-4, 2)\) and \((4, -2)\).

At \((-4, 2)\);

\[z = 24(-4) - 6(-4)2^2 -8(2)^3 = -64\]

and at \((4, -2)\);

\[z = 24(4) - 6(4)(-2)^2 -8(-2)^3 = 64\]

So our two critical points are \((-4, 2, -64)\) and \((4, -2, 64)\), but to find out if they are minima, maxima or saddle points, we need to find the determinant, D.

Find the Determinant

\[ \begin{align} f_{xx} = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) &= \frac{\partial}{\partial x} ( 24 - 6y^2 ) \\ &= 0\\ \\ f_{yy} = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) &= \frac{\partial}{\partial y} ( -12xy - 24y^2 ) \\ &= -12x - 48y\\ \\ f_{xy} = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) &= \frac{\partial}{\partial x} ( -12xy - 24y^2 ) \\ &= -12y \\ \\ D = f_{xx} f_{yy} - f_{xy}^2 &= 0(-12x - 48y) -(-12y)^2 \\ &= -144y^2 \end{align} \]

At our two critical points \((-4, 2, -64)\) and \((4, -2, 64)\), \(D = -144(2)^2\) and \(D = -144(-2)^2\) respectively, and since for both points \(D \leq 0\) both of our critical points are saddle points of f.

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 \((x, y)\).

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

Step 2

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

R <- function(x,y) {
  81*x - 21*(x^2) + 28*y*x + 40*y - 23*(y^2)
}
Revenue <- R(2.3, 4.1)
Revenue
## [1] 117

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

\[ x+y=96 = \text{total units to produce each week}\\ y=96-x \]

Given \(y=96-x\) we can substitute \(96-x\) every place we see y in or original function and turn this into a univariate problem.

\[ \begin{align} C(x,y) &= \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700 \\ &= \frac{1}{6}x^2 + \frac{1}{6}(96-x)^2 + 7x + 25(96-x) + 700 \\ &= \frac{1}{6}x^2 + 1536 -32x +\frac{1}{6}x^2 + 7x + 2400 -25x + 700 \\ C_1(x) &= \frac{1}{3}x^2 -50x + 4636 \\ \end{align} \]

Find the derivative of \(C_1(x)\)

\[ \begin{align} C_1'(x) &= \frac{2}{3}x -50 \\ \end{align} \]

Find the critical points:

\[ \begin{align} C_1'(x) = \frac{2}{3}x -50 &= 0\\ \frac{2}{3}x &= 50 \\ x &= 75 \end{align} \]

Since \(y=96-x\), when \(x=75\), then \(y=21\).

The company needs to produce 75 units in LA and 21 units in Denver each week in order to meet their production commitment and minimize costs.

C <- function(x){1/3*(x^2) -50*x + 4636}
plot(C(0:96), type='l')
points(75, 2761, pch=19, col="red")

Problem 5

Evaluate the double integral on the given region.

\[ \iint_R (e^{8x+3y}) dA \ ; R:2 \leq x \leq 4 \text{ and } 2 \leq y \leq 4 \] Write your answer in exact form without decimals.

\[ \begin{align} \iint_R (e^{8x+3y}) dA &= \int_2^4 \left[ \int_2^4 (e^{8x+3y}) dx \right] \ dy \\ &= \int_2^4 \left[ \frac{1}{8} \int_2^4 (e^{8x+3y}) \ 8 \ dx \right] \ dy \\ &= \int_2^4 \left[ \left. \frac{1}{8} (e^{8x+3y}) \right|_2^4 \right] \ dy \\ &= \int_2^4 \left[ \frac{1}{8} \left[ (e^{8(4)+3y}) - (e^{8(2)+3y})\right] \right] \ dy \\ &= \int_2^4 \left[ \frac{1}{8} (e^{32}e^{3y}) - (e^{16} e^{3y}) \right] \ dy \\ &= \int_2^4 \left[ \frac{1}{8} (e^{32} - e^{16}) e^{3y} \right] \ dy \\ &= \frac{1}{8} (e^{32} - e^{16}) \int_2^4 \frac{1}{3} \left[ e^{3y} \right] 3 \ dy \\ &= \frac{1}{8} (e^{32} - e^{16}) \frac{1}{3} \left[ \left. e^{3y} \right|_2^4 \right]\\ &= \frac{1}{8} (e^{32} - e^{16}) \frac{1}{3} \left( e^{3(4)} - e^{3(2)} \right) \\ &= \frac{1}{24} (e^{32} - e^{16}) \left( e^{12} - e^6 \right) \\ &= \frac{1}{24} (e^{44} - e^{38} - e^{28} + e^{22}) \end{align} \]