HW15

I. 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. Separate the points to components x (dependent) and y (independent) variables -

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

2. Generate regression model for the set of points above -

reg_lm = lm(y~x)

3. List the summary of the regression model generated -

summary(reg_lm)

## 
## 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

4. Formulate the regression line equation from the infomration (slope and y intercept) provided in the summary -

y = -14.80 + 4.26x

II. 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 -

1. Start by finding the first derivative of the function with respect to x (fx) and y (fy) -

fx = 24 - 6y$^2$

fy = 12xy - 24y$^2$

2. Find the crtical points (may be a minimum, maximum or saddle point) for fx and fx -

solving for y

0 = 24 - 6y$^2$

y = sqrt(24/6)

y = +/- 2

solving for x where y = 2

0 = -12(2)x - 24(2)$^2$

0 = -24x - 96

x = 96/-24 = -4

solving for x where y = -2

0 = -12(-2)x - 24(-2)$^2$

0 = 24x - 96

x = 96/24 = 4

Critical points for this function are (2,-4) and (-2,4)

3. Find the possible concativitity of the crtical points by solving for the second derivative fxx and fxx as well as mixed derivative fxy -

fxx = 0 (no info provided about concativity, possible saddle point)

fyy = -12x - 48y

for critical point (2,-4)

0 = -12(2) - 48(-4) = + (upward concativity - possible minimum)

for critical point (-2-,4)

0 = -12(-2) - 48(4) = - (downward concativity - possible maximum)

fxy = -12y

for critical point (2,-4)

fxy = -12(-4) = 48

for critical point (2,4)

fxy = -12(4) = -48

4. Perform the second partial derivative test to determine the maximum (H>0), minimum (H also > 0) and saddle point (H < 0)

H = fxx * fyy - (fxy)$^2$

for critical point (2,-4)

0 * fyy - (48)$^2$ = - (negative) at (2,-4) conclusion: saddle point

for critical point (-2,-4)

0 * fyy - (-48)$^2$ = (negative) at (-2,4) conclusion: saddle point

III.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.

Solution

Step 1. Find the revenue function R (x, y).

R(x,y) = x(81-21x+17y) + y(40+11x-23y)

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

Substituting amount for house brand (x) and name brand (y)

Revenue = 2.3*(81-21*(2.3)+17*(4.1)) + 4.1*(40+11*(2.3)-23*(4.1))
Revenue

## [1] 116.62

IV. 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) = 1/6 x$^2$ + 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 -

1. Set up the equations for total cost and total units produced.

x - number of units to be produced in LA

y - number of units to be produced in Denver

x + y = 96 (total products to be produced by both cities)

C(x, y) = 1/6 x$^2$ + 1/6 y$^2$ + 7x + 25y + 700

2. Set up the cost equation as a single variable equation so we can get the derivative of the cost function wtih respect to x

y = 96- x

C(x,y) = $\frac{1}{6}x^2 + \frac{1}{6}y^2 +7x +25y +700$

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

C(x) = $\frac{1}{6}x^2 + 1536-32x + \frac{1}{6}x^2 +7x +2400-25x +700$

C(x) = $\frac{1}{3}x^2 -50x +4636$

$\frac{dC}{dx} = \frac{2}{3}x -50 =0$

x = 3/2*50 = 75 units when $\frac{dC}{dx}$ is 0, y = 96 - (75) = 21

3. Determine if (75,21) is a minimum point by getting the second derivative of x with respect to Cost (concativity of curve) -

$\frac{d^2C}{dx^2} = \frac{2}{3}$

Since $\frac{d^2C}{dx^2}$ is positive, the point (75,21) is a minimum point.

Conclusion: produce 75 units in LA and 21 in Denver to keep cost at a minimum and meet the constraint of 96 units to be produced per day

V. 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.

Solution

1. Set up double integral with inner integral with respect to x and outer integral with respect y for the given values of x and y.

