Untitled

library(knitr)

## Warning: package 'knitr' was built under R version 3.5.3

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:

xvalue <- c(5.6, 6.3, 7, 7.7, 8.4)
yvalue <- c(8.8, 12.4, 14.8, 18.2, 20.8)
lm_m1 <- lm(yvalue ~ xvalue)
print(lm_m1)

## 
## Call:
## lm(formula = yvalue ~ xvalue)
## 
## Coefficients:
## (Intercept)       xvalue  
##     -14.800        4.257

So, y = (-14.8) + 4.257x

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:

First derivatives:

\(f_x(x, y) = 24 - 6y^2\)

\(f_y(x, y) = -12xy - 24y^2\)

If \(24-6y^2=0\), then \(y^2 = 4\) and \(y = \pm2\).

If \(y=2\) and \(-12xy - 24y^2=0\), then \(-24x = 24\times 4\) and \(x=-4\).

If \(y=-2\) and \(-12xy - 24y^2=0\), then \(24x = 24\times 4\) and \(x=4\).

Calculate \(f(x, y)\).

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

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

The critical points are \((4,-2,64)\) and \((-4, 2, -64)\).

Second Derivatives:

\(f_{xx}=0\)

\(f_{yy}=-12x-48y\)

\(f_{xy}=-12y\)

\(D(x,y) = f_{xx} f_{yy}-f^2_{xy} = -(-12y)^2 = -144y^2\).

\(D(x,y)<0\) for all \((x, y)\)

Conclusion: D<0 for any point, so we can conclude that our two points are saddle points.

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 ).

Solution:

\[ R(x,y) = (81- 21x + 17y)x + (40 + 11x - 23y)y\\ R(x, y)=81x - 21x^2 + 28yx + 40y - 23y^2 \]

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

Solution:

\(R(2.3, 4.1)=81\times 2.3+40\times 4.1+28\times 2.3\times 4.1-21\times (2.3)^2-23\times (4.1)^2 = 116.62\)

Total revenue: $116.62.

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:

Cost function \[ c(x,y)=\frac{1}{6}x^{2}+\frac{1}{6}y^{2}+7x+25y+700\\ \]

We can assume that x+y=total number of units \[ x+y=96\\ x=96-y \]

This relationship allows us to convert C(x,y) into a univariate function. Substitute the above relationship and simplify using algebra. \[ c(x,y)=\frac{1}{6}x^{2}+\frac{1}{6}y^{2}+7x+25y+700\\ =\frac{1}{6}(96-y)^{2}+\frac{1}{6}y^{2}+7(96-y)+25y+700\\ C(y)=\frac{1}{3}y^{2}-14y+2908 \]

For minimal value \(C_1'(y)=\frac{2}{3}y-14=0\), so \(y=21\) and \(x=96-y=75\).

5, Evaluate the double integral on the given region

\[\begin{split} \int\int_R (e^{8x+3y}) dA, R:2\le x\le4\ and\ 2 \le y \le 4 \end{split}\]

Write your answer in exact form without decimals.

Solution:

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