Data 605 Assignment 15
Question One
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)
lm(y~x)##
## Call:
## lm(formula = y ~ x)
##
## Coefficients:
## (Intercept) x
## -14.800 4.257\[y = -14.800 + 4.257*x\]
Question Two
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
Solve 1st and 2nd Partial Derivatives of fxfy:
\[f_x = 24-6y^2\] \[f_y = -12xy-24y^2\]
Second Derivatives
\[f_{xy} = -12y\] \[f_{xx} = 0\] \[f_{yy} = -12x-48y\]
Determine Critical Points by seeing where partial derivatives are equal to zero
\[f_x = 24 - 6y^2 = 0 \quad y=2,-2\] \[f_y = -12xy - 24y^2 = 0 \quad points = (-4,2) and(4,-2)\]
Solve the equation
\[f(x,y) = 24*4-6*4*(-2)^2-8(-2)^3 = 64\] \[f(x,y) = 24*-4-(6*-4*(2)^2)-8(2)^3 = -64\]
Question Three
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. 3. ###Step 1. Find the revenue function R ( x, y ).
\[R(x) = (81-21x+17y)*x\] \[R(y) = (40+11x-23y)*y\] \[R(x,y)= 28xy-23y^2-21x^2+40y+81x\]
Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?
Revenue = function(x,y){
rev = 28*x*y-23*y^2-21*x^2+40*y+81*x
return(rev)
}
(rev = Revenue(2.3,4.1))## [1] 116.62Question Four
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) = 1/6X^2 + 1/6y^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?
we know \[x + y = 96\] and therefore \[y = 96 - x\]
so if change the cost function to \[C(x, y) = C(x, 96 - x)\]
then we get \[= \frac{1}{6}x^2 + \frac{1}{6}(96 - x)^2 + 7x + 25(96 - x) + 700\]
this then simplifies to \[= \frac{1}{3}x^2 - 50x + 4636\]
Derivative of the Cost Function and Local Minimum
\[C^{'} = (2/3) x - 50 = 0\]
\[x = 75\]
since \[x+y = 96\] then
\[y = 21\]
Hence 75 units produced in Los Angeles and 21 in Denver is optimal to have the minimum cost
Question Five
Evaluate the double integral on the given region. \[\int \int_R (e^{8x+3y}) \quad ; R:2 \le x \le 4,2 \le y \le 4\]
\[\int e^{nx}dx = e^{nx}/n\]
\[\int_2^4 (\int_2^4 (e^{8x+3y}) dx) dy\]
\[= \int_2^4 e^{3y} (\int_2^4 (e^{8x} ) dx) dy\]
\[= \int_2^4 e^{3y} \left. ((1/8)e^{8x}) \right|_{2}^{4} dy\]
\[= \int_2^4 (e^{3y}/8) (e^{32} - e^{16}) dy\]
\[= ((e^{32} - e^{16})/8) \int_2^4 e^{3y} dy\]
\[= ((e^{32} - e^{16})/8) \left. ((1/3)e^{3y}) \right|_{2}^{4}\]
\[= (1/24) (e^{32} - e^{16}) (e^{12} - e^{6}) \]