$= \int_{2}^{4} \int_{2}^{4} (e^{8x}e^{3y}) \ dx \ dy$

2. Integrated function with respect to x keeping y constant

$=\int_{2}^{4} \frac{1}{8}e^{3y+32}-\frac{1}{8}e^{3y+16} \ dy$

3. Integrate the resulting outer function with respect to y

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

integrand <- function(y) {(1/8)*(exp(1)^16-1)*exp(1)^(3*y+16)}
volume = integrate(integrand, lower = 2, upper = 4)
print(volume,digits=17)

## 534155947497083840 with absolute error < 5930

HW15

Antonio Bayquen

December 11, 2018

I. 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. Separate the points to components x (dependent) and y (independent) variables -

2. Generate regression model for the set of points above -

3. List the summary of the regression model generated -

4. Formulate the regression line equation from the infomration (slope and y intercept) provided in the summary -

y = -14.80 + 4.26x

II. 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 -

1. Start by finding the first derivative of the function with respect to x (fx) and y (fy) -

fx = 24 - 6y\(^2\)

fy = 12xy - 24y\(^2\)

2. Find the crtical points (may be a minimum, maximum or saddle point) for fx and fx -

solving for y

0 = 24 - 6y\(^2\)

y = sqrt(24/6)

y = +/- 2

solving for x where y = 2

0 = -12(2)x - 24(2)\(^2\)

0 = -24x - 96

x = 96/-24 = -4

solving for x where y = -2

0 = -12(-2)x - 24(-2)\(^2\)

0 = 24x - 96

x = 96/24 = 4

Critical points for this function are (2,-4) and (-2,4)

3. Find the possible concativitity of the crtical points by solving for the second derivative fxx and fxx as well as mixed derivative fxy -

fxx = 0 (no info provided about concativity, possible saddle point)

fyy = -12x - 48y

for critical point (2,-4)

0 = -12(2) - 48(-4) = + (upward concativity - possible minimum)

for critical point (-2-,4)

0 = -12(-2) - 48(4) = - (downward concativity - possible maximum)

fxy = -12y

for critical point (2,-4)

fxy = -12(-4) = 48

for critical point (2,4)

fxy = -12(4) = -48

4. Perform the second partial derivative test to determine the maximum (H>0), minimum (H also > 0) and saddle point (H < 0)

H = fxx * fyy - (fxy)\(^2\)

for critical point (2,-4)

0 * fyy - (48)\(^2\) = - (negative) at (2,-4) conclusion: saddle point

for critical point (-2,-4)

0 * fyy - (-48)\(^2\) = (negative) at (-2,4) conclusion: saddle point

Solution

Step 1. Find the revenue function R (x, y).

R(x,y) = x(81-21x+17y) + y(40+11x-23y)

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

Substituting amount for house brand (x) and name brand (y)

Solution -

1. Set up the equations for total cost and total units produced.

x - number of units to be produced in LA

y - number of units to be produced in Denver

x + y = 96 (total products to be produced by both cities)

C(x, y) = 1/6 x\(^2\) + 1/6 y\(^2\) + 7x + 25y + 700

2. Set up the cost equation as a single variable equation so we can get the derivative of the cost function wtih respect to x

y = 96- x

C(x,y) = \(\frac{1}{6}x^2 + \frac{1}{6}y^2 +7x +25y +700\)

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

C(x) = \(\frac{1}{6}x^2 + 1536-32x + \frac{1}{6}x^2 +7x +2400-25x +700\)

C(x) = \(\frac{1}{3}x^2 -50x +4636\)

\(\frac{dC}{dx} = \frac{2}{3}x -50 =0\)

x = 3/2*50 = 75 units when \(\frac{dC}{dx}\) is 0, y = 96 - (75) = 21

3. Determine if (75,21) is a minimum point by getting the second derivative of x with respect to Cost (concativity of curve) -

\(\frac{d^2C}{dx^2} = \frac{2}{3}\)

Since \(\frac{d^2C}{dx^2}\) is positive, the point (75,21) is a minimum point.

Conclusion: produce 75 units in LA and 21 in Denver to keep cost at a minimum and meet the constraint of 96 units to be produced per day

V. 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.

Solution

1. Set up double integral with inner integral with respect to x and outer integral with respect y for the given values of x and y.

\(= \int_{2}^{4} \int_{2}^{4} (e^{8x}e^{3y}) \ dx \ dy\)

2. Integrated function with respect to x keeping y constant

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

3. Integrate the resulting outer function with respect to y