Problem 1: Equation of the Regression Line

Find the equation of the regression line for the given points. Round any final values to the nearest hundredth, if necessary.

\[ (5.6, 8.8), (6.3, 12.4), (7, 14.8), (7.7, 18.2), (8.4, 20.8) \] The solution can be obtained by organizing the data points into two vectors of \(X\) and \(Y\) values and then applying a simple linear regression model on them and observing the cofficients of the slope and intercept. We use R to drive the output and then plot the scatter plot with the fitted regression line.

(X = c( 5.6,  6.3,  7,  7.7,  8.4) )

## [1] 5.6 6.3 7.0 7.7 8.4

( Y = c( 8.8,  12.4,  14.8,  18.2,  20.8 ) )

## [1]  8.8 12.4 14.8 18.2 20.8

mod1 = lm( Y ~ X)
summary(mod1)

## 
## 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

plot(X,Y)
abline(mod1, col='red')

We conclude that the equation for the regression line rounded up to the nearest hundredth is:

\[ Y = 4.26 X - 14.80 \]

Problem 2: Critical Points, Local extrema and saddle points of a function

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

We first compute all the first and second partial derivatives and then use the 2nd Derivative Test to classify any critical points.

\[ \begin{align} f_x & = 24 - 6y^2 & \text{ and } & f_y & = -12xy -24y^2 \\ f_{xx} & = 0 & \text{ and } & f_{yy} & = -12x - 48y \\ f_{xy} & = -12y & \text{ and } & f_{yx} & = -12y \\ \end{align} \]

The critical points are wherever \(f_{x} = 0\) and \(f_{y} = 0\). Use the first order condition to obtain y: \[ f_{x} = 24-6y^2=0 \implies y=2 \text{ or } y=-2 \] Plug each solution for \(y\) into the other first order condition \(f_{y}=0\) to solve for \(x\).

\[\begin{align} y& =2 & \text{implies } \\ f_{y} & = 0 \\ -12xy-24y^2 & = 0 \\ -12x(2) - 24(2)^2 & = 0 & \text{ plugging in y=2}\\ -24x - 24\cdot 4 & = 0 \\ -24x & = 96 \\ x & = -4 \\ \end{align} \]

\[\begin{align} y& =-2 & \text{implies } \\ f_{y} & = 0 \\ -12xy-24y^2 & = 0 \\ -12x(-2) - 24(-2)^2 & = 0 & \text{ plugging in y=-2}\\ 24x - 24\cdot 4 & = 0 \\ 24x & = 96 \\ x & = 4 \\ \end{align} \] Thus, the critical points are at \((4,-2), (-4,2)\).

Evaluating the function at these critical points gives:

f = function(x,y){
   24 * x - 6 * x * (y^2)  - 8 * (y^3)
}

(W = f(4,-2) )

## [1] 64

(V = f(-4,2 ))

## [1] -64

Next, we calculate the discriminant \(D\) in order to apply the Second Derivative Test.

\[ \begin{align} D & = f_{xx}f_{yy} - f_{xy}^2 & \text{ } & \text{ by definition } \\ & = 0 - (-12y)^2 & & \\ & = -144y^2 & & \text{which implies} \\ D & \leq 0 & & \text{ for all y} \end{align} \] This discriminant implies all critical points are saddle points.

The entire set of local extrema and saddle points of \(f\) is the following list:

\[(4,-2, 64) , (-4, 2, -64) \text{ are saddle points} \]

Problem 3: Revenue of Two Brands

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:

The revenue is the product of the sale price times the number of units sold summed over both products. This requires:

\[ \begin{align} R(x,y) & = x ( 81 -21x + 17y) + y ( 40 + 11x -23y) \\ & = 81x - 21x^2 + 17xy + 40y + 11xy - 23y^2 \\ & = 81x + 40y -21x^2 -23y^2 + 28xy \end{align} \] The last line gives the formula for our revenue function \(R(x,y)\).

To solve step 2, we use R to evaluate the revenue function. If the house brand is sold for \(x=2.3\) and the name brand is sold for \(y=4.10\) then the revenue is: \[R(x,y)=R(2.3,4.1)\]

R = function(x,y){
   81 * x + 40* y - 21 * (x^2) - 23*(y^2) + 28 * x* y
}

x = 2.3
y = 4.1

(R(x,y))

## [1] 116.62

The total revenue is \(R(2.3, 4.1) = 116.62\) dollars.

Problem 4: Cost Minimization Of Two Plants

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:

One method to solve this problem is to write the weekly cost as a function of one variable \(x\) and apply univariate tests for relative extrema.

We will eliminate \(y\) in the cost function \(C(x,y)\) to obtain a univariate function \(f(x) = C(x,y(x))\).

\[\begin{align} C(x,y) & = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700 & \text{ } \\ f(x) & = \frac{1}{6}x^2 + \frac{1}{6}(96-x)^2 + 7x + 25(96-x) + 700 & \text{substitute y=96-x} \\ f'(x)& = \frac{1}{6}2x + \frac{1}{6}2(96-x)(-1) + 7 + 25(-1) & \\ f'(x)& = \frac{1}{3}x - \frac{1}{3}(96-x) - 18 & \\ f'(x)& = \frac{2}{3}x - 32 - 18 & \\ f'(x)& = \frac{2}{3}x - 50 & \\ f''(x) &= \frac{2}{3} > 0 \end{align} \] Clearly \(f'(x) = 0\) implies \(x = 75\). The second derivative \(f''\) is always positive so the solution is a relative minimum.

The meaningful values of \(x\) are between \[0 \leq x \leq 96\]. Since the function \(f\) is twice differentiable over the interval \([0,96]\) and always positive, the relative minimum is an absolute minimum.

Lastly, \(y= 96-x = 21\). Thus, to minimize the total weekly cost, we should produce \(x=75\) units in Los Angeles and \(y=21\) units in Denver.

Problem 5: Double Integral

Evaluate the double integral on the given region.

\[ \int \int_{R} e^{8x + 3y} dA \text{ where } R:=\bigg \lbrace 2 \leq x \leq 4; 2 \leq y \leq 4\bigg \rbrace \]

Write your answer in exact form without decimals.

Solution

The double integral is separable into the product of two single integrals in x and y respectively.

\[ \begin{align} \int \int_{R} e^{8x + 3y} dA & = & \bigg(\int_{2}^{4} e^{8x}dx \bigg)\bigg( \int_{2}^{4}e^{3y}dy \bigg) \\ & = & \bigg( \frac{1}{8}e^{8x} \bigg \lvert_{x=2}^{4}\bigg) \bigg( \frac{1}{3} e^{3y} \bigg \lvert_{y=2}^{4} \bigg) \\ & = & \bigg( \frac{ e^{32} - e^{16}}{8} \bigg) \bigg( \frac{ e^{12} - e^{6} }{ 3} \bigg) \\ & = & \frac{1}{24}(e^{32}-e^{16})(e^{12}- e^{6}) \end{align} \]

There is no way to simplify the exponential polynomial further so that is solution in algebraic form.