library(tidyverse)

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 1 

x = c(5.6, 6.3, 7, 7.7, 8.4)
y = c(8.8, 12.4, 14.8, 18.2, 20.8)
regression_model = lm(y ~ x)
regression_model
## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257

From the results of the regression model, the equation of the regression line for the given points is: \(y = -14.8 + 4.26x\)

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 2

Find the partial derivatives of \(f(x,y):\)

\(\frac{\partial}{\partial x}f(x, y) = 24 - 6y^2\)

\(\frac{\partial}{\partial y}f(x, y) = -12xy - 24y^2\)

Let, \(24-6y^2=0; \ then,\ y^2 = 4 \ => y = \pm \ 2\)

For \(y=2, \ then\ -12xy - 24y^2=0; \ -24x = 24 \times 4; \ => x=-4\)

For \(y=-2, then\ -12xy - 24y^2=0; \ 24x = 24\times 4; \ => x=4\).

Calculating \(f(x, y)\) using these values of x and y

\(f(4,-2) = 24\times 4 - 6\times4\times (-2)^2 - 8 \times (-2)^3 = 64\)

\(f(-4,2) = 24\times (-4) - 6\times(-4)\times 2^2 - 8 \times 2^3 = -64\)

Therefore, the two critical points are: \((4,-2,64)\ and\ (-4, 2, -64)\) which are two points in a 3-dimensional space/surface.

Find the second partial derivatives and use that to test if points are minimum, maximum or saddle points.

Second partial derivatives:

\(\frac{\partial^2}{\partial x}f(x, y) = 0\)

\(\frac{\partial^2}{\partial y}f(x, y) = -12x-48y\)

$f(x, y) = -12y $. This is the mmixed partial derivative.

Then find \(D(x,y) = f_{xx} f_{yy} - f^2_{xy} = 0 - (-12y)^2 = -144y^2\). This will always result in D < 0 for any value of y.

According to the second derivative test, any critical point will be a saddle point since
\(D(x, y)<0\ \forall \ (x, y)\ \epsilon \ R\).

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?

Solution 3

Step 1
Combine the two functions by multiplying the price with the number of units.

\[ \begin{split} R(x,y) &= (81 - 21x + 17y)x + (40 + 11x - 23y)y \\ &= 81x - 21x^2 + 17xy + 40y + 11xy - 23y^2\\ &=81x + 40y + 28xy - 21x^2 - 23y^2 \end{split} \] Therefore, the revenue function is given by: \(R ( x, y ) = 81x + 40y + 28xy - 21x^2 - 23y^2\).

Step 2
Substitute, \(x = 2.3, \ and\ y = 4.10\)
\(R(2.3, 4.1) = 81\times 2.3 + 40\times 4.1 + 28\times 2.3\times 4.1 - 21\times (2.3)^2 - 23\times (4.1)^2 = 116.62\)

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 4

From the given problem, we can see that sum of units in Los Angeles and Denver is 96 units.
Therefore, \(x + y = 96; => x = 96 - y\). Substituting this into the cost function, we have:
\[ \begin{split} C(x,y) = C(96-y,y) &= \frac{1}{6} x^2 + \frac{1}{6} y^2 + 7x + 25y + 700 \\ & = \frac{1}{6} (96-y)^2 + \frac{1}{6} y^2 + 7\times (96-y) + 25y + 700 \\ & = \frac{1}{6}(y^2 - 192 y + 9216) + \frac{1}{6}y^2 + 672 - 7y + 25y + 700\\ & = \frac{1}{6}y^2 - 32y + 1536 + \frac{1}{6}y^2 + 18y + 1372\\ & = \frac{1}{3}y^2 - 14y + 2908\\ & = f(y) \end{split} \]

Now we have the cost as a function of y only. To minimize the cost, we find the first derivative of the cost function wrt y and set the derivative to zero.
\(f'(y) = \frac{2}{3}y-14; \ y = 21\). Therefore, \(x = 96-y = 75\).
Hence, the company should manufacture 75 units in Los Angeles and 21 units in Denver to minimize cost.

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 the exact form without decimals.

Solution 5

For the double integral, we integrate first with respect to y and apply the integral limits for y and then do the same for x as well as shown below:

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