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)
m <- cor(x,y)*((sd(y)/sd(x)))
b <- mean(y)-m*mean(x)
m## [1] 4.257143b## [1] -14.8The equation of the regression line is y = 4.257143x-14.8
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.
\(f ( x, y ) = 24x - 6xy^2 - 8y^3\)
\(f_x(x,y) = 24 - 6y^2\) \(f_y(x,y) = -12xy -24y^2\)
\(f_{xy}(x,y) = -12y\) \(f_{xx}(x,y) = 0\) \(f_{yy}=-12x-48y\)
Setting above to 0 gives us (2,-4),(0,0) and (-2,+4) as critical values plugging into D will give us classification.
\(D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2\)
\(D(x,y)=(0)(12x-48y)-(-12y)^2\)
\(D(x,y)=-144y^2\)
D(2,-4) = -144 D(-2,4) = -144
Therefore both points are saddle points.
D(0,0) = 0 Therfore inconclusive.
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.
\(Revenue = x(f(x))+y(f(y))\) \(Revenue = x(81-21x+17y) + y(40+11x-23y)\)
\(Revenue =
81x-21x^2+17xy+40y+11xy-23y^2\)
Revenue for house
brand = 2.30 and name brand for $4.10
81*2.30 -21*2.30^2+17*4.10*2.30+40*4.10+11*2.30*4.10-23*4.10^2## [1] 116.62A 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?
x + y = 96
y= 96-x
\(C(x)=\frac{1}{6}x^2 + \frac{1}{6}(96-x)^2 + 7x + 25(96-x) + 700\)
Minimize by setting derivative to 0
\(C'(x)=\frac{1}{3}x-\frac{1}{3}(96-x)+7-25\)
\(0=\frac{2}{3}x-50\)
\(50=\frac{2}{3}x\)
\(x = 75\) so \(y = 21\)
Los Angeles should make 75 units and Denver should produce 14
\(\int_R\int(e^{8x+3y}dA; R:2\le x\le 4 and 2 \le y \le 4)\)
\(\int_2 ^4\int_2^ 4\)(e^{8x+3y}dxdy)
\(\int_2^4 \frac{1}{8}e^{8x+3y}|^4_2 dy\)
\(\int_2^4 \frac{1}{8}e^{32+3y}-\frac{1}{8}e^{16+3y}\)
\(\frac{1}{24}e^{32+3y}-\frac{1}{24}e^{16+3y}|^4_2\)
$e^44 - e^22 - e^38 + e^22
\(\frac{1}{24}e^{44}- \frac{1}{24}e^{38}\)