HW15

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)

rm <- lm(y~x)
rm

## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##     -14.800        4.257

The equation is y = 4.26x - 14.80.

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.

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

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

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

\[f_{xx}(x,y) = 0\]

\[f_{yy}=-12x-48y\]

\[f_{xy}(x,y) = -12y\]

Set each partial derivative equal to zero:

\[f_{x}(x, y)=24-6y^{2}=0\\ 6y^{2} =24\\ y^{2} = 4\\ y=-2\ or\ y =2\]

Solve partial derivative by substitute y=-2 or y =2

\[f_{y}(x, y)=-12xy-24y^{2}=0\\ -12xy=24y^{2}\\ x=-2y\\ subsitute\ y=-2\ and\ y=2\\ x=4\ or\ x =-4\\ \]

The following points for (x, y, z) is (4, -2, z) and (-4, 2, z)

To find the value of z:

#(4, -2, z)
#24x-6xy^2-8y^3
x = 4
y = -2
z = 24*x-6*x*y^2-8*y^3

z

## [1] 64

#(-4, 2, z)
#24x-6xy^2-8y^3
x = -4
y = 2
z = 24*x-6*x*y^2-8*y^3

z

## [1] -64

The critical points are (4, -2, 64) & (-4, 2, -64)

Use Second Derivative Test to find out if f has saddle points

\[D=f_(xx)(x,y)f_{yy}(x,y)-[f_{xy}(x, y)]^{2}\]

\[f_{x}(x, y)=24-6y^{2}\\ f_{xx}(x, y)=0\] \[f_{y}(x, y)=-12xy-24y^{2}\\ f_{yy}(x, y)=-12x-48y\] \[f_{x}(x, y)=24-6y^{2}\\ f_{xy}(x, y)=-12y\]

\[D=(0)(-12x-48y)-[-12y]^{2}\\ D=-144y^{2}\]

Since D<0, f has a saddle point at critical points

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). We can perform some algebra by distributing and combining similar terms where possible

R(x, y) can be obtained by combining “house” & “name” function

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

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

The revenue is $116.62

#R(x, y) = R(2.30, 4.10)
x = 2.30
y = 4.10

r = 81*x-21*x^2+28*y*x+40*y-23*y^2

r

## [1] 116.62

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), 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?

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

\[x + y = 96\\ y = 96-x\]

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

\[C'(x)=\frac{1}{3}x-\frac{1}{3}(96-x)+7-25 \]

set C’(x) equal to 0

\[\frac{1}{3}x-\frac{1}{3}(96-x)+7-25 = 0\]

\[0 = \frac{2}{3}x-50 \] \[x = 75\] \[y = 96-75 = 21\]

Number of units should be produced in Los Angeles is 75

Number of units should be produced in Denver is 21

\[\int_R\int(e^{8x+3y}dA; R:2\le x\le 4\ and\ 2 \le y \le 4)\]

\[ \int_{2}^{4} e^{8x+3y} \, dy = \left[\frac{1}{3}e^{8x+3y}\right]_{2}^{4} = \frac{1}{3}(e^{8x+12} - e^{8x+6}) \]

\[ \int_{2}^{4} \frac{1}{3}(e^{8x+12} - e^{8x+6}) \, dx \]

\[ \frac{1}{3}\left(\frac{1}{8}e^{8x+12} - \frac{1}{8}e^{8x+6}\right) \bigg|_{2}^{4} \]

\[ = \frac{1}{3}\left(\frac{1}{8}e^{44} - \frac{1}{8}e^{28} - \frac{1}{8}e^{36} + \frac{1}{8}e^{22}\right) \]

\[ = \frac{1}{24}(e^{44} - e^{36} - e^{28} + e^{22}) \